Date: Monday 2026-09-07, 21:30 ET · Cycle 3 of the Homework 1 loop · built from the cycle-2 exam grade · 95 pages at the 18pt reading size · about 228 minutes of reading at your measured pace (Guide #1's 25 pages took you 60 minutes), in two parts you can read separately

Built from: the cycle-2 comprehension exam, graded 2026-09-07 at 21:05 ET (7 of 10 on 9 answered, mean confidence 6.44, zero dangerous - every definition and one-step move of Guide #2 §1-§3 at 10; the one gap is carrying a chain of shapes through three or four factors), your notes on Guide #2 §0-§3 (15 lines, all correct, the shape names in your own hand), the sections of Guide #2 you did not read (§4-§7, carried here whole), and the example bank written for the rest of the homework (18 blocks, two independent checkers). Aimed at: the WHOLE of Homework 1 - 1.1, 1.2, 1.3a, 1.3b, 1.4a, 1.4b, 1.4c, 1.5, 1.6a, 1.6b, 1.6c - due Tuesday 2026-09-08 night. Your order, verbatim: "I want for study guide 3 to have the remaining parts I left out on study guide 2 plus what I need to finish the homework with simple examples of the ideas and symbols and everything leading up to the questions ... but just slightly different with the goal of me being able to complete homework [1] with little to no help. ... while I'm at work I'll be reading and reviewing over and over study guide 3: so make it to where I can comprehend and understand what's happening with no ability to write what's there, but when I get home I'll go back and write what's needed before starting homework 1."

Stated purpose (ASSIGNMENT_PREP §4): so you can master this material and complete this assignment without any help. This guide is written to be READ without a pen: every arithmetic step is printed, every ladder is worked to its last number, and nothing is left "for you to check" until the write-at-home list at the end.

0. Where you are, what this guide is for, what you will be able to do

Where you are. Two hours after Guide #2's first three sections and one page of notes, the ten-item exam on exactly those sections came back 7 of 10 on the 9 you answered, mean confidence 6.44, zero dangerous - the third instrument in a row with nothing wrong at high confidence, and your confidence tracked the truth: 7.7 on the items you got, 2.0 on the two you missed. What you own, at confidence 10: the shape names (the column is $4 \times 1$ - the "1 x n" of your cycle-1 teach-back is gone, on the exam and in your notes), whether two grids can be multiplied and what shape comes out ("my brain snapped right into place on it"), which of $p^{⊤} q$ , $p q$ , $p + q$ , $p^{⊤}$ is a number, and the product on numbers. Two of the three cycle-1 misses reversed. What is still missing is one thing, in two costumes: item 8 asked why $K$ must be $2 \times 3$ in $a^{⊤} K c$ - the forced-sizes method, which is Exercise 1.3a - and item 10 asked which of four chains ends $1 \times 1$ ; both wrong, both at low confidence. You can decide the shape of TWO factors; you have not yet carried a chain through three or four. And item 6, the transpose of a product on numbers, you left blank with "I know that it's like L and then K transpose" - which is exactly right as a sentence ( $L^{⊤} K^{⊤}$ ), and not yet a thing you have done. Those two - the chain and the flip of a product - are §2 and §3 of this guide, worked to the last digit, before anything new opens. Everything else from §1-§3 of Guide #2 gets one line in §1. The teach-back you did not write is on the write-at-home list. The goal of this guide is narrow: put every idea, symbol and equation Homework 1 asks for into your hands, on other letters and other numbers, so you can write the homework yourself. When you finish you should be able to carry a chain of shapes through four factors and say whether it is a number; flip a product on numbers and say why the order reverses; read a double sum as a vector-matrix-vector product and go back the other way; take the gradient of a linear term and of a quadratic form and say where symmetry is used; say what invertible means and what full column rank buys; sandwich a column around a grid and read the sign; find eigenvalues and rebuild a symmetric grid from them; run a three-way equivalence as a ring; complete the square around a candidate to show it is the only bottom; and read a projector, a trace, a variance and an unbiased estimator. One object anchors all of it, as before: a row of the Nazareth board is a vector, the 30-name board is a matrix, a scoring rule is a transpose, the backtest is a hat.

The plan, and the honest arithmetic about length. You asked for the whole homework and you asked me not to make this a hundred pages. Both of those cannot be true at once, so here is the count up front rather than as a surprise: the homework is eleven sub-parts for 15 points, and 9 of those points stand on machinery no lecture taught. This guide is long. It is built so that you can stop halfway and still be in good shape.

Part A, §1-§7 - 39 pages, about 94 minutes. The fix for what the exam found, then the four sections of Guide #2 you never reached. On its own Part A gets you 1.3a, 1.3b, 1.4a, 1.4b and 1.4c, and all of 1.1 except its last two words, which §24 finishes: 6 of the 15 points, five of the six marked approachable on the sheet, with 1.4c marked stretch - that is the Part A item that will take real time. If your day holds only one sitting, make it this one - these are the points you can actually bank.

Part B, §8 onward - 56 pages, about 134 minutes. The machinery for 1.2, 1.5, 1.6a, 1.6b and 1.6c: the other 9 points. That is more points than Part A holds, and every one of them is marked hard on the sheet - which is exactly why Part A comes first: it is fewer points at far better odds. Read it in order - §8 is the one trick everything after it uses, §9-§13 build 1.2, §14-§15 build 1.5, §16-§23 build 1.6, and §24 closes the last two words 1.1 needs. Each section stands on the one before, so do not skip ahead.

§25 is the write-at-home list and §26 is the coverage table - which section answers which row of the homework. Both are short, and §26 is worth two minutes before you start so you know where everything is.

How to read it at work, with no pen. Every example is printed with every step, so your job is to follow, not to compute: read the numbers, see that each line comes from the one above, and read the ladders in order - a ladder is a run of examples, simple and harder, each shown working or failing. Part A puts the examples first and the rule after them. Part B, from §8 on, states the rule first and then works the ladder - by then you are reading new machinery, and seeing the claim before the numbers is faster; if a rule paragraph loses you, jump straight to the first example under it and come back. When a paragraph is bold and starts Write this down, skip it at work; those lines are collected at the end, in order, for the evening. If a paragraph loses you, do not re-read it - read the next example, which shows the same thing on numbers, and come back.

What to do at home. Open to the write-at-home list. Copy it by hand in order, doing each piece of arithmetic it names yourself (they are the ones the guide already printed, so you can check each against the page). Then the homework, in the exact order §26 gives - cheapest points first - with the guide open beside you. One thing to know before you copy anything down. Part B (§8 on) is written entirely on letters the homework does not use, so nothing there can be transcribed by accident. Part A is not - §2, §3, §5 and §7 still call the board $X$ , the outcome $y$ and the weights $θ$ , because they are Guide #2's sections carried across whole and re-lettering them the night before would have cost you more than it bought. So in Part A, when a line looks like it could go straight onto your sheet, it probably could - re-derive it on your own instead, and use §14, §15 and §16, which redo the same arguments on clean letters.

1. What you own from Guide #2 §1-§3 - one line each, and that is all the review they get

These were at confidence 10 on the exam. Read them once; do not study them.

The alphabet. $ℝ$ is one number; $ℝ^{d}$ is a column of $d$ numbers; $ℝ^{n \times d}$ is a grid with $n$ rows and $d$ columns; $\in$ says "is in". A vector is a column.
The shape names. A shape is rows × columns, rows first. A column of $n$ is $n \times 1$ (tall); a row of $n$ is $1 \times n$ (wide). Your notes, line 6, in your hand.
The product. Every entry of a product is one row of the left grid against one column of the right: multiply matching entries, add. $(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 1 + 2 \\ 3 + 4 \end{matrix}) = (\begin{matrix} 3 \\ 7 \end{matrix})$ .
Legality and the output shape. $(r \times s) (s \times t) = r \times t$ : the inner sizes must match, the outer sizes survive. $(2 \times 3) (3 \times 2)$ is legal and gives $2 \times 2$ ; $(2 \times 3) (2 \times 3)$ cannot be formed, inner $3 \neq 2$ .
Which of two is a number. Of $p^{⊤} q$ , $p q$ , $p + q$ , $p^{⊤}$ for columns $p, q$ in $ℝ^{3}$ : only $p^{⊤} q = (1 \times 3) (3 \times 1) = 1 \times 1$ is a number; $p q$ cannot be formed; $p + q$ is a column; $p^{⊤}$ is a row.
Transpose. Flip over the diagonal: rows become columns. A column's transpose is a row; $(u^{⊤})^{⊤} = u$ ; a $1 \times 1$ is its own transpose, so $u^{⊤} v = v^{⊤} u$ .
Symmetric - the one word here that was right at only 5, so it gets a sentence. A square grid is symmetric when it equals its own transpose, $S^{⊤} = S$ : the entry in row $i$ , column $j$ equals the entry in row $j$ , column $i$ . $S = (\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix})$ is (flip it: unchanged); $N = (\begin{matrix} 2 & 4 \\ 0 & 3 \end{matrix})$ is not (the $4$ above the diagonal faces a $0$ below). §4 and §5 lean on this word; §9 makes it the first thing 1.2 checks.

2. The chain of shapes - the one gap - unlocks Exercise 1.3a and every "is this a number?" in 1.5 and 1.6

Two factors, which you own. $p, q \in ℝ^{3}$ . Write the shape under each symbol and pair the inner sizes: $p^{⊤} q$ is $(1 \times 3) (3 \times 1)$ , inner $3 = 3$ , legal, and the outer sizes $1$ and $1$ survive: $1 \times 1$ , a number. That is the whole method; the only new thing below is doing it more than once in a row.

Three factors, works. $a \in ℝ^{2}$ , $K$ a $2 \times 3$ grid, $c \in ℝ^{3}$ . The chain $a^{⊤} K c$ with the shapes written under it: $(1 \times 2) (2 \times 3) (3 \times 1)$ . Walk left to right, one pair at a time. First pair: $(1 \times 2) (2 \times 3)$ , inner $2 = 2$ , legal, result $1 \times 3$ . Now that result meets the last factor: $(1 \times 3) (3 \times 1)$ , inner $3 = 3$ , legal, result $1 \times 1$ . A number. On numbers, so you see it is not a trick of the notation: $a = (\begin{matrix} 1 \\ 2 \end{matrix})$ , $K = (\begin{matrix} 1 & 0 & 2 \\ 0 & 1 & 1 \end{matrix})$ , $c = (\begin{matrix} 1 \\ 1 \\ 1 \end{matrix})$ . $K c$ first (row against column): $(\begin{matrix} 1 + 0 + 2 \\ 0 + 1 + 1 \end{matrix}) = (\begin{matrix} 3 \\ 2 \end{matrix})$ , a $2 \times 1$ . Then $a^{⊤}$ against it: $1 \cdot 3 + 2 \cdot 2 = 7$ . One number, as the shapes promised.

Three factors, fails - and where it fails. Same letters, the chain $a^{⊤} c K$ : $(1 \times 2) (3 \times 1) (2 \times 3)$ . First pair: inner $2$ and $3$ , not equal - stop. The chain cannot be formed, and it does not matter what comes after the failure. The order of the factors is part of the answer; $a^{⊤} K c$ and $a^{⊤} c K$ are not two spellings of one thing.

Four factors, works - the exam's item 10. The board $X \in ℝ^{30 \times 5}$ (thirty names, five features) and the weights $θ \in ℝ^{5}$ . The chain $θ^{⊤} X^{⊤} X θ$ , shapes under: $(1 \times 5) (5 \times 30) (30 \times 5) (5 \times 1)$ - remember $X^{⊤}$ flips $30 \times 5$ to $5 \times 30$ . Walk: $(1 \times 5) (5 \times 30)$ , inner $5 = 5$ , gives $1 \times 30$ . Then $(1 \times 30) (30 \times 5)$ , inner $30 = 30$ , gives $1 \times 5$ . Then $(1 \times 5) (5 \times 1)$ , inner $5 = 5$ , gives $1 \times 1$ . A number - so it can be a loss. The other three chains from that item, walked the same way: $X X^{⊤} = (30 \times 5) (5 \times 30) = 30 \times 30$ , a grid, name against name; $X θ = (30 \times 5) (5 \times 1) = 30 \times 1$ , a column, one score per name; $X^{⊤} X = (5 \times 30) (30 \times 5) = 5 \times 5$ , a grid, feature against feature. Three real objects, none of them a number, and only a number can be set equal to a loss.

Four factors, legal but not a number. $X θ θ^{⊤} X^{⊤}$ : $(30 \times 5) (5 \times 1) (1 \times 5) (5 \times 30)$ . Walk: $30 \times 1$ , then $(30 \times 1) (1 \times 5) = 30 \times 5$ , then $(30 \times 5) (5 \times 30) = 30 \times 30$ . Legal at every step, and a $30 \times 30$ grid at the end - the same four letters as $θ^{⊤} X^{⊤} X θ$ in a different order, and a completely different object. The test is where the chain ENDS: $1 \times 1$ is a number; anything else is a grid, a column or a row.

Middle-school version. A chain of shapes is a line of dominoes: each piece's right-hand number must equal the next piece's left-hand number, or the line breaks there. If the line holds all the way, the answer's shape is the very first number and the very last one. A line that starts with $1$ and ends with $1$ is a single number.

The forced-sizes method - Exercise 1.3a in other letters, worked in full. This is the item you missed. A function is written $h (a, c) = a^{⊤} K c + c^{⊤} H c + g^{⊤} a + k$ with $a \in ℝ^{2}$ , $c \in ℝ^{3}$ , and you are told $h$ returns one number. Nobody tells you the sizes of $K$ , $H$ , $g$ , $k$ : the neighbours force them. One term at a time, shapes under every symbol, unknowns as question marks:

term	shapes	what is forced	size
$a^{⊤} K c$	$(1 \times 2) (? \times ?) (3 \times 1)$	$a^{⊤}$ is $1 \times 2$ , so $K$ needs $2$ rows; $c$ is $3 \times 1$ , so $K$ needs $3$ columns	$K$ is $2 \times 3$
$c^{⊤} H c$	$(1 \times 3) (? \times ?) (3 \times 1)$	$3$ rows to meet $c^{⊤}$ , $3$ columns to meet $c$	$H$ is $3 \times 3$
$g^{⊤} a$	$(?) (2 \times 1)$	the term is a number, so $g^{⊤}$ must be $1 \times 2$ - a row of $2$ - so $g$ is a column of $2$	$g$ is $2 \times 1$
$k$	added to numbers	a lone letter added to numbers is a number	$k$ is $1 \times 1$

Check each term ends $1 \times 1$ : $(1 \times 2) (2 \times 3) (3 \times 1) = 1 \times 1$ ; $(1 \times 3) (3 \times 3) (3 \times 1) = 1 \times 1$ ; $(1 \times 2) (2 \times 1) = 1 \times 1$ ; $k$ . Four numbers added: one number. The exam's distractors, so you see them fail: " $K$ is $2 \times 3$ because $a$ is $1 \times 2$ and $c$ is $1 \times 3$ " lands on the right size by calling two columns rows - the old inversion; " $K c$ is $2 \times 3$ , the shape of $K$ " - no, $K c = (2 \times 3) (3 \times 1) = 2 \times 1$ , a column; "the whole term comes out $2 \times 3$ " - no, it is a number, the stem said so.

The same method, second instance, different sizes - so the pattern is the method and not the numbers. $q (p, r) = p^{⊤} M r + r^{⊤} N r + w^{⊤} r + z^{⊤} p + c_{0}$ , with $p \in ℝ^{4}$ , $r \in ℝ^{2}$ , $q$ a number. $p^{⊤} M r$ : $(1 \times 4) (? \times ?) (2 \times 1)$ , so $M$ is $4 \times 2$ . $r^{⊤} N r$ : $(1 \times 2) (? \times ?) (2 \times 1)$ , so $N$ is $2 \times 2$ . $w^{⊤} r$ : $w^{⊤}$ must be $1 \times 2$ , so $w$ is $2 \times 1$ . $z^{⊤} p$ : $z^{⊤}$ must be $1 \times 4$ , so $z$ is $4 \times 1$ . $c_{0}$ : a number. Notice the middle grid of a term with two DIFFERENT vectors around it ( $M$ , between $p$ and $r$ ) is rectangular, $4 \times 2$ ; the middle grid of a term with the SAME vector on both sides ( $N$ , between $r^{⊤}$ and $r$ ) is square, $2 \times 2$ . That is not a coincidence; it is the shapes.

The same method, fails - when the sizes contradict. $a^{⊤} K c + a^{⊤} K a$ with the same $a \in ℝ^{2}$ , $c \in ℝ^{3}$ : the first term forces $K$ to $2 \times 3$ ; the second, $(1 \times 2) (? \times ?) (2 \times 1)$ , forces $K$ to $2 \times 2$ . One letter, two sizes - the expression is not well formed and no answer exists. The method finds that too: a size forced twice must agree.

HOW TO CHECK: a chain of shapes

Write the shape under every symbol; a transpose flips $r \times s$ to $s \times r$ ; an unknown grid gets $? \times ?$ .
Walk left to right, one pair at a time: inner sizes must match, the outer sizes survive, the result meets the next factor.
The first mismatch ends it: the chain cannot be formed.
Where it ends is what it is: $1 \times 1$ is a number; $n \times 1$ a column; $1 \times n$ a row; anything else a grid.
For an unknown grid, its left neighbour forces its rows and its right neighbour forces its columns; a lone letter added to numbers is a number; a size forced twice must agree.

Fund read. Every object in the least-squares story has a shape you can now read: $X θ$ is the column of thirty scores, $X^{⊤} X$ is the $5 \times 5$ feature-against-feature grid, $X X^{⊤}$ is the $30 \times 30$ name-against-name grid, and $θ^{⊤} X^{⊤} X θ$ is one number - the squared length of the score column, as §8 will show. When a formula on a slide looks like alphabet soup, write the shapes under it; the soup sorts itself.

Write this down: shapes under every symbol, walk left to right, inner sizes match, outer survive · $a^{⊤} K c$ : $(1 \times 2) (2 \times 3) (3 \times 1) = 1 \times 1$ , checked $= 7$ on numbers · $θ^{⊤} X^{⊤} X θ$ : $(1 \times 5) (5 \times 30) (30 \times 5) (5 \times 1) = 1 \times 1$ · the other three chains: $30 \times 30$ , $30 \times 1$ , $5 \times 5$ - none a number · forced sizes: left neighbour forces rows, right neighbour forces columns; a lone letter added to numbers is a number · the four-row table for $h (a, c)$ : $K$ $2 \times 3$ , $H$ $3 \times 3$ , $g$ $2 \times 1$ , $k$ $1 \times 1$ .

3. The transpose of a product, on numbers - the move you named and did not do

What you wrote on item 6: "I know that it's like L and then K transpose." That is the rule, $(K L)^{⊤} = L^{⊤} K^{⊤}$ , in words. Here it is done, every entry, so you have seen it happen once before you do it at home.

Middle-school version. Getting dressed, then undressed. Socks go on first and shoes second; to undo it you take the shoes off first and the socks second. The undo runs the list backwards - and so does the flip: the last thing in becomes the first thing out. That is the whole of why the order reverses.

Simple, works - the exam's own numbers. $K = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ , $L = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ . First the product $K L$ , row against column, four entries: row 1 of $K$ against column 1 of $L$ is $1 \cdot 1 + 2 \cdot 1 = 3$ ; row 1 against column 2 is $1 \cdot 0 + 2 \cdot 1 = 2$ ; row 2 against column 1 is $3 \cdot 1 + 4 \cdot 1 = 7$ ; row 2 against column 2 is $3 \cdot 0 + 4 \cdot 1 = 4$ . So $K L = (\begin{matrix} 3 & 2 \\ 7 & 4 \end{matrix})$ , and flipping it over its diagonal, $(K L)^{⊤} = (\begin{matrix} 3 & 7 \\ 2 & 4 \end{matrix})$ . Now the other side of the rule. $L^{⊤} = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ and $K^{⊤} = (\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix})$ ; their product $L^{⊤} K^{⊤}$ , four entries: $1 \cdot 1 + 1 \cdot 2 = 3$ ; $1 \cdot 3 + 1 \cdot 4 = 7$ ; $0 \cdot 1 + 1 \cdot 2 = 2$ ; $0 \cdot 3 + 1 \cdot 4 = 4$ . So $L^{⊤} K^{⊤} = (\begin{matrix} 3 & 7 \\ 2 & 4 \end{matrix})$ - the same grid. The keyed option was this grid; you had the rule and stopped before the arithmetic.

Simple, fails - the same order. $K^{⊤} L^{⊤}$ , four entries: $1 \cdot 1 + 3 \cdot 0 = 1$ ; $1 \cdot 1 + 3 \cdot 1 = 4$ ; $2 \cdot 1 + 4 \cdot 0 = 2$ ; $2 \cdot 1 + 4 \cdot 1 = 6$ . So $K^{⊤} L^{⊤} = (\begin{matrix} 1 & 4 \\ 2 & 6 \end{matrix})$ , not $(\begin{matrix} 3 & 7 \\ 2 & 4 \end{matrix})$ . The order must reverse; the same order gives a different grid, even for square grids where it is legal.

Harder, works - and why the order MUST reverse when the grids are not square. $G = (\begin{matrix} 1 & 0 \\ 2 & 1 \\ 0 & 3 \end{matrix})$ is $3 \times 2$ and $H = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ is $2 \times 2$ . $G H$ is $(3 \times 2) (2 \times 2) = 3 \times 2$ , six entries: row 1: $1 \cdot 1 + 0 \cdot 0 = 1$ , $1 \cdot 2 + 0 \cdot 1 = 2$ ; row 2: $2 \cdot 1 + 1 \cdot 0 = 2$ , $2 \cdot 2 + 1 \cdot 1 = 5$ ; row 3: $0 \cdot 1 + 3 \cdot 0 = 0$ , $0 \cdot 2 + 3 \cdot 1 = 3$ . So $G H = (\begin{matrix} 1 & 2 \\ 2 & 5 \\ 0 & 3 \end{matrix})$ and $(G H)^{⊤} = (\begin{matrix} 1 & 2 & 0 \\ 2 & 5 & 3 \end{matrix})$ , a $2 \times 3$ . Now $H^{⊤} G^{⊤}$ : $H^{⊤} = (\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix})$ is $2 \times 2$ , $G^{⊤} = (\begin{matrix} 1 & 2 & 0 \\ 0 & 1 & 3 \end{matrix})$ is $2 \times 3$ , the chain $(2 \times 2) (2 \times 3)$ is legal and gives $2 \times 3$ . Entries: row 1: $1 \cdot 1 + 0 \cdot 0 = 1$ , $1 \cdot 2 + 0 \cdot 1 = 2$ , $1 \cdot 0 + 0 \cdot 3 = 0$ ; row 2: $2 \cdot 1 + 1 \cdot 0 = 2$ , $2 \cdot 2 + 1 \cdot 1 = 5$ , $2 \cdot 0 + 1 \cdot 3 = 3$ . $H^{⊤} G^{⊤} = (\begin{matrix} 1 & 2 & 0 \\ 2 & 5 & 3 \end{matrix})$ - the same. And the same order, $G^{⊤} H^{⊤}$ , is $(2 \times 3) (2 \times 2)$ : inner $3 \neq 2$ , it cannot even be formed - that is item 7, which you got at confidence 7: the shapes rule out the same order whenever the grids are not square.

Harder, works - three factors, the rule used twice. $(A B C)^{⊤}$ : group it as $((A B) C)^{⊤}$ , apply the rule once to the two pieces $(A B)$ and $C$ : $C^{⊤} (A B)^{⊤}$ ; apply it again inside: $C^{⊤} B^{⊤} A^{⊤}$ . The whole line reverses. On shapes with $A$ $1 \times 2$ , $B$ $2 \times 3$ , $C$ $3 \times 1$ : $A B C$ is $1 \times 1$ , and $C^{⊤} B^{⊤} A^{⊤}$ is $(1 \times 3) (3 \times 2) (2 \times 1) = 1 \times 1$ - the transpose of a number is that number, which is why $a^{⊤} K c = c^{⊤} K^{⊤} a$ : a $1 \times 1$ equals its own transpose, and its transpose is the reversed chain.

Two consequences the homework uses on every page. First, $(X^{⊤} X)^{⊤} = X^{⊤} (X^{⊤})^{⊤} = X^{⊤} X$ - the grid $X^{⊤} X$ is symmetric, always, for any board $X$ ; the same for $K^{⊤} K$ with any $K$ . Second, the transpose of a difference: $(X θ - y)^{⊤} = θ^{⊤} X^{⊤} - y^{⊤}$ - transpose each piece, reverse each product, keep the minus. §5 uses both.

Fund read. $X^{⊤} X$ being symmetric is why the gradient of your backtest loss has the clean form §5 derives (Fact B needs a symmetric grid), and why 1.2's positive-definite question is even askable about it.

HOW TO CHECK: did I flip a product correctly?

Flip each factor on its own: an $r \times s$ becomes $s \times r$ .
Reverse their order - the last factor written becomes the first.
Check the chain is legal: the inner sizes of the reversed order must match. If they do not, you reversed the wrong way.
Sanity: $K = (1, 2; 3, 4)$ and $L = (1, 0; 1, 1)$ give $(K L)^{⊤} = L^{⊤} K^{⊤} = (3, 7; 2, 4)$ , while the same order $K^{⊤} L^{⊤}$ gives $(1, 4; 2, 6)$ - a different grid.

Write this down: $(K L)^{⊤} = L^{⊤} K^{⊤}$ , checked on $K = (1, 2; 3, 4)$ , $L = (1, 0; 1, 1)$ : both sides $(3, 7; 2, 4)$ ; the same order gives $(1, 4; 2, 6)$ · for non-square grids the same order is not even legal: $G^{⊤} H^{⊤}$ is $(2 \times 3) (2 \times 2)$ · $(A B C)^{⊤} = C^{⊤} B^{⊤} A^{⊤}$ · $X^{⊤} X$ is symmetric for every $X$ · $(X θ - y)^{⊤} = θ^{⊤} X^{⊤} - y^{⊤}$ .

4. The double sum, from zero - unlocks Exercise 1.3b

One sum first, on numbers. The tall $\sum$ (read "sum") is a for-loop with a counter. $\sum_{i = 1}^{3} i$ means: start $i$ at $1$ , stop at $3$ , add up every value of the thing after the $\sum$ - here the thing is $i$ itself:

\sum_{i = 1}^{3} i = 1 + 2 + 3 = 6

With a column in play, $u = (\begin{matrix} 2 \\ 5 \\ 7 \end{matrix})$ , the thing after the sum can be an entry: $\sum_{i = 1}^{3} u_{i} = u_{1} + u_{2} + u_{3} = 2 + 5 + 7 = 14$ . Read: $u_{i}$ is "the $i$ -th entry of $u$ "; the loop visits each entry once. And the row-against-column of §1 is one loop too: $u^{⊤} v = \sum_{i} u_{i} v_{i}$ , on $u = (1; 2)$ and $v = (3; 5)$ : $1 \cdot 3 + 2 \cdot 5 = 13$ .

A double sum is one loop inside another. Definition: $\sum_{i = 1}^{2} \sum_{j = 1}^{2}$ means "for each $i$ from 1 to 2, and for each $j$ from 1 to 2, add up the thing after both sums" - every pair $(i, j)$ gets visited once, so there are $2 \times 2 = 4$ terms. On the simplest thing, $i \cdot j$ :

\sum_{i = 1}^{2} \sum_{j = 1}^{2} i \cdot j = 1 + 2 + 2 + 4 = 9

Read: the four pairs, in order, are $(1, 1)$ , $(1, 2)$ , $(2, 1)$ , $(2, 2)$ , and their products are $1 \cdot 1$ , $1 \cdot 2$ , $2 \cdot 1$ , $2 \cdot 2$ - hold $i = 1$ and run $j$ through $1, 2$ ; then hold $i = 2$ and run $j$ again. The two loops can have different lengths: $\sum_{i = 1}^{2} \sum_{j = 1}^{3} (i + j)$ has $2 \times 3 = 6$ terms: for $i = 1$ : $2 + 3 + 4 = 9$ ; for $i = 2$ : $3 + 4 + 5 = 12$ ; total $21$ . That is all "double sum" means: a grid of terms, one per $(i, j)$ pair, all added. Figure 1 draws the grid.

Now put a matrix in the middle. Let $S = (\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix})$ and $u = (\begin{matrix} 1 \\ 2 \end{matrix})$ . $S_{i j}$ (read "S i j") is the entry of $S$ in row $i$ , column $j$ : $S_{11} = 2$ , $S_{12} = 1$ , $S_{21} = 1$ , $S_{22} = 3$ . $S$ is symmetric (§1): flip it and it is unchanged - the $1$ above the diagonal matches the $1$ below. The double sum the homework cares about is

\sum_{i = 1}^{2} \sum_{j = 1}^{2} u_{i} S_{i j} u_{j}

Read: for each pair $(i, j)$ , multiply the $i$ -th entry of $u$ , the $(i, j)$ entry of $S$ , and the $j$ -th entry of $u$ ; add the four products. Written out, one row of the table per pair:

pair $(i, j)$	$u_{i}$	$S_{i j}$	$u_{j}$	term
$(1, 1)$	$1$	$2$	$1$	$1 \cdot 2 \cdot 1 = 2$
$(1, 2)$	$1$	$1$	$2$	$1 \cdot 1 \cdot 2 = 2$
$(2, 1)$	$2$	$1$	$1$	$2 \cdot 1 \cdot 1 = 2$
$(2, 2)$	$2$	$3$	$2$	$2 \cdot 3 \cdot 2 = 12$

Total: $2 + 2 + 2 + 12 = 18$ . Each term is three ordinary numbers multiplied, so their order inside a term does not matter: $u_{i} S_{i j} u_{j}$ , $u_{i} u_{j} S_{i j}$ and $S_{i j} u_{i} u_{j}$ are the same term, and $\sum_{i} \sum_{j} u_{i} u_{j} S_{i j}$ is the same $18$ . A homework can write the three factors in any order; count the copies of the vector, not their position.

Middle-school version. A double sum is a times table with the total written under it. Row headers down the side, column headers along the top, one number in every cell, then add up the whole table. Two loops means two headers; one table; one total.

Figure 1. The four terms of the double sum with S in the middle, laid out as the grid the two loops walk. The accent cells are the two cross terms; they match because S is symmetric.

The same 18 by matrix multiplication - this is the point of the section. First $S u$ , by §1 (row against column): $S u = (\begin{matrix} 2 \cdot 1 + 1 \cdot 2 \\ 1 \cdot 1 + 3 \cdot 2 \end{matrix}) = (\begin{matrix} 4 \\ 7 \end{matrix})$ . Then $u^{⊤}$ against that column: $u^{⊤} (S u) = 1 \cdot 4 + 2 \cdot 7 = 18$ . Same number. So the four-term double sum and the two-step product $u^{⊤} S u$ are two ways of writing one quantity.

Why they are always equal - shown, not asserted. The $i$ -th entry of $S u$ is row $i$ of $S$ against $u$ , which as a sum is $(S u)_{i} = \sum_{j = 1}^{2} S_{i j} u_{j}$ . Check it on row 1: $S_{11} u_{1} + S_{12} u_{2} = 2 \cdot 1 + 1 \cdot 2 = 4$ , the $4$ above. Then $u^{⊤} (S u)$ is $u$ against that column, $\sum_{i = 1}^{2} u_{i} (S u)_{i}$ ; put the inner sum in for $(S u)_{i}$ and you have $\sum_{i = 1}^{2} u_{i} \sum_{j = 1}^{2} S_{i j} u_{j}$ , which is the double sum with $u_{i}$ pulled out front of the inner loop - the same four terms. On the table: the two rows with $i = 1$ add to $u_{1} \times (S_{11} u_{1} + S_{12} u_{2}) = 1 \times (2 + 2) = 4 = u_{1} \times (S u)_{1}$ ; the two rows with $i = 2$ add to $2 \times (1 + 6) = 14 = u_{2} \times (S u)_{2}$ ; $4 + 14 = 18$ . "Pulled out front" means exactly that: the $u_{1}$ shared by the first two rows is factored out once. Every step is a line you can check on the table above.

Two new words, defined now that you have the thing. The number $u^{⊤} S u$ is called a quadratic form in $u$ : "quadratic" because every term holds two copies of $u$ ( $u_{i} u_{j}$ , the way $x^{2}$ holds two copies of $x$ ), and "form" is an old word for a formula that eats a vector and returns a number. The simplest one is $S$ = the grid with $1$ s on the diagonal and $0$ s elsewhere: then $u^{⊤} S u = u_{1}^{2} + u_{2}^{2} = 1 + 4 = 5$ , the squared length of $u$ . That squared length has a name and a symbol: $‖ u ‖^{2}$ (read "norm of $u$ , squared"; norm = length), and it always equals $u^{⊤} u$ , the column against itself - $1 \cdot 1 + 2 \cdot 2 = 5$ . So a squared length is a quadratic form too; §5 and §7 use the symbol, and Part B lives on it. This is cycle-1 exam item 10, which you missed: with $x = (1; 2; 3)$ , $x^{⊤} x = 1 + 4 + 9 = 14$ , one number, because a row against a column always ends $1 \times 1$ . The answer you chose, $(\begin{matrix} 1 \\ 4 \\ 9 \end{matrix})$ , keeps the squares apart as a column; it is not $x^{⊤} x$ , and no product of $x$ with its own transpose gives it - it is entry-by-entry squaring, which this course never writes.

The ladder for the reverse direction - reading a sum and naming it - which is what 1.3b asks.

Simple, fails - not a quadratic form. $\sum_{i = 1}^{2} \sum_{j = 1}^{2} S_{i j} u_{i}$ : two loops, but only one copy of $u$ . On the numbers: $2 \cdot 1 + 1 \cdot 1 + 1 \cdot 2 + 3 \cdot 2 = 11$ . It is a legitimate number, but it is linear in $u$ (double $u$ and it doubles), so it is not $u^{⊤} S u$ and no quadratic form matches it. The tell is the count of $u$ 's.

Simple, works. $\sum_{j = 1}^{2} S_{1 j} u_{j}$ : one loop, row 1 of $S$ against $u$ . That is $(S u)_{1}$ , the first entry of $S u$ - the $4$ .

Harder, fails - two copies, wrong subscripts. $\sum_{i = 1}^{2} \sum_{j = 1}^{2} u_{i} S_{i j} u_{i}$ - two copies of $u$ , so the count passes, but both carry $i$ . On the numbers: $i = 1$ : $1 \cdot 2 \cdot 1 + 1 \cdot 1 \cdot 1 = 3$ ; $i = 2$ : $2 \cdot 1 \cdot 2 + 2 \cdot 3 \cdot 2 = 16$ ; total $19$ , not $18$ . It is not $u^{⊤} S u$ : the inner loop adds up a row of $S$ on its own instead of pairing row $i$ against $u$ . That disagreement is what step 4 of the box below catches.

Harder, works. $\sum_{i = 1}^{2} \sum_{j = 1}^{2} u_{i} u_{j} S_{i j}$ - two loops, two $u$ 's carrying one counter each, a matrix entry carrying both, in whatever order the factors are printed. Regroup each term as $u_{i} (S_{i j} u_{j})$ and take the inner loop first: $\sum_{j} S_{i j} u_{j} = (S u)_{i}$ . Outer loop: $\sum_{i} u_{i} (S u)_{i} = u^{⊤} (S u) = u^{⊤} S u$ . Homework 1.3b is this recognition with its own letters.

Harder, works, one size up - so you trust it past $2 \times 2$ . $T = (\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{matrix})$ , $u = (\begin{matrix} 1 \\ 1 \\ 2 \end{matrix})$ . Matrix route: $T u = (\begin{matrix} 1 + 2 + 0 \\ 2 + 0 + 2 \\ 0 + 1 + 6 \end{matrix}) = (\begin{matrix} 3 \\ 4 \\ 7 \end{matrix})$ , then $u^{⊤} T u = 1 \cdot 3 + 1 \cdot 4 + 2 \cdot 7 = 21$ . Loop route, nine terms: $1 + 2 + 0 + 2 + 0 + 2 + 0 + 2 + 12 = 21$ . Same number at $3 \times 3$ ; the pattern is the same at any size.

HOW TO CHECK: turning a sum into matrix notation

Count the copies of the vector. Two copies → a quadratic form is possible. One copy → it is linear, look for $b^{⊤} u$ or $(S u)_{i}$ instead.
Find the innermost sum. Ask: "is this row $i$ of some matrix against some vector?" If yes, it is $(S u)_{i}$ .
Find the outer sum. Ask: "is this a vector dotted against that column?" If yes, it is $u^{⊤} (\cdot)$ .
Test on $2 \times 2$ with numbers. If the loop total and the matrix total disagree, a subscript is on the wrong letter.

Fund read. Lecture 3, slide 17, prices a wrong weight vector with a quadratic form: the weight error in the middle of a sandwich whose matrix is $\frac{1}{n} X^{⊤} X$ , the grid that records, for every pair of features, how much they move together. Two features that always move together make some directions of error cheap and others expensive; the matrix carries that. For the homework you need only its shape: error-transpose, matrix, error, one number - a chain of three that ends $1 \times 1$ , by §2.

Write this down: $\sum$ = a for-loop; a double sum = a loop inside a loop, one term per $(i, j)$ pair · $\sum_{i} \sum_{j} u_{i} S_{i j} u_{j} = u^{⊤} S u$ , checked: $18$ both ways for $S = (2, 1; 1, 3)$ , $u = (1; 2)$ · quadratic form = two copies of $u$ with a matrix between them = one number · $(S u)_{i} = \sum_{j} S_{i j} u_{j}$ is the bridge · $x^{⊤} x$ = the squares ADDED, one number ( $1 + 4 + 9 = 14$ ; never a column of squares)

5. The gradient, from zero - unlocks Exercises 1.4a and 1.4b

One knob first. A function of one number, $f (u) = u^{2}$ , at $u = 2$ : $f (2) = 4$ . Nudge $u$ by $0.01$ : $f (2.01) = 4.0401$ . The rise was $0.0401$ for a nudge of $0.01$ - about $4$ per unit. That rate is the derivative (read "the slope of $f$ at $u = 2$ "), and Calc 1 gives it exactly: the derivative of $u^{2}$ is $2 u$ , which at $u = 2$ is $4$ . Slope = how fast $f$ rises per unit nudge. That is the only calculus this section uses: the derivative of $u$ is $1$ , of $u^{2}$ is $2 u$ , of a constant is $0$ , a number in front stays in front ( $5 u$ has slope $5$ - the nudge below shows it, $0.05$ per $0.01$ - and $3 u^{2}$ has slope $3 \cdot 2 u = 6 u$ ), and the derivative of a sum is the sum of the derivatives.

Two knobs. Now $f$ eats a column of two numbers, $u = (\begin{matrix} u_{1} \\ u_{2} \end{matrix})$ , and returns one number: $f (u) = 5 u_{1} - 3 u_{2}$ . The course writes that as $f : ℝ^{2} \to ℝ$ (read "f from R two to R" - eats a 2-vector, returns a number). At $u = (1; 1)$ , $f = 5 - 3 = 2$ . There are now two things you can nudge:

Nudge $u_{1}$ only, to $1.01$ , holding $u_{2} = 1$ : $f = 5.05 - 3 = 2.05$ , a rise of $0.05$ per $0.01$ - slope $5$ .
Nudge $u_{2}$ only, to $1.01$ , holding $u_{1} = 1$ : $f = 5 - 3.03 = 1.97$ , a fall of $0.03$ per $0.01$ - slope $- 3$ .

Definition. The slope of $f$ when you nudge only $u_{1}$ and hold everything else still is the partial derivative of $f$ with respect to $u_{1}$ , written $\frac{\partial f}{\partial u_{1}}$ (read "partial f, partial u-one"; the curly $\partial$ is just a $d$ that says "one knob at a time"). Computed the Calc-1 way: treat $u_{2}$ as a constant and differentiate $5 u_{1} - 3 u_{2}$ in $u_{1}$ ; the $5 u_{1}$ gives $5$ , the $- 3 u_{2}$ is a constant and gives $0$ . So $\frac{\partial f}{\partial u_{1}} = 5$ and, the same way, $\frac{\partial f}{\partial u_{2}} = - 3$ - exactly the two nudges.

Definition. The gradient of $f$ is the column of all its partial derivatives, one per knob, written $\nabla f (u)$ (read "grad f of u" or "nabla f" - nabla is the name of the upside-down triangle symbol, nothing more):

\nabla f (u) = (\begin{matrix} \frac{\partial f}{\partial u_{1}} \\ \frac{\partial f}{\partial u_{2}} \end{matrix}) = (\begin{matrix} 5 \\ - 3 \end{matrix})

Read: two knobs, two slopes, stacked as a column - so $\nabla f (u)$ lives in $ℝ^{2}$ , the same set as $u$ . With $d$ knobs it has $d$ entries. (The homework writes the same definition with $x$ and $n$ .)

Middle-school version. You are standing on a hillside holding two knobs. Turn the first a hair and you climb a little; turn the second a hair and you climb a different amount. Write the two climb rates on a card, one under the other. That card is the gradient - one rate per knob, and the card is as long as you have knobs. Figure 2 draws the hill from above.

Figure 2. The bowl g(u) seen from above, drawn schematically (the true rings of this S are tilted ellipses; the arrows keep the 8 : 14 ratio). At u = (1; 2) the two knob slopes are 8 and 14 (computed below); the gradient stacks them into one arrow that points uphill. Fund read: the knobs are your weights, the bowl is the backtest loss, and the arrow says which way to turn each weight to make the backtest worse, so you turn the other way.

Fact A - the gradient of a fixed vector against $u$ - the ladder.

Simple, works (done above). $c = (\begin{matrix} 5 \\ - 3 \end{matrix})$ and $f (u) = c^{⊤} u = 5 u_{1} - 3 u_{2}$ . Gradient: $(\begin{matrix} 5 \\ - 3 \end{matrix})$ , which is $c$ itself. Nothing depended on where $u$ was.

Simple, fails - the shape trap. Someone answers "the gradient is $c^{⊤}$ ", the row $(5, - 3)$ . Shape check: the gradient has one entry per knob, stacked - a $2 \times 1$ column, the same shape as $u$ . A row is $1 \times 2$ . Wrong shape, wrong answer, and on a graded proof it costs the point. The gradient is $c$ , the column.

Harder, works, three knobs. $c = (\begin{matrix} 2 \\ - 1 \\ 4 \end{matrix})$ , $f (u) = c^{⊤} u = 2 u_{1} - u_{2} + 4 u_{3}$ . Knob 1: the only term holding $u_{1}$ is $2 u_{1}$ , slope $2$ ; knob 2: slope $- 1$ ; knob 3: slope $4$ . Stack: $\nabla f = (\begin{matrix} 2 \\ - 1 \\ 4 \end{matrix}) = c$ again.

Harder, fails - when the "fixed" vector is not fixed. $f (u) = u^{⊤} u = u_{1}^{2} + u_{2}^{2}$ looks like $c^{⊤} u$ with $c = u$ , so someone answers "the gradient is $c$ , which is $u$ ". Nudge at $(1; 2)$ : $f = 5$ ; $f (1.01, 2) = 1.0201 + 4 = 5.0201$ , a rise of $0.0201$ per $0.01$ - slope $2$ , not the $1$ that answer predicts for knob 1. Fact A needs $c$ to stay put while $u$ moves; when both copies move you are in Fact B below, and the answer is $2 u$ .

The general method, which is Exercise 1.4a in its own letters. The same three steps at general $d$ are what 1.4a asks you to write: un-compress $c^{⊤} u$ into a sum over an index, differentiate with respect to one knob (every term without that knob is a constant and dies), then stack the $d$ answers back into a column. The algebra is yours.

Fact B - the gradient of a quadratic form - the ladder, and the trap that costs the point. Take $S = (\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix})$ from §4 and $g (u) = u^{⊤} S u$ . §4 un-compressed it to four terms; two of them are the same cross term, and they merge:

g (u) = 2 u_{1}^{2} + 1 \cdot u_{1} u_{2} + 1 \cdot u_{2} u_{1} + 3 u_{2}^{2}

and the two cross terms merge:

g (u) = 2 u_{1}^{2} + 2 u_{1} u_{2} + 3 u_{2}^{2}

Read: the cross term $u_{1} u_{2}$ appears twice - once from $S_{12}$ , once from $S_{21}$ - and the two are equal because $S$ is symmetric ( $S_{12} = S_{21} = 1$ ), so they add to $2 u_{1} u_{2}$ . At $u = (1; 2)$ : $g = 2 + 4 + 12 = 18$ , the §4 number.

Simple, fails - the trap. Differentiate $u^{⊤} S u$ as if only the left $u$ moved: "the slope is $S u$ ", which at $(1; 2)$ is $(\begin{matrix} 4 \\ 7 \end{matrix})$ , so knob 1's slope would be $4$ . Nudge test: $g (1.01, 2) = 2 (1.0201) + 2 (1.01) (2) + 12 = 18.0802$ ; the rise was $0.0802$ per $0.01$ - slope $8$ , not $4$ . The answer $S u$ fails the arithmetic. Both $u$ 's move when you nudge $u_{1}$ ; the formula only counted one.

Simple, works. Differentiate the un-compressed line knob by knob, collecting every term that holds the knob's letter. Knob 1: from $2 u_{1}^{2}$ comes $4 u_{1}$ ; from $2 u_{1} u_{2}$ comes $2 u_{2}$ ( $u_{2}$ is a constant here); $3 u_{2}^{2}$ gives $0$ . Knob 2: $2 u_{1}^{2}$ gives $0$ ; $2 u_{1} u_{2}$ gives $2 u_{1}$ ; $3 u_{2}^{2}$ gives $6 u_{2}$ .

\nabla g (u) = (\begin{matrix} 4 u_{1} + 2 u_{2} \\ 2 u_{1} + 6 u_{2} \end{matrix})

At $(1; 2)$ : $(\begin{matrix} 8 \\ 14 \end{matrix})$ . The nudge test said $8$ for knob 1; for knob 2, $g (1, 2.01) = 2 + 4.02 + 12.1203 = 18.1403$ , a rise of $0.1403$ per $0.01$ - slope $14$ . Both match. Now compare with $S u = (\begin{matrix} 2 u_{1} + u_{2} \\ u_{1} + 3 u_{2} \end{matrix})$ : the gradient is exactly twice it, $\nabla g (u) = 2 S u$ . The 2 is the two cross terms.

Harder, fails - what "symmetric" is doing in 1.4b. Take a matrix that is not symmetric, $N = (\begin{matrix} 2 & 4 \\ 0 & 3 \end{matrix})$ , so $N_{12} = 4$ but $N_{21} = 0$ . Un-compress: $u^{⊤} N u = 2 u_{1}^{2} + 4 u_{1} u_{2} + 0 \cdot u_{2} u_{1} + 3 u_{2}^{2}$ , that is $2 u_{1}^{2} + 4 u_{1} u_{2} + 3 u_{2}^{2}$ . Gradient, knob by knob: $(\begin{matrix} 4 u_{1} + 4 u_{2} \\ 4 u_{1} + 6 u_{2} \end{matrix})$ ; at $(1; 1)$ that is $(\begin{matrix} 8 \\ 10 \end{matrix})$ . The formula $2 N u$ at $(1; 1)$ gives $(\begin{matrix} 12 \\ 6 \end{matrix})$ . Different - so $2 N u$ is false for a non-symmetric matrix. What works is $(N + N^{⊤}) u$ : $N + N^{⊤} = (\begin{matrix} 4 & 4 \\ 4 & 6 \end{matrix})$ , and $(\begin{matrix} 4 & 4 \\ 4 & 6 \end{matrix}) (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 8 \\ 10 \end{matrix})$ , the true gradient. When the matrix is symmetric, $N + N^{⊤} = 2 N$ , and the formula collapses to $2 N u$ . That is why Exercise 1.4b says its matrix is symmetric: it is the one assumption that turns $(N + N^{⊤}) u$ into $2 N u$ , and it is the assumption the exercise hands you - your proof must name the line where it is used.

Harder, works, one size up. $T = (\begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 0 & 1 & 3 \end{matrix})$ from §4 is symmetric (flip it: unchanged). At $u = (1; 1; 2)$ , §4 found $T u = (3; 4; 7)$ , so $2 T u = (6; 8; 14)$ . From the nine terms, merging the equal cross terms: $u^{⊤} T u = u_{1}^{2} + 4 u_{1} u_{2} + 2 u_{2} u_{3} + 3 u_{3}^{2}$ . Knob 1: $2 u_{1} + 4 u_{2} = 2 + 4 = 6$ . Knob 2: $4 u_{1} + 2 u_{3} = 4 + 4 = 8$ . Knob 3: $2 u_{2} + 6 u_{3} = 2 + 12 = 14$ . Stacked: $(6; 8; 14) = 2 T u$ . The rule survives the size.

Harder, works - the general method, on the four terms first, letters last. Back to $S$ and knob 1, one term of the §4 table at a time. $S_{11} u_{1}^{2}$ gives $2 S_{11} u_{1} = 4 u_{1}$ . $S_{12} u_{1} u_{2}$ gives $S_{12} u_{2} = u_{2}$ . $S_{21} u_{2} u_{1}$ gives $S_{21} u_{2} = u_{2}$ . $S_{22} u_{2}^{2}$ gives $0$ . Total $4 u_{1} + 2 u_{2}$ - the slope found above. Now split the $4 u_{1}$ as $2 u_{1} + 2 u_{1}$ and group the four pieces into two brackets:

(S_{11} u_{1} + S_{12} u_{2}) + (S_{11} u_{1} + S_{21} u_{2})

The first bracket is row 1 of $S$ against $u$ - that is $(S u)_{1} = 2 u_{1} + u_{2}$ . The second is column 1 of $S$ against $u$ - also $2 u_{1} + u_{2}$ , because $S$ is symmetric: row 1 and column 1 hold the same numbers. Two equal brackets, so the slope at knob 1 is $2 (S u)_{1}$ . Knob 2 the same way gives $2 (S u)_{2}$ . Now $N = (2, 4; 0, 3)$ , the non-symmetric one: row 1 against $u$ is $2 u_{1} + 4 u_{2}$ , column 1 against $u$ is $2 u_{1} + 0 \cdot u_{2}$ - different - and their sum, $4 u_{1} + 4 u_{2}$ , is exactly the non-symmetric answer above. The same steps at general $d$ are what 1.4b asks you to write: un-compress the double sum, differentiate with respect to one knob, notice that the terms holding that knob are row $k$ of the matrix against the vector plus column $k$ against the vector, use symmetry to say the two are equal, then stack. The algebra is yours. The check on your own version: at knob 1 on $S = (2, 1; 1, 3)$ , $u = (1; 2)$ , your line must give $8$ ; and on the non-symmetric $N$ it must fail, or the symmetry line is missing.

HOW TO CHECK: computing a gradient of a matrix expression

Un-compress to an explicit sum with subscripts (§4).
Pick one knob $u_{k}$ . Differentiate in $u_{k}$ alone, every other $u_{j}$ a constant. Hunt every place $u_{k}$ appears - in a quadratic form it appears in two families.
Re-compress into matrix form. Shape check: $\nabla f (u)$ is $d \times 1$ , the same shape as $u$ - never a row.
Nudge test at $d = 2$ with numbers: change one entry by $0.01$ and confirm the rise matches slope $\times 0.01$ .

Toward 1.4c - expanding a squared length, on numbers, in other letters (different from your homework's). Let $Z = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , $v = (\begin{matrix} 1 \\ 1 \end{matrix})$ , $r = (\begin{matrix} 3 \\ 2 \end{matrix})$ , and look at the squared length of $Z v - r$ . First the direct way: $Z v = (\begin{matrix} 3 \\ 1 \end{matrix})$ , so $Z v - r = (\begin{matrix} 0 \\ - 1 \end{matrix})$ and $‖ Z v - r ‖^{2} = 0 + 1 = 1$ . Now the expansion. A squared length is the thing transposed against itself, $(Z v - r)^{⊤} (Z v - r)$ ; by the reversal rule of §3, $(Z v - r)^{⊤} = v^{⊤} Z^{⊤} - r^{⊤}$ , and multiplying out the two brackets gives four terms:

(Z v - r)^{⊤} (Z v - r) = v^{⊤} Z^{⊤} Z v - v^{⊤} Z^{⊤} r - r^{⊤} Z v + r^{⊤} r

On the numbers: $v^{⊤} Z^{⊤} Z v = ‖ Z v ‖^{2} = 9 + 1 = 10$ ; $Z^{⊤} r = (\begin{matrix} 1 \cdot 3 + 0 \cdot 2 \\ 2 \cdot 3 + 1 \cdot 2 \end{matrix}) = (\begin{matrix} 3 \\ 8 \end{matrix})$ , so $v^{⊤} Z^{⊤} r = 3 + 8 = 11$ ; $r^{⊤} Z v = 3 \cdot 3 + 2 \cdot 1 = 11$ ; $r^{⊤} r = 9 + 4 = 13$ . Total: $10 - 11 - 11 + 13 = 1$ , the direct answer. The two middle terms are the same number - §3 said why: $(v^{⊤} Z^{⊤} r)^{⊤} = r^{⊤} Z v$ by the reversal rule, and a $1 \times 1$ is its own transpose - so they merge, and the expansion has three terms: $v^{⊤} Z^{⊤} Z v$ , minus $2 r^{⊤} Z v$ , plus $r^{⊤} r$ . For the gradient in $v$ : Fact B handles the first term, because $Z^{⊤} Z = (\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix}) (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix}) = (\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix})$ is symmetric (§3 showed that any $K^{⊤} K$ is); Fact A handles the second, whose fixed vector is the column $Z^{⊤} r$ ; the third has no $v$ in it and differentiates to zero. The expansion and its assembly in the homework's letters are yours to write at home; check the shape ( $d \times 1$ ) and compare with Lecture 3, slide 15, remembering the slide carries a $\frac{1}{2 n}$ that the exercise does not.

Fund read. Your backtest loss as a function of the weights is a squared norm of $X θ - y$ : expand it and it is a quadratic form in $θ$ plus a linear term plus a constant, and Facts A and B give its gradient - the arrow in Figure 2 that says which way to turn each weight. Gradient descent (Lecture 4) is nothing but "step against the arrow, repeat."

Write this down: partial derivative = the slope from nudging ONE knob · gradient = the column of all partial slopes, same shape as $u$ , never a row · $\nabla (c^{⊤} u) = c$ (checked: $(5, - 3)$ and $(2, - 1, 4)$ ) · $\nabla (u^{⊤} S u) = 2 S u$ only for symmetric $S$ ; the 2 is the two cross terms (checked: slope 8 by nudge, $S u$ said 4) · for non-symmetric $N$ it is $(N + N^{⊤}) u$ · nudge test: $0.01$ in, rise $\approx$ slope $\times 0.01$ out.

6. ℛ versus ℛ̂ - light on the hat, which the cycle-1 exam verified; the Bayes floor, from zero

Your paragraph is the definition, and it held under exam conditions (cycle-1 items 4, 16, 19, 20, all right at confidence 8 to 10 - the diagnostic's "the hat is a vector" is gone). ℛ is your true strikeout rate against every pitcher alive; ℛ̂ is your strikeout rate over the games you have played. You can only count the games you played and hope they were representative. Nothing to add to that. The hat half of this section is light; the Bayes half, which your notes could not write down, is built again from the same story.

Numbers, once. Five games, two strikeouts: $\hat{ℛ} = \frac{2}{5} = 0.4$ . That is a number you can print today. $ℛ$ is what the rate would be over every game you will ever play - a number that exists and that you cannot print. The hat says "estimated from the sample," on every letter that wears it: $\hat{ℛ}$ , $\hat{θ}$ , $\hat{f}$ (your line 16). Fund read: the backtest is $\hat{ℛ}$ - the games you have played; live is $ℛ$ .

The floor, in the same story. Even a batter who knew every pitcher's tendencies perfectly would still strike out sometimes - some pitches are unhittable. The best possible plan, and the strikeout rate it would have, is the Bayes predictor and the Bayes risk: the floor no plan gets under. One word first: the loss is the penalty for announcing $z$ when the truth was $y$ ; the course's is half the squared miss, $\frac{1}{2} (y - z)^{2}$ - miss by $2$ and pay $2$ , miss by $1$ and pay $\frac{1}{2}$ , miss by $0$ and pay nothing. Lecture 2, slide 10, then defines the plan one input at a time: at a fixed $x$ , the conditional risk $r_{x} (z)$ is the average loss of announcing $z$ when the input is that $x$ , and the Bayes predictor announces the $z$ that makes it smallest.

One computation, so "the best $z$ " is not a bare statement (this illustrates 1.1; it is not its proof). Suppose at some fixed $x$ the outcome $y$ is $1$ half the time and $3$ half the time, with the course's loss $\frac{1}{2} (y - z)^{2}$ . A ladder of announcements. Fails: announce $z = 1$ : the loss is $0$ when $y = 1$ and $\frac{1}{2} \cdot 4 = 2$ when $y = 3$ ; average $1.0$ . Fails the same way: announce $z = 3$ : the loss is $\frac{1}{2} \cdot 4 = 2$ when $y = 1$ and $0$ when $y = 3$ ; average $1.0$ . Works: announce $z = 2$ , the average of the two outcomes: the loss is $\frac{1}{2} \cdot 1$ either way; average $0.5$ . Fails, less badly: announce $z = 2.5$ : losses $\frac{1}{2} \cdot 2.25 = 1.125$ and $\frac{1}{2} \cdot 0.25 = 0.125$ ; average $0.625$ . The middle wins, and the pattern you are seeing - the announcement that minimises average squared loss is the average outcome - is the claim Exercise 1.1 asks you to prove in general.

The symbol 1.1 opens by asking you to write. The best announcement is not a number you read off; it is the announcement at which the conditional risk is smallest, and mathematics has a symbol that says exactly that. Write the conditional risk at input $s$ , for announcement $c$ , as $ρ (c | s)$ - read "rho of c given s", the average loss you take if you announce $c$ whenever the input is $s$ . Then the best announcement at $s$ is written

{argmin}_{c} ρ (c | s)

read "arg min over c of rho of c given s". The important half is the word arg: $min$ would be the smallest VALUE the risk takes, a number; $argmin$ is the ARGUMENT that achieves it - the announcement itself. On the two-outcome ladder above, the smallest average loss was $0.5$ , so the $min$ is $0.5$ and the $argmin$ is $z = 2$ . Those are different objects and the exercise asks for the second. Your homework writes this same object with a starred letter of its own; the symbol and the reading are what you need.

The technique, shown on the same two numbers (this is the move 1.1 needs; the proof in general is yours). Call the average outcome $m$ ; here $m = 2$ . The trick is to add and subtract $m$ inside the square: $y - z = (y - m) + (m - z)$ , so

(y - z)^{2} = (y - m)^{2} + 2 (y - m) (m - z) + (m - z)^{2}

Now average over the two outcomes $y = 1$ and $y = 3$ , term by term. First term: $(1 - 2)^{2}$ and $(3 - 2)^{2}$ are both $1$ , average $1$ - and $z$ is not in it. Middle term: $2 (m - z)$ is a fixed number for a fixed $z$ , so it comes out front, and what is left to average is $y - m$ : $(1 - 2) = - 1$ and $(3 - 2) = 1$ , average $0$ . The middle term averages to zero, whatever $z$ is. Last term: $(2 - z)^{2}$ , the only piece that depends on $z$ , and it is smallest - zero - at $z = 2 = m$ . Put the $\frac{1}{2}$ of the loss back: average loss $= \frac{1}{2} (1 + (2 - z)^{2})$ , which gives $0.5$ at $z = 2$ , $1.0$ at $z = 1$ , $0.625$ at $z = 2.5$ - the numbers above. The same three steps with an average over a whole distribution instead of two outcomes are Exercise 1.1; Part B's last section (conditional expectation and the tower property) gives the two remaining words 1.1 uses. Two facts it leans on, with numbers. (i) Reading a normal distribution's parameters: $N (m, s^{2})$ says the outcomes average $m$ and scatter with variance $s^{2}$ - variance is the average squared distance from the average, so variance $4$ means a typical miss of $2$ . The first number, $m$ , is all the announcement needs; the variance only sets how big the floor is. (ii) Averages pass through a fixed multiply-and-add - linearity of expectation: outcomes $1$ and $3$ average $2$ ; the outcomes $5 y + 1$ are $6$ and $16$ , and they average $11 = 5 \cdot 2 + 1$ . That is the rule that let $2 (m - z)$ come out front of the middle term above.

Write this down: hat = estimated from the sample, never direction · $\hat{ℛ} = \frac{2}{5}$ from five games is printable, $ℛ$ is not · the Bayes predictor = the best announcement at each $x$ ; its risk is the floor · at a fixed $x$ , minimising average squared loss over $z$ is a one-variable problem · add and subtract the average $m$ : the cross term averages to zero, only $(m - z)^{2}$ depends on $z$ (checked on $y = 1, 3$ ).

7. Invertible, from zero - the floor under Exercise 1.5

Undo, on one number. $2 \times 3 = 6$ ; to get the $3$ back, divide by $2$ : $\frac{6}{2} = 3$ . Multiplying by $2$ is undoable - there is an operation that takes you back. Now $0 \times 3 = 0$ and $0 \times 5 = 0$ : from the $0$ you cannot tell whether you started at $3$ or $5$ . Multiplying by $0$ is not undoable. That is the whole idea; the rest is the same idea for grids.

Undo, on a grid - simple, works. $K = (\begin{matrix} 2 & 0 \\ 0 & 3 \end{matrix})$ doubles the first entry and triples the second: $K (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 2 \\ 3 \end{matrix})$ , $K (\begin{matrix} 5 \\ 7 \end{matrix}) = (\begin{matrix} 10 \\ 21 \end{matrix})$ . To undo: halve the first entry, third the second - the grid $K^{- 1} = (\begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{3} \end{matrix})$ (read "K inverse"). Check: $K^{- 1} (\begin{matrix} 10 \\ 21 \end{matrix}) = (\begin{matrix} 5 \\ 7 \end{matrix})$ , back where you started. And $K^{- 1} K = (\begin{matrix} \frac{1}{2} \cdot 2 + 0 \cdot 0 & \frac{1}{2} \cdot 0 + 0 \cdot 3 \\ 0 \cdot 2 + \frac{1}{3} \cdot 0 & 0 \cdot 0 + \frac{1}{3} \cdot 3 \end{matrix}) = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ - do $K$ , then undo it, and nothing has changed.

Definition (two words). $I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})$ is the identity: $1$ s on the diagonal, $0$ s elsewhere, the grid that changes nothing ( $I (\begin{matrix} 5 \\ 7 \end{matrix}) = (\begin{matrix} 5 \\ 7 \end{matrix})$ ). A square grid $K$ is invertible when there is a grid $K^{- 1}$ with $K^{- 1} K = I$ - an undo. Equivalently: every output comes from exactly one input, so you can always trace back. A square grid that is not invertible is called singular.

Simple, fails. $N = (\begin{matrix} 2 & 4 \\ 1 & 2 \end{matrix})$ . $N (\begin{matrix} 1 \\ 1 \end{matrix}) = (\begin{matrix} 2 \cdot 1 + 4 \cdot 1 \\ 1 \cdot 1 + 2 \cdot 1 \end{matrix}) = (\begin{matrix} 6 \\ 3 \end{matrix})$ and $N (\begin{matrix} 3 \\ 0 \end{matrix}) = (\begin{matrix} 2 \cdot 3 + 0 \\ 3 + 0 \end{matrix}) = (\begin{matrix} 6 \\ 3 \end{matrix})$ . Two different inputs, the same output - from $(6; 3)$ there is no way back, exactly like the $0 \times 3 = 0 \times 5$ above. Not invertible. The tell: the second column of $N$ is twice the first, so $N$ squashes two directions into one. Figure 3 draws both cases.

Figure 3. Two inputs, two outputs (left): the grid can be undone. Two inputs, one output (right): it cannot. The ad minus bc test decides a 2 by 2 in one line.

Harder, works. $K = (\begin{matrix} 2 & 1 \\ 1 & 3 \end{matrix})$ , the $S$ of §4. The $2 \times 2$ test: $a d - b c = 2 \cdot 3 - 1 \cdot 1 = 5 \neq 0$ , so it is invertible, and the recipe for a $2 \times 2$ is: swap the diagonal, negate the other two, divide by that number: $K^{- 1} = \frac{1}{5} (\begin{matrix} 3 & - 1 \\ - 1 & 2 \end{matrix})$ . Why $a d - b c$ : when $a$ and $c$ are not zero, the second column $(b; d)$ is a multiple of the first $(a; c)$ exactly when $\frac{b}{a} = \frac{d}{c}$ ; cross-multiplying gives $a d = b c$ , and that form is the test that keeps working even when a column has a zero in it - for the failing $N$ above, $\frac{4}{2} = \frac{2}{1}$ , so $a d - b c = 0$ and the two columns squash into one direction. Check the inverse, row against column: $K K^{- 1} = \frac{1}{5} (\begin{matrix} 2 \cdot 3 + 1 \cdot (- 1) & 2 \cdot (- 1) + 1 \cdot 2 \\ 1 \cdot 3 + 3 \cdot (- 1) & 1 \cdot (- 1) + 3 \cdot 2 \end{matrix}) = \frac{1}{5} (\begin{matrix} 5 & 0 \\ 0 & 5 \end{matrix}) = I$ .

Harder, fails - one size up, and the word for it. $N = (\begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{matrix})$ : the third column is the first column plus the second. Then $N (\begin{matrix} 1 \\ 1 \\ - 1 \end{matrix}) = (\begin{matrix} 1 + 0 - 1 \\ 0 + 1 - 1 \\ 1 + 1 - 2 \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix})$ , and of course $N$ sends $(0; 0; 0)$ to $(0; 0; 0)$ too - two inputs, one output, singular. The set of all inputs $v$ with $N v = 0$ is the null space of $N$ (Lecture 1, slide 23); it always holds $v = 0$ , and a grid is invertible exactly when it holds nothing else - when $N v = 0$ forces $v = 0$ . And a set of columns where no column is a combination of the others is called linearly independent; $N$ 's columns are not (the third is a combination), which is the same failure seen from the columns' side.

Why the fund cares - the normal equation, on numbers. Least squares means: choose the weights $θ$ that make the squared misses $‖ X θ - y ‖^{2}$ as small as possible. §5 gives the gradient, and a candidate bottom is where it is zero - §15 shows why a zero slope alone is not enough. Do that to this bowl (its gradient is the three-term expansion of §5 - Exercise 1.4c) and, after the 2s cancel, what is left is the normal equation $X^{⊤} X θ = X^{⊤} y$ (Lecture 3, slide 15) - a system whose grid is $X^{⊤} X$ . A tiny board: three names, two features (a constant $1$ and one signal), one graded outcome each:

X = (\begin{matrix} 1 & 1 \\ 1 & 2 \\ 1 & 3 \end{matrix}), y = (\begin{matrix} 2 \\ 3 \\ 5 \end{matrix})

Then $X^{⊤} X = (\begin{matrix} 3 & 6 \\ 6 & 14 \end{matrix})$ (row against column: $1 + 1 + 1$ , $1 + 2 + 3$ , $1 + 4 + 9$ ) and $X^{⊤} y = (\begin{matrix} 10 \\ 23 \end{matrix})$ ( $2 + 3 + 5$ ; $2 + 6 + 15$ ). The test: $3 \cdot 14 - 6 \cdot 6 = 6 \neq 0$ , invertible, so the system has exactly one solution: $\hat{θ} = (X^{⊤} X)^{- 1} X^{⊤} y = \frac{1}{6} (\begin{matrix} 14 & - 6 \\ - 6 & 3 \end{matrix}) (\begin{matrix} 10 \\ 23 \end{matrix}) = \frac{1}{6} (\begin{matrix} 2 \\ 9 \end{matrix}) = (\begin{matrix} 0.333 \\ 1.5 \end{matrix})$ . Read: theta-hat equals the inverse of X-transpose-X, times X-transpose-y; Check it against the equation: $X^{⊤} X \hat{θ} = (\begin{matrix} 3 \cdot \frac{1}{3} + 6 \cdot 1.5 \\ 6 \cdot \frac{1}{3} + 14 \cdot 1.5 \end{matrix}) = (\begin{matrix} 1 + 9 \\ 2 + 21 \end{matrix}) = (\begin{matrix} 10 \\ 23 \end{matrix}) = X^{⊤} y$ . The fitted rule is "score $= 0.333 + 1.5 \times$ signal", and its three predictions $1.83$ , $3.33$ , $4.83$ miss the outcomes by $0.17$ , $- 0.33$ , $0.17$ .

The same board with a duplicated signal - fails. Replace the constant column with a second copy of the signal, scaled: $X = (\begin{matrix} 1 & 2 \\ 2 & 4 \\ 3 & 6 \end{matrix})$ (column 2 is twice column 1 - a signal beside a second copy of itself, doubled; RVOL on two nearly identical volume baselines does the same thing less visibly). Then $X^{⊤} X = (\begin{matrix} 14 & 28 \\ 28 & 56 \end{matrix})$ , and the test: $14 \cdot 56 - 28 \cdot 28 = 784 - 784 = 0$ . Singular. What that means on the board: $θ = (2; 0)$ and $θ = (0; 1)$ give the identical scores $X θ = (2; 4; 6)$ - two different weight vectors, the same backtest, so the backtest cannot choose between them, nor between the infinitely many others on the line joining them. The fit does not fail loudly; it hands you one of them, and the weights mean nothing. Rank - the largest number of columns you can pick with none of the picked ones a combination of the others (equivalently, the number of columns minus the nullity) - is the diagnostic: full column rank (the columns of $X$ are linearly independent - $X v = 0$ only for $v = 0$ ) is exactly the condition that makes $X^{⊤} X$ invertible, which is what Lecture 3 asserts and Exercise 1.5 asks you to prove.

What 1.5 asks, and where this guide gives the tools. 1.5 asks you to prove that full column rank makes $X^{⊤} X$ invertible and that $\hat{θ}$ is the unique minimiser. The two tools it turns on - left-multiply $X^{⊤} X v = 0$ by $v^{⊤}$ to get $‖ X v ‖^{2} = 0$ , and completing the square around the candidate to show nothing else is as low - are not in the four lectures; they are Part B's 1.5 sections, on a parallel board with other letters. Read this section as the floor: what invertible means, the $2 \times 2$ test, the null space, and why the fund cares.

HOW TO CHECK: is this matrix invertible?

Is it square? If not, it has no inverse; ask instead whether $K^{⊤} K$ is.
For a $2 \times 2$ $(a, b; c, d)$ : invertible exactly when $a d - b c \neq 0$ .
Any size: can you find a nonzero $v$ with $K v = 0$ - a column that is a combination of the others? If yes, singular.
Fund version: before trusting a fitted $θ$ , check the feature columns for proportional pairs.

Write this down: invertible = undoable = every output has one input = $K v = 0$ only for $v = 0$ · $I$ changes nothing; $K^{- 1} K = I$ · $2 \times 2$ test: $a d - b c \neq 0$ (checked: $(2, 0; 0, 3)$ yes, $(2, 4; 1, 2)$ no) · a column that is a combination of the others → singular · least squares = make $‖ X θ - y ‖^{2}$ smallest, gradient to zero gives $X^{⊤} X θ = X^{⊤} y$ · proportional feature columns → $X^{⊤} X$ singular → many $θ$ , one backtest · full column rank → $X^{⊤} X$ invertible → $\hat{θ} = (X^{⊤} X)^{- 1} X^{⊤} y$ .

PART B - what the homework still needs, taught on other letters

Part A closed at §7. Part B is every demand of Homework 1 that no earlier guide taught, in the order the ideas depend on each other: the one trick (§8), the three legs of a ring (§9-§13), the two halves of a uniqueness proof (§14-§15), the projector and its numbers (§16-§18), and the noise (§19-§24). Read it at work with no pen: every arithmetic step is printed. The pen work is the one line at the foot of each section, collected at the end for the evening.

The letters of Part B

One letter, one meaning from here to the end of the guide. Six letters are re-used: $S$ , $H$ , $G$ , $Z$ , $u$ and $K$ named other objects in Part A (§1-§7), and those Part A objects do not appear again after §7 - so from §8 on, read every letter off this table and not off Part A. From §8 on, no letter of the homework's appears as an object. Part A is different and you should know it: §2, §3, §5 and §7 still write the board $X$ , the outcome $y$ and the weights - the homework's own symbols for those same objects - because they are Guide #2's sections carried across whole. So in Part A, re-letter before you copy anything onto your sheet; from §8 on you cannot copy by accident because nothing here shares a name with the exercise.

Letter	Read	What it is	First used
$Q$	"Q"	the running symmetric grid $(\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$	§8
$J$	"J"	the running three-by-three grid $(\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix})$	§8
$H$	"H"	$(\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix})$ , the second symmetric grid; it is also $Q$ times $Q$	§10
$G$	"G"	a tall grid, more rows than columns; with numbers it is $(\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{matrix})$ , and $G^{⊤} G = Q$	§8
$G_{1}$ , $G_{4}$ , $G_{5}$ , $G_{6}$	"G one", "G four" ...	the other tall grids, each printed where it appears: $G_{1}$ has a column that is minus the other; $G_{4}$ and $G_{5}$ have four rows; $G_{6}$ has one row	§8, §14, §15
$Q_{1}$	"Q one"	the symmetric grid that fails the sandwich test	§9
$F$	"F"	the three-name grid $(\begin{matrix} 1 & - 1 \\ 1 & 0 \\ 1 & 1 \end{matrix})$ : a constant column and a centred signal	§16
$Z$	"Z"	the projector built from $F$ ; $Z_{1}, Z_{2}$ are two toy grids	§16
$I$	"the identity"	ones on the diagonal, zeros elsewhere; $I_{2}$ says the size	§10
$W$	"W"	the grid whose columns are the unit eigenvectors of $Q$ (and of $H$ ); $W_{1}$ the unscaled version	§11
$Λ$	"capital lambda"	the diagonal grid of eigenvalues; $Λ^{1 / 2}$ its entrywise root	§11
$S$	"S"	the factor $Λ^{1 / 2} W^{⊤}$ built for $H$ , with $S^{⊤} S = H$	§12
$K, L, R$	"K, L, R"	$(\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ , $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ , $diag (1, 2)$ - Guide #2's $K$ and $L$	§18, §19
$Y$	"Y"	$(\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , where the transpose-of-an-inverse rule is shown	§16
$v$	"v"	the column you feed to a grid; $v_{1}, v_{2}$ its entries; $v'$ a second such column	§8
$λ$	"lambda"	an eigenvalue, a stretch factor	§10
$ω_{1}, ω_{2}, ω_{3}$	"omega one ..."	unit eigenvectors, the columns of $W$	§11
$α, κ, ζ$	"alpha, kappa, zeta"	plain real numbers in the ring toy	§13
$r, \hat{r}$	"r, r hat"	the weight column being fitted, the least-squares candidate	§15
$r_{0}$	"r zero"	the true weights behind the outcome	§23
$q$	"q"	a perturbation, $q = r - \hat{r}$	§15
$t$	"t"	the outcome: a column of three in §15 and §23, a single number in §24	§15
$k$	"k"	a single knob, in the one-knob rungs	§15
$ℓ$	"ell"	the objective being minimised	§15
$p_{i}$	"p i"	the $i$ -th probe: a one in slot $i$ , zeros elsewhere	§17
$η, δ$	"eta, delta"	the noise column, each entry a coin showing $2$ or $- 2$ ; and a fresh independent copy	§19, §22
$τ^{2}$	"tau squared"	the common variance of the noise, here $4$	§20
$u$	"u"	a random column, §19 only	§19
$a$	"a"	a fixed weight column	§21
$γ, β$	"gamma, beta"	$γ = (Z η)_{2}$ , a random number built from the coins; $β$ any second random number	§22
$s, c, m$	"s, c, m"	the input, the announcement, the conditional mean	§24
$d, n$ ; $i, j$		columns and rows of a grid; slot counters	§14

$𝔼$ , $ℙ$ , $Var$ , $Cov$ , $tr$ , $det$ , $diag$ , $null$ , $rank$ and $nullity$ are each defined where they first appear.

8. Left-multiply by v-transpose: a matrix equation becomes a squared length

You cannot read a sign off a column. Put its transpose in front and everything collapses to one number - and when the grid is something-transpose-times-something, that number is a squared length.

Symbols, defined before use. $G$ : tall grid, $3 \times 2$ ; $G^{⊤} G$ ("G transpose G"): square, $2 \times 2$ . $v \in ℝ^{2}$ : column with entries $v_{1}, v_{2}$ ; $G v$ : column in $ℝ^{3}$ . Dot product of same-length columns, $v^{⊤} v'$ : multiply matching entries, add, one number. $‖ G v ‖^{2}$ , squared length of $G v$ : sum of squares of its entries, i.e. $(G v)^{⊤} (G v)$ . Sandwich $v^{⊤} Q v$ : row, grid, column, one number. Reversal (§3): $(G v)^{⊤} = v^{⊤} G^{⊤}$ . Point-separating (Lecture 1, slide 22): a squared length is zero only for the zero column.

Formal. For any $G$ and matching column $v$ ,

v^{⊤} (G^{⊤} G) v = (G v)^{⊤} (G v) = ‖ G v ‖^{2}

Read: the sandwich of $v$ around G-transpose-G is the squared length of G v.

Middle-school version. The only bag of coins weighing nothing is the empty one. A squared length adds squares, none negative: a zero reading means every square was zero.

Simple, works. $G = (\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{matrix})$ , so $G^{⊤} G = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$ . Call it $Q$ (§9). $v = (1; 2)$ : $G v = (1; 2; 3)$ , $‖ G v ‖^{2} = 1 + 4 + 9 = 14$ ; $Q v = (2 + 2; 1 + 4) = (4; 5)$ , $v^{⊤} Q v = 1 \cdot 4 + 2 \cdot 5 = 14$ .

Harder, fails. The trick works; the CONCLUSION fails. $G_{1} = (\begin{matrix} 1 & - 1 \\ 2 & - 2 \\ 0 & 0 \end{matrix})$ gives $G_{1^{⊤}} G_{1} = (\begin{matrix} 5 & - 5 \\ - 5 & 5 \end{matrix})$ , and $v = (1; 1)$ gives $G_{1^{⊤}} G_{1} v = (5 - 5; - 5 + 5) = (0; 0)$ . Left-multiply: $‖ G_{1} v ‖^{2} = 0$ , so $G_{1} v = (1 - 1; 2 - 2; 0) = (0; 0; 0)$ . But $v \neq 0$ : " $G v = 0$ forces $v = 0$ " is a SEPARATE hypothesis (§14), false here.

Harder, works. $G_{4} = (\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{matrix})$ , so $G_{4^{⊤}} G_{4} = (\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix})$ , called $J$ . $v = (1; 2; - 1)$ : $G_{4} v = (1; 2; - 1; 2)$ , $‖ G_{4} v ‖^{2} = 1 + 4 + 1 + 4 = 10$ ; $J v = (3; 4; 1)$ , $v^{⊤} J v = 3 + 8 - 1 = 10$ . Also $J v = 0$ gives $‖ G_{4} v ‖^{2} = 0$ , so $G_{4} v = 0$ , whose first three entries are $v_{1}, v_{2}, v_{3}$ , so $v = 0$ .

Parallel worked example (different letters and sizes from your homework). A sign: for $Q = G^{⊤} G$ and $v \neq 0$ , $v^{⊤} Q v = ‖ G v ‖^{2} > 0$ , since $G v \neq 0$ - the entries $v_{1}$ , $v_{2}$ , $v_{1} + v_{2}$ cannot all vanish; in your 1.2 invertibility supplies $G v \neq 0$ . An equation: from $G^{⊤} G v = 0$ to $G v = 0$ , the four lines of §14. In your homework's letters this is one leg of Exercise 1.2 and the opening line of Exercise 1.5 - yours to write.

HOW TO CHECK: can I use the left-multiply trick here?

If the grid is not $G^{⊤} G$ , the sandwich need not be a squared length.
Left-multiply by $v^{⊤}$ ; regroup as $(G v)^{⊤} (G v) = ‖ G v ‖^{2}$ .
Zero squared length forces $G v = 0$ always; $v = 0$ needs full column rank.
Sanity: $G$ , $v = (1; 2)$ give $14$ both routes; $G_{1}$ fails the last step.

Fund read. Sandwiching a weight-error direction around the board's board-transpose-times-board grid gives the squared length of the thirty score changes it causes.

Write this down: $v^{⊤} (G^{⊤} G) v = ‖ G v ‖^{2}$ ; from $G^{⊤} G v = 0$ , left-multiply by $v^{⊤}$ , get $G v = 0$ ; then $v = 0$ needs full column rank.

9. Positive definite and semidefinite

A square grid is positive definite when it turns every nonzero column into a positive number through the sandwich. Take the grid with two on the diagonal and one off it, and the column one-on-top-of-one: the sandwich gives $2 + 1 + 1 + 2 = 6$ , positive.

Symbols, defined before use. $Q$ (read "Q") is a square symmetric grid - equal to its own flip - here $Q = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$ . $v \in ℝ^{2}$ is a column with entries $v_{1}, v_{2}$ , and $v^{⊤} Q v$ is the sandwich of §8. $\neq$ ("is not equal to"); $>$ ("is greater than"); $\geq$ ("is at least"). $v \neq 0$ means $v$ is not the column of zeros.

Formal. A symmetric grid $Q$ is positive definite when

v^{⊤} Q v > 0

for every column $v \neq 0$ ; it is positive semidefinite when $v^{⊤} Q v \geq 0$ for every $v$ . Read: whichever nonzero column you sandwich, the number that comes out is positive (definite), or at least never negative (semidefinite).

Middle-school version. A scale that only ever reads a positive weight for anything standing on it is the definite kind; one that sometimes reads exactly zero for something real is semidefinite but not definite. The test is never one object; it is every object you could ever put on it.

Simple, fails. $Q_{1} = (\begin{matrix} 1 & 2 \\ 2 & 1 \end{matrix})$ and $v = (1; - 1)$ . $Q_{1} v = (1 - 2; 2 - 1) = (- 1; 1)$ , and $v^{⊤} Q_{1} v = 1 \cdot (- 1) + (- 1) \cdot 1 = - 2$ . Negative, so $Q_{1}$ is not positive definite and not even semidefinite.

Simple, works. $Q = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$ . At $v = (1; - 1)$ : $Q v = (2 - 1; 1 - 2) = (1; - 1)$ and $v^{⊤} Q v = 1 \cdot 1 + (- 1) (- 1) = 2$ , positive where $Q_{1}$ gave $- 2$ . Un-compress the sandwich as in §4: $v^{⊤} Q v = 2 v_{1^{2}} + 2 v_{1} v_{2} + 2 v_{2^{2}}$ . Complete the square: $2 v_{1^{2}} + 2 v_{1} v_{2} + 2 v_{2^{2}} = (v_{1} + v_{2})^{2} + v_{1^{2}} + v_{2^{2}}$ . Three squares added; the sum is zero only when $v_{1} = 0$ and $v_{2} = 0$ . So the sandwich is positive for every nonzero $v$ : positive definite.

Harder, works. $J = (\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix})$ . Un-compress, nine terms, cross terms merged into: $v^{⊤} J v = 2 v_{1^{2}} + 2 v_{2^{2}} + 2 v_{3^{2}} + 2 v_{1} v_{2} + 2 v_{1} v_{3} + 2 v_{2} v_{3}$ . Complete the square: this equals $(v_{1} + v_{2} + v_{3})^{2} + v_{1^{2}} + v_{2^{2}} + v_{3^{2}}$ . Four squares, three of which are the coordinates themselves, so zero only at $v = 0$ . Positive definite.

Parallel worked example (different letters and sizes from your homework). A second proof about $Q$ , the one that scales, on the §8 numbers - $Q = G^{⊤} G$ , so $v^{⊤} Q v = ‖ G v ‖^{2}$ - a squared length, never negative, so $Q$ is at least semidefinite. It is zero only when $G v = 0$ , that is $v_{1} = 0$ , $v_{2} = 0$ , $v_{1} + v_{2} = 0$ - only $v = 0$ . The step " $G v = 0$ only for $v = 0$ " is what will carry the weight in your 1.2 (there the factor is invertible) and 1.5 (there it is full column rank).

HOW TO CHECK: is this symmetric grid positive definite?

Un-compress $v^{⊤} Q v$ into squares and merged cross terms.
Complete the square: squares covering every coordinate on its own means definite.
Before any algebra, try $(1; 0)$ , $(0; 1)$ , $(1; 1)$ , $(1; - 1)$ . One negative kills it; four positives prove nothing.
Sanity: $Q$ at $(1; - 1)$ must give $2$ , and $Q_{1} = (1, 2; 2, 1)$ at $(1; - 1)$ must give $- 2$ .

Fund read. The five-by-five feature grid of the board is always semidefinite, and definite exactly when no feature column is a mix of the others.

Write this down: positive definite: $v^{⊤} Q v > 0$ for EVERY nonzero $v$ (semidefinite: $\geq 0$ ); one negative column kills it, positives prove nothing; prove it by completing the square.

10. Eigenvalues and eigenvectors

Most columns come out of a grid pointing a new way; a few come out pointing the way they went in, only stretched. Those columns are the grid's eigenvectors, and the stretch factors are its eigenvalues.

Symbols, defined before use. $λ$ (read "lambda") is a number, the stretch factor. $Q v = λ v$ (read "Q v equals lambda v"): pushing $v$ through $Q$ gives back $v$ scaled by $λ$ . $I$ (read "the identity") is the grid with ones on the diagonal, zeros elsewhere; it hands back whatever it is given (§7). $‖ v ‖^{2} = v^{⊤} v$ is the squared length of §8.

Formal. A number $λ$ is an eigenvalue of the square grid $Q$ , and a column $v \neq 0$ is an eigenvector for it, when

Q v = λ v

Read: Q does nothing to v except scale it by lambda.

Middle-school version. Pull a rubber sheet twice as much left-to-right as up-and-down. Almost every arrow on it tilts, but two do not - the one straight across, which doubles, and the one straight up, which keeps its length.

Simple, fails. $Q = (\begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix})$ , $v = (1; 0)$ : $Q v = (2; 1)$ . For this to be $λ v = (λ; 0)$ the second entry would have to be $0$ ; it is $1$ .

Simple, works. $v = (1; 1)$ : $Q v = (2 + 1; 1 + 2) = (3; 3) = 3 \cdot (1; 1)$ . An eigenvector, $λ = 3$ . And $v = (1; - 1)$ : $Q v = (2 - 1; 1 - 2) = (1; - 1) = 1 \cdot (1; - 1)$ : an eigenvector, $λ = 1$ . The same two columns work for $H = (\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix})$ : $H (1; 1) = (9; 9)$ , so $λ = 9$ ; $H (1; - 1) = (5 - 4; 4 - 5) = (1; - 1)$ , so $λ = 1$ .

Harder, works. $J$ with $v = (1; 1; 1)$ : $J v = (2 + 1 + 1; 1 + 2 + 1; 1 + 1 + 2) = (4; 4; 4) = 4 v$ , so $λ = 4$ . With $v = (1; - 1; 0)$ : $J v = (2 - 1; 1 - 2; 1 - 1) = (1; - 1; 0) = 1 \cdot v$ , so $λ = 1$ . With $v = (1; 1; - 2)$ : $J v = (2 + 1 - 2; 1 + 2 - 2; 1 + 1 - 4) = (1; 1; - 2)$ , $λ = 1$ again. Repeated eigenvalues are allowed; §11 needs all three.

Parallel worked example (different letters and sizes from your homework). Feed an eigenvector into the sandwich of §9. Numbers: $Q$ , $v = (1; 1)$ , $λ = 3$ : $v^{⊤} Q v = v^{⊤} (Q v) = v^{⊤} (3 v) = 3 \cdot v^{⊤} v = 3 \cdot 2 = 6$ . Directly: $2 + 1 + 1 + 2 = 6$ . Letters: if $Q v = λ v$ with $v \neq 0$ , then

v^{⊤} Q v = v^{⊤} (λ v) = λ v^{⊤} v = λ ‖ v ‖^{2}

and $‖ v ‖^{2} > 0$ because $v \neq 0$ (§8: a squared length is zero only for the zero column). So the sandwich value at an eigenvector has the same sign as lambda; if every sandwich value is positive (§9), that hands you the sign of every eigenvalue. In your homework's letters this is the (a) implies (b) leg of Exercise 1.2 - yours to write.

HOW TO CHECK: is v an eigenvector of Q, and what is lambda?

Compute $Q v$ .
Every ratio $(Q v)_{i} / v_{i}$ must be the SAME number - that is lambda - and where $v_{i} = 0$ , $(Q v)_{i}$ must be $0$ too.
Confirm $v \neq 0$ .

Fund read. The board's five-feature grid has five eigenvectors, with five eigenvalues saying how much variation each carries.

Write this down: $Q v = λ v$ with $v \neq 0$ : $v$ an eigenvector, $λ$ its eigenvalue; at an eigenvector the sandwich is $λ ‖ v ‖^{2}$ , the same sign as $λ$ .

11. The spectral theorem and orthogonal grids

A symmetric grid comes apart into three pieces: a turn, a stretch along the axes, and the turn undone. For two-one-one-two the stretch is by one along one axis and three along the other, onto §10's eigenvectors.

Symbols, defined before use. $W$ (read "W") is the grid whose columns $ω_{1}, ω_{2}$ (read "omega one, omega two") are eigenvectors of $Q$ at length one. Columns are perpendicular when their dot product (§8) is zero; orthonormal columns are perpendicular, each of length one. $W^{⊤} W = I$ (read "W transpose W equals the identity") is the property orthogonal. $Λ$ (read "capital lambda") is a diagonal grid of eigenvalues; $diag (1, 3)$ (read "diag one, three") has $1, 3$ on the diagonal, zeros off it. $\sqrt{2}$ (read "root two") squares to $2$ ; $\frac{1}{\sqrt{2}}$ (read "one over root two") scales a root-two column to length one. $W^{- 1}$ (read "W inverse") is the undo of $W$ (§7). A column times a row, $ω_{1} ω_{1^{⊤}}$ , is a grid ( $(2 \times 1) (1 \times 2) = 2 \times 2$ ): an outer product.

Formal. $W$ is orthogonal when $W^{⊤} W = I$ ; then $W^{- 1} = W^{⊤}$ , the undo is the flip. The spectral theorem (Grosse, CSC2541 ch. 1 §2.2): every symmetric $Q$ can be written

Q = W Λ W^{⊤}

with $W$ orthogonal, its columns orthonormal eigenvectors of $Q$ , and $Λ$ diagonal, holding the matching eigenvalues, all real. Read: a symmetric grid is a turn, a stretch along the axes, then the turn back.

Middle-school version. Squash a lump of dough. There is always one viewpoint from which it was pure stretching along straight lines - turn your head there, read the stretches, turn back.

Simple, fails. Unscaled: $W_{1} = (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix})$ . Then $W_{1^{⊤}} W_{1} = (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix}) = (\begin{matrix} 1 + 1 & 1 - 1 \\ 1 - 1 & 1 + 1 \end{matrix}) = (\begin{matrix} 2 & 0 \\ 0 & 2 \end{matrix})$ , which is $2 I$ , not $I$ : perpendicular, but length root two, so $W_{1} diag (1, 3) W_{1^{⊤}}$ comes out as $2 Q$ .

Simple, works. Divide by root two: $W = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix})$ , columns $ω_{1} = \frac{1}{\sqrt{2}} (1; - 1)$ for $λ = 1$ , $ω_{2} = \frac{1}{\sqrt{2}} (1; 1)$ for $λ = 3$ , $Λ = diag (1, 3)$ , and $W^{⊤} W = \frac{1}{2} \cdot 2 I = I$ : orthogonal.

Harder, works. $J$ , eigenvalues $1, 1, 4$ (§10), unit eigenvectors $ω_{1} = \frac{1}{\sqrt{2}} (1; - 1; 0)$ , $ω_{2} = \frac{1}{\sqrt{6}} (1; 1; - 2)$ , $ω_{3} = \frac{1}{\sqrt{3}} (1; 1; 1)$ . $W^{⊤} W = I$ ; rebuild as outer products, $1 \cdot ω_{1} ω_{1^{⊤}} + 1 \cdot ω_{2} ω_{2^{⊤}} + 4 \cdot ω_{3} ω_{3^{⊤}}$ . Entry $(1, 1)$ : $\frac{1}{2} + \frac{1}{6} + \frac{4}{3} = \frac{3}{6} + \frac{1}{6} + \frac{8}{6} = 2$ ; $J$ is back.

Parallel worked example (different letters and sizes from your homework). $H = (\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix})$ has eigenvalues $1, 9$ and the eigenvectors of $Q$ (§10): same $W$ , $Λ = diag (1, 9)$ . $W Λ = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 9 \\ - 1 & 9 \end{matrix})$ , so $W Λ W^{⊤} = \frac{1}{2} (\begin{matrix} 1 + 9 & - 1 + 9 \\ - 1 + 9 & 1 + 9 \end{matrix}) = (\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix})$ . §12 uses $W^{- 1} = W^{⊤}$ to build the factor your 1.2 asks for.

HOW TO CHECK: is W orthogonal, and does W Lambda W-transpose rebuild Q?

Dot each pair of columns of $W$ : zero; each with itself: one. That is $W^{⊤} W = I$ .
Check each column is an eigenvector, $Q ω_{k} = λ_{k} ω_{k}$ , $λ_{k}$ in slot $k$ .
Sanity: for $Q$ , entry $(1, 2)$ rebuilds as $- \frac{1}{2} + \frac{3}{2} = 1$ .

Fund read. The columns of $W$ turn the five features into uncorrelated principal components, along each the board's variation is one eigenvalue.

Write this down: orthogonal: $W^{⊤} W = I$ , so $W^{- 1} = W^{⊤}$ ; spectral theorem: symmetric $Q = W Λ W^{⊤}$ , columns of $W$ unit eigenvectors, $Λ$ the eigenvalues on a diagonal.

12. The square root of a diagonal grid, and building a factor

A diagonal grid with positive entries has an obvious square root: root each entry where it sits, so $diag (1, 3) diag (1, 3) = diag (1, 9)$ . That root lets you write a positive definite grid as something-transpose times something.

Symbols, defined before use. $Λ^{1 / 2}$ (read "lambda to the one half") is the diagonal grid of the square roots of $Λ$ 's diagonal entries. $\sqrt{λ_{k}}$ (read "root lambda k") is the non-negative number whose square is $λ_{k}$ , existing only when $λ_{k} \geq 0$ . $S$ (read "S") is the factor to be built; $S^{⊤} S$ (read "S transpose S") is what it must multiply back to. Invertible is §7: undoable, determinant not zero.

Formal. For $Λ = diag (λ_{1}, \dots, λ_{d})$ with every $λ_{k} > 0$ ,

Λ^{1 / 2} = diag (\sqrt{λ_{1}}, \dots, \sqrt{λ_{d}}), Λ^{1 / 2} Λ^{1 / 2} = Λ

Read: root each diagonal entry in place; squaring the result squares each entry back to Lambda, because diagonal grids multiply entry by entry. Also $(Λ^{1 / 2})^{⊤} = Λ^{1 / 2}$ , and $Λ^{1 / 2}$ is invertible because no diagonal entry is zero.

Middle-school version. A square rug nine feet in area has a side of three feet, and side times side gives the rug back. A diagonal grid is a row of rugs, and its root is the row of their sides - as long as every area is positive, since minus nine square feet has no side.

Simple, fails. $diag (4, - 1)$ . The first slot roots to $2$ ; the second needs a real number whose square is $- 1$ , and there is none. Positivity of the eigenvalues is where this construction dies, and it is why the middle statement of your 1.2 asks for positive eigenvalues, not merely nonzero ones.

Simple, works. $diag (4, 9)$ roots to $diag (2, 3)$ , and $diag (2, 3) diag (2, 3) = (\begin{matrix} 2 \cdot 2 + 0 & 0 \\ 0 & 3 \cdot 3 \end{matrix}) = diag (4, 9)$ .

Harder, works. From §11, $H = W Λ W^{⊤}$ with $W = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & 1 \\ - 1 & 1 \end{matrix})$ and $Λ = diag (1, 9)$ , so $Λ^{1 / 2} = diag (1, 3)$ . Set $S = Λ^{1 / 2} W^{⊤} = (\begin{matrix} 1 & 0 \\ 0 & 3 \end{matrix}) \cdot \frac{1}{\sqrt{2}} (\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}) = \frac{1}{\sqrt{2}} (\begin{matrix} 1 & - 1 \\ 3 & 3 \end{matrix})$ . Then $S^{⊤} S = \frac{1}{2} (\begin{matrix} 1 & 3 \\ - 1 & 3 \end{matrix}) (\begin{matrix} 1 & - 1 \\ 3 & 3 \end{matrix})$ , whose four entries are $\frac{1 + 9}{2}$ , $\frac{- 1 + 9}{2}$ , $\frac{- 1 + 9}{2}$ , $\frac{1 + 9}{2}$ - that is $(\begin{matrix} 5 & 4 \\ 4 & 5 \end{matrix}) = H$ . And $S$ is invertible: $det S = \frac{1}{2} (1 \cdot 3 - (- 1) \cdot 3) = \frac{6}{2} = 3 \neq 0$ .

Parallel worked example (different letters and sizes from your homework). The same construction in letters, $S = Λ^{1 / 2} W^{⊤}$ :

S^{⊤} S = (Λ^{1 / 2} W^{⊤})^{⊤} (Λ^{1 / 2} W^{⊤}) = W (Λ^{1 / 2})^{⊤} Λ^{1 / 2} W^{⊤}

and a diagonal grid is its own flip, so the two roots merge:

W Λ^{1 / 2} Λ^{1 / 2} W^{⊤} = W Λ W^{⊤} = H

Three rules, one per equals sign after the first: the reversal rule with the flip of a flip giving $W$ back (§3); a diagonal grid is its own flip; root times root is Lambda. Invertible, since a product of undoable grids is undoable: $S^{- 1} = W Λ^{- 1 / 2}$ , where $Λ^{- 1 / 2} = diag (1, \frac{1}{3})$ . In your homework's letters this is the (b) implies (c) leg of Exercise 1.2 - yours to write.

HOW TO CHECK: does my factor S satisfy S-transpose S equals the target?

Every eigenvalue must be positive before you root; a zero or a negative stops it.
Multiply $S^{⊤} S$ out by the reversal rule - the two roots merge into $Λ$ .
Invertibility: the undo is $S^{- 1} = W Λ^{- 1 / 2}$ .

Fund read. A whitening grid: if the board's five-feature grid is positive definite, an $S$ with $S^{⊤} S$ equal to it turns the five raw features into five uncorrelated ones of unit spread.

Write this down: $Λ^{1 / 2}$ = root each diagonal entry (needs every $λ_{k} > 0$ ); $S = Λ^{1 / 2} W^{⊤}$ gives $S^{⊤} S = W Λ W^{⊤}$ , and $S$ is invertible.

13. Proving three statements equivalent: the ring

Three statements are equivalent when each is true exactly when the others are. Proving it takes three arguments, not six: if the first forces the second, the second the third, and the third the first, you can walk from any one to any other.

Symbols, defined before use. $\to$ ("implies"): if the left statement is true, so is the right. "If and only if" means each forces the other. (a), (b), (c) label statements. $α$ ("alpha"), $κ$ ("kappa") and $ζ$ ("zeta") are real numbers; $\sqrt{α}$ ("root alpha") is the non-negative number whose square is $α$ .

Formal. To prove that (a), (b), (c) are all equivalent, it suffices to prove

(a) \to (b), (b) \to (c), (c) \to (a)

Read: a forces b, b forces c, c forces a. Implications chain, so (b) $\to$ (a) is the walk (b) $\to$ (c) $\to$ (a).

Middle-school version. Three towns on a one-way ring road: drive first to second, second to third, third back to first, and you can reach any town from any other.

Simple, fails. Whole number $α$ : (a) $α = 2$ ; (b) $α$ is even; (c) $α$ is whole. (a) $\to$ (b): $2$ is even, true. (b) $\to$ (c): every even number is whole, true. But (c) $\to$ (a) is false: $α = 6$ is whole and is not $2$ .

Harder, fails. (a) $α = 4$ ; (b) $α > 0$ ; (c) $α$ is even. Both (a) $\to$ (b) and (a) $\to$ (c) hold: $4$ is positive, $4$ is even. But (b) $\to$ (c) fails at $α = 3$ and (c) $\to$ (a) fails at $α = 6$ . A hub with spokes is not a ring.

Harder, works. Claim. For a real number $α \neq 0$ the following are equivalent: (a) $α > 0$ ; (b) $α = κ^{2}$ for some real $κ \neq 0$ ; (c) $α \cdot ζ > 0$ for every $ζ > 0$ .

Proof. We show (a) $\to$ (b) $\to$ (c) $\to$ (a). (a) $\to$ (b). Assume $α > 0$ . Then $\sqrt{α}$ is a real number, nonzero because $α$ is nonzero. Set $κ = \sqrt{α}$ ; then $κ^{2} = α$ . So (b). (b) $\to$ (c). Assume $α = κ^{2}$ with $κ \neq 0$ . Let $ζ > 0$ be given. Then $α \cdot ζ = κ^{2} \cdot ζ$ ; $κ^{2} > 0$ since $κ \neq 0$ , and $ζ > 0$ , so the product is positive. So (c). (c) $\to$ (a). Assume (c). It holds for every positive $ζ$ , so apply it with $ζ = 1$ : $α \cdot 1 = α > 0$ . So (a). The three implications close a cycle, so any one of the statements implies each of the others, and the three are equivalent. That closes the proof.

Parallel worked example (different letters and sizes from your homework). For a SYMMETRIC grid $Q$ , statement (a): every sandwich value is positive. Statement (b): every stretch factor is positive. Statement (c): the grid is $S^{⊤} S$ for some invertible $S$ . The legs: (a) $\to$ (b) is §10's probe line, $v^{⊤} Q v = λ ‖ v ‖^{2}$ with an eigenvector as $v$ ; (b) $\to$ (c) is §12's factor, $S = Λ^{1 / 2} W^{⊤}$ , invertible because its undo is written down, and available at all only because symmetry is what lets §11 write the grid as $W Λ W^{⊤}$ ; (c) $\to$ (a) is §8's first job, $v^{⊤} S^{⊤} S v = ‖ S v ‖^{2} > 0$ for $v \neq 0$ , since an invertible $S$ sends only the zero column to zero. In your homework's letters this ring is Exercise 1.2 - yours to write.

HOW TO CHECK: does my equivalence proof close?

Count the statements: $n$ .
Count the arrows: at least $n$ , one closed ring back to the start.
Each leg assumes its tail and concludes its head, using nothing else.
Sanity: (a) $α = 2$ , (b) even, (c) whole never returns; $α = 6$ shows why.

Fund read. A mixed feature column, a singular five-by-five feature grid, and two weight vectors with the identical backtest travel together by a ring of three legs, one of which is §14.

Write this down: (a) iff (b) iff (c) is proved by the ring (a) $\to$ (b) $\to$ (c) $\to$ (a); each leg "Assume [tail] ... So [head]"; say why the ring suffices.

14. Full column rank, the null space, rank-nullity, and why G-transpose G is invertible

A grid has full column rank when no column can be built by scaling and adding the others. Such a grid, when it is tall (more rows than columns), has no inverse of its own, but its transpose times itself does.

Symbols, defined before use. A linear combination of columns is numbers times columns, added. Columns are linearly independent when only all-zero numbers give the zero column; $d$ is the columns, $n$ the rows. $null (G)$ ("null of G") is the null space: every column $v$ with $G v = 0$ . $rank (G)$ is the number of independent columns; $nullity (G)$ is the number of independent directions in the null space - zero when only the zero column is there. $(G^{⊤} G)^{- 1}$ ("G transpose G, inverse") is the undo whose existence is the claim.

Formal. For any grid $G$ with $d$ columns (rank-nullity):

rank (G) + nullity (G) = d

and full column rank means $rank (G) = d$ , equivalently nullity zero, equivalently $G v = 0$ only for $v = 0$ . The claim: then $G^{⊤} G$ is invertible, and the proof establishes

G^{⊤} G v = 0 \to v = 0

Read: columns that count plus ways to combine columns to nothing is the columns you started with; only zero goes to zero under G-transpose-G, so it undoes.

Middle-school version. If no paint tube's colour can be mixed from the others, every tube counts; tubes that count plus recipes for no paint comes to the tubes you started with.

Simple, fails. $G_{1} = (\begin{matrix} 1 & - 1 \\ 2 & - 2 \\ 0 & 0 \end{matrix})$ , column two $(- 1)$ times column one. $G_{1} (1; 1) = (1 - 1; 2 - 2; 0) = (0; 0; 0)$ . The null space is every multiple of $(1; 1)$ - nullity $1$ , rank $1$ , and $1 + 1 = 2 = d$ . And $G_{1^{⊤}} G_{1} = (\begin{matrix} 5 & - 5 \\ - 5 & 5 \end{matrix})$ (§8) has determinant $25 - 25 = 0$ : singular.

Simple, works. $G = (\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{matrix})$ : $G v = (v_{1}; v_{2}; v_{1} + v_{2})$ , zero only when $v_{1} = 0$ and $v_{2} = 0$ . Nullity $0$ , rank $2 = d$ . Its square grid $G^{⊤} G = Q$ has determinant $3 \neq 0$ .

Harder, works. $G_{5} = (\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{matrix})$ . $G_{5} v = 0$ row by row: $v_{1} + v_{2} = 0$ , $v_{2} + v_{3} = 0$ , $v_{1} + v_{3} = 0$ , $v_{1} + v_{2} + v_{3} = 0$ . From the first two, $v_{1} = - v_{2}$ and $v_{3} = - v_{2}$ ; into the third: $- v_{2} - v_{2} = - 2 v_{2} = 0$ , so $v_{2} = 0$ , then $v_{1} = v_{3} = 0$ . Nullity $0$ , rank $3 = d$ .

Parallel worked example (different letters and sizes from your homework).

Letters ( $G$ any $n \times d$ grid of full column rank). At general size the same four moves are what 1.5's first half asks you to write, and the algebra is yours: (1) assume the square grid $G^{⊤} G$ sends some column $v$ to zero; (2) left-multiply by that column's transpose and name the result a squared length, by section 8; (3) a squared length is zero only for the zero column, so the tall grid sends $v$ to zero - and here, exactly once, full column rank forces the column itself to be zero; (4) a square grid whose null space holds only the zero column has nullity zero, so by rank-nullity its rank is full and it is invertible. In your homework's letters this is the first half of Exercise 1.5 - yours to write.

HOW TO CHECK: full column rank, and a real proof?

Solve $G v = 0$ row by row; only $v = 0$ means full column rank.
Did the proof start from $G^{⊤} G v = 0$ , not from $G v = 0$ ?
Did it left-multiply by $v^{⊤}$ and use rank-nullity once?

Fund read. The board's feature grid has full column rank when no feature is a mix of the others, so its normal equation has one solution.

Write this down: full column rank = only the zero column in the null space = rank $d$ ; from $G^{⊤} G v = 0$ get $‖ G v ‖^{2} = 0$ , $v = 0$ , so $G^{⊤} G$ is invertible.

15. Uniqueness of a minimiser: completing the square around the candidate

A zero slope says the ground is flat, not that the bottom is there or is unique. To prove a point is the only bottom, write any height as the candidate's height plus something never negative and zero only there.

Symbols, defined before use. $ℓ$ (read "ell") is the objective on the unknown column $r$ (read "r"): $ℓ (r) = ‖ G r - t ‖^{2}$ , $G$ the $3 \times 2$ grid of §8, $t$ a fixed target in $ℝ^{3}$ . $\hat{r}$ (read "r hat") is the candidate minimiser from the zero-gradient equation. $q$ (read "q") is any perturbation: any $r$ is $r = \hat{r} + q$ . The normal equation (§7) is $G^{⊤} G \hat{r} = G^{⊤} t$ , that is $G^{⊤} (G \hat{r} - t) = 0$ . A stationary point has zero gradient; a saddle is stationary, neither top nor bottom. $k$ (read "k") is one knob.

Formal (completing the square). If $\hat{r}$ solves the normal equation, then for every $r$ , with $q = r - \hat{r}$ ,

‖ G r - t ‖^{2} = ‖ G \hat{r} - t ‖^{2} + ‖ G q ‖^{2}

Read: any height is the candidate's height plus the squared length of G q, since the normal equation zeroes the cross term $2 q^{⊤} G^{⊤} (G \hat{r} - t)$ . Equality needs $G q = 0$ , which under full column rank means $r = \hat{r}$ .

Middle-school version. A ledge halfway up a hill is flat too, so flat does not mean lowest. Show every step away goes up, by an amount zero only for no step.

Simple, works. $ℓ (k) = (k - 3)^{2} + 5$ . Slope $2 (k - 3)$ , zero at $k = 3$ . The square is completed: $(k - 3)^{2} \geq 0$ with equality only at $k = 3$ , so $ℓ \geq 5$ everywhere and $ℓ = 5$ only at $3$ .

Harder, fails. $ℓ (r) = (r_{1} - r_{2})^{2}$ ; gradient $(2 (r_{1} - r_{2}); - 2 (r_{1} - r_{2}))$ , zero along the line $r_{1} = r_{2}$ : not unique. Here $ℓ (r) = ‖ G_{6} r ‖^{2}$ , $G_{6} = (\begin{matrix} 1 & - 1 \end{matrix})$ ; $G_{6^{⊤}} G_{6} = (\begin{matrix} 1 & - 1 \\ - 1 & 1 \end{matrix})$ has determinant $0$ : no full column rank, and $q = (1; 1)$ gives $G_{6} q = 0$ .

Parallel worked example (different letters and sizes from your homework). Setup. $G = (\begin{matrix} 1 & 0 \\ 0 & 1 \\ 1 & 1 \end{matrix})$ , $t = (2; 3; 2)$ , $ℓ (r) = ‖ G r - t ‖^{2}$ . From $G^{⊤} G r = G^{⊤} t$ : $G^{⊤} t = (2 + 2; 3 + 2) = (4; 5)$ , so $Q r = (4; 5)$ and, with §14's undo, $\hat{r} = \frac{1}{3} (\begin{matrix} 2 & - 1 \\ - 1 & 2 \end{matrix}) (4; 5) = \frac{1}{3} (8 - 5; - 4 + 10) = (1; 2)$ . Residual $G \hat{r} - t = (1; 2; 3) - (2; 3; 2) = (- 1; - 1; 1)$ , so $ℓ (\hat{r}) = 1 + 1 + 1 = 3$ , and $G^{⊤} (- 1; - 1; 1) = (- 1 + 1; - 1 + 1) = (0; 0)$ .

Letters. At general size the same four moves are what 1.5's second half asks you to write, and the algebra is yours: (1) write any candidate as the minimiser plus a perturbation; (2) expand the squared length of the sum into three terms; (3) the cross term carries the normal equation's expression and is therefore zero - this is the one line where the candidate's defining property is used, and it is why the square completes around that point and nowhere else; (4) what is left is the height at the candidate plus a squared length, so nothing is lower, and equality forces the perturbation through the tall grid to be zero, which under full column rank forces the perturbation itself to be zero. In your homework's letters this is the second half of Exercise 1.5 - yours to write.

HOW TO CHECK

Say that the normal equation gives stationarity only.
Write $r = \hat{r} + q$ and expand.
Kill the cross term by the normal equation; name full column rank where $‖ G q ‖^{2} = 0$ forces $q = 0$ .

Fund read. The backtest picks one weight vector, and this proves no other ties it when the features are independent.

One more thing, because your exercise carries it. The objective you will be handed has a positive constant in front of it. That constant multiplies every height by the same factor - the height at the candidate and the leftover squared length alike - so it moves no minimiser and it breaks no tie. You can carry it through untouched, or set it aside and put it back at the end; the argument is identical either way.

Write this down: zero gradient = flat, not bottom; around the candidate $ℓ (\hat{r} + q) = ℓ (\hat{r}) + ‖ G q ‖^{2}$ because the normal equation kills the cross term; unique under full column rank.

16. Projectors: symmetric, idempotent, Z F = F, and the complementary projector

A projector flattens every column onto a floor and leaves it there: apply it twice, and the second time changes nothing. The floor of least squares is everything the feature grid can make, and its projector is its own flip, its own square, and fixes the grid's columns.

Symbols, defined before use. $F$ (read "F") is a tall full-column-rank grid, $(\begin{matrix} 1 & - 1 \\ 1 & 0 \\ 1 & 1 \end{matrix})$ : a constant column and a centred signal column. $Z$ (read "Z") is the $3 \times 3$ projector built from it, $Z = F (F^{⊤} F)^{- 1} F^{⊤}$ (the course uses a different letter). Idempotent (read "eye-dem-potent") means $Z^{2} = Z$ ; symmetric means $Z^{⊤} = Z$ ; $I - Z$ (read "I minus Z") is the complementary projector. The transpose of an inverse is the inverse of the transpose: on the grid $Y = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , the inverse is $(\begin{matrix} 1 & - 2 \\ 0 & 1 \end{matrix})$ , whose flip $(\begin{matrix} 1 & 0 \\ - 2 & 1 \end{matrix})$ is also the inverse of $Y^{⊤} = (\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix})$ . So $((F^{⊤} F)^{- 1})^{⊤} = (F^{⊤} F)^{- 1}$ here.

Formal (Bach, chapter 3, section 3.3.2, Proposition 3.2). For $F$ of full column rank, $Z = F (F^{⊤} F)^{- 1} F^{⊤}$ satisfies

Z^{⊤} = Z, Z^{2} = Z, Z F = F

and $I - Z$ satisfies $(I - Z)^{⊤} = I - Z$ , $(I - Z)^{2} = I - Z$ , and $(I - Z) F = 0$ . Read: $Z$ is symmetric, idempotent, and fixes $F$ 's columns; the complement wipes them out, and every column splits as $v = Z v + (I - Z) v$ .

Middle-school version. Your shadow at noon: a shadow's shadow is the same shadow, and anything already flat is its own shadow.

Simple, fails. $Z_{1} = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ . Squared: $(\begin{matrix} 1 \cdot 1 + 1 \cdot 0 & 1 \cdot 1 + 1 \cdot 0 \\ 0 & 0 \end{matrix}) = Z_{1}$ : idempotent. But $Z_{1^{⊤}} = (\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}) \neq Z_{1}$ : not symmetric.

Simple, works. $Z_{2} = \frac{1}{2} (\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix})$ , the projector onto the line through $(1; 1)$ . Squared: $\frac{1}{4} (\begin{matrix} 1 + 1 & 1 + 1 \\ 1 + 1 & 1 + 1 \end{matrix}) = Z_{2}$ .

Harder, fails. Symmetric but not idempotent: $Q$ . $Q^{2} = (\begin{matrix} 4 + 1 & 2 + 2 \\ 2 + 2 & 1 + 4 \end{matrix}) = H \neq Q$ .

Parallel worked example (different letters and sizes from your homework). Numbers. $F^{⊤} F = (\begin{matrix} 3 & 0 \\ 0 & 2 \end{matrix})$ , with inverse $diag (\frac{1}{3}, \frac{1}{2})$ , so $(F^{⊤} F)^{- 1} F^{⊤} = (\begin{matrix} \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{matrix})$ and

Z = F (F^{⊤} F)^{- 1} F^{⊤} = \frac{1}{6} (\begin{matrix} 5 & 2 & - 1 \\ 2 & 2 & 2 \\ - 1 & 2 & 5 \end{matrix})

Idempotent, with $\frac{1}{36}$ out front: row $1$ against column $1$ , $\frac{25 + 4 + 1}{36} = \frac{5}{6}$ ; row $1$ against column $2$ , $\frac{5 \cdot 2 + 2 \cdot 2 + (- 1) \cdot 2}{36} = \frac{12}{36} = \frac{2}{6}$ , which is $Z$ 's entry in row $1$ , column $2$ . Same grid.

Fixing the grid, $Z F = F$ , one column at a time. Column $1$ of $F$ is $(1; 1; 1)$ , and $Z$ carries a $\frac{1}{6}$ out front. Row $1$ against it: $\frac{5 \cdot 1 + 2 \cdot 1 + (- 1) \cdot 1}{6} = \frac{5 + 2 - 1}{6} = \frac{6}{6} = 1$ . Row $2$ : $\frac{2 \cdot 1 + 2 \cdot 1 + 2 \cdot 1}{6} = \frac{2 + 2 + 2}{6} = \frac{6}{6} = 1$ . Row $3$ : $\frac{(- 1) \cdot 1 + 2 \cdot 1 + 5 \cdot 1}{6} = \frac{- 1 + 2 + 5}{6} = \frac{6}{6} = 1$ . So $Z (1; 1; 1) = (1; 1; 1)$ - column $1$ comes back unchanged.

Column $2$ of $F$ is $(- 1; 0; 1)$ . Row $1$ : $\frac{5 \cdot (- 1) + 2 \cdot 0 + (- 1) \cdot 1}{6} = \frac{- 5 + 0 - 1}{6} = \frac{- 6}{6} = - 1$ . Row $2$ : $\frac{2 \cdot (- 1) + 2 \cdot 0 + 2 \cdot 1}{6} = \frac{- 2 + 0 + 2}{6} = \frac{0}{6} = 0$ . Row $3$ : $\frac{(- 1) \cdot (- 1) + 2 \cdot 0 + 5 \cdot 1}{6} = \frac{1 + 0 + 5}{6} = \frac{6}{6} = 1$ . So $Z (- 1; 0; 1) = (- 1; 0; 1)$ - column $2$ comes back unchanged too. Both columns of $F$ are already on the floor, which is exactly $Z F = F$ .

The complement on numbers. Subtract $Z$ from the identity entry by entry. On the diagonal the identity contributes $1$ : $1 - \frac{5}{6} = \frac{1}{6}$ , $1 - \frac{2}{6} = \frac{4}{6}$ , $1 - \frac{5}{6} = \frac{1}{6}$ . Off it the identity contributes $0$ , so each entry is just minus $Z$ 's: $0 - \frac{2}{6} = - \frac{2}{6}$ in slots $(1, 2)$ , $(2, 1)$ , $(2, 3)$ , $(3, 2)$ , and $0 - (- \frac{1}{6}) = \frac{1}{6}$ in slots $(1, 3)$ and $(3, 1)$ . That grid is

I - Z = \frac{1}{6} (\begin{matrix} 1 & - 2 & 1 \\ - 2 & 4 & - 2 \\ 1 & - 2 & 1 \end{matrix})

Apply it to column $1$ of $F$ , $(1; 1; 1)$ . Row $1$ : $\frac{1 \cdot 1 + (- 2) \cdot 1 + 1 \cdot 1}{6} = \frac{1 - 2 + 1}{6} = \frac{0}{6} = 0$ . Row $2$ : $\frac{(- 2) \cdot 1 + 4 \cdot 1 + (- 2) \cdot 1}{6} = \frac{- 2 + 4 - 2}{6} = \frac{0}{6} = 0$ . Row $3$ : $\frac{1 \cdot 1 + (- 2) \cdot 1 + 1 \cdot 1}{6} = \frac{1 - 2 + 1}{6} = \frac{0}{6} = 0$ . So $(I - Z) (1; 1; 1) = (0; 0; 0)$ , the zero column.

Now column $2$ , $(- 1; 0; 1)$ . Row $1$ : $\frac{1 \cdot (- 1) + (- 2) \cdot 0 + 1 \cdot 1}{6} = \frac{- 1 + 0 + 1}{6} = \frac{0}{6} = 0$ . Row $2$ : $\frac{(- 2) \cdot (- 1) + 4 \cdot 0 + (- 2) \cdot 1}{6} = \frac{2 + 0 - 2}{6} = \frac{0}{6} = 0$ . Row $3$ : $\frac{1 \cdot (- 1) + (- 2) \cdot 0 + 1 \cdot 1}{6} = \frac{- 1 + 0 + 1}{6} = \frac{0}{6} = 0$ . So $(I - Z) (- 1; 0; 1) = (0; 0; 0)$ as well, which is $(I - Z) F = 0$ : the standing-up part of each column of $F$ is nothing at all.

Letters. The three properties at general size are what 1.6a's first half asks you to write, and the algebra is yours - each is one line: for the fixing property, the inverse meets the grid it inverts and leaves the identity behind; for idempotence, the same cancellation happens in the middle of the doubled product; for symmetry, flip the three factors, reverse their order, and use that the middle factor is its own flip. For the complement, subtract from the identity and use each property you just proved. In your homework's letters this is the first half of Exercise 1.6a - yours to write.

HOW TO CHECK: is Z a projector of the least-squares kind?

Flip it: unchanged?
Square it: check one diagonal and one off-diagonal entry.
Multiply it into its own grid: unchanged? And $I - Z$ into it: zero?
Sanity: $Z$ has diagonal $\frac{5}{6}, \frac{1}{3}, \frac{5}{6}$ .

Fund read. The board's hat grid turns graded outcomes into fitted scores, $I - Z$ into residuals.

Write this down: $Z = F (F^{⊤} F)^{- 1} F^{⊤}$ : $Z^{⊤} = Z$ , $Z^{2} = Z$ , $Z F = F$ ; $I - Z$ symmetric, idempotent, $(I - Z) F = 0$ ; every column is $Z v + (I - Z) v$ ; on the numbers here, $Z$ leaves both columns of $F$ where they are, $Z (1; 1; 1) = (1; 1; 1)$ and $Z (- 1; 0; 1) = (- 1; 0; 1)$ , while $I - Z = \frac{1}{6} (1, - 2, 1; - 2, 4, - 2; 1, - 2, 1)$ sends both of them to the zero column.

17. Probes, and pulling out a diagonal entry - light

A column with a single one and zeros everywhere else is a probe: multiply a grid by it and you get one column of the grid; put the probe's transpose in front as well and you get one entry. For the grid two-one-one-two, the probe one-on-zero pulls out the first column, $(2; 1)$ , and the sandwich then pulls out the top-left entry, $1 \cdot 2 + 0 \cdot 1 = 2$ .

Symbols, defined before use. $p_{i}$ (read "p i") is the $i$ -th probe: a $1$ in slot $i$ , $0$ elsewhere - in $ℝ^{3}$ , $p_{2} = (0; 1; 0)$ . Your homework writes these with a different letter and calls them standard basis vectors; same objects. $Z p_{i}$ is column $i$ of $Z$ , $p_{i^{⊤}} Z$ is row $i$ , and $Z_{i i}$ (read "Z i i") is the entry in row $i$ , column $i$ .

Formal. For any square grid $Z$ ,

p_{i^{⊤}} Z p_{i} = Z_{i i}

Read: probe on the right picks a column, probe on the left picks that column's $i$ -th entry.

Middle-school version. A spreadsheet: point at a row and a column and you have named one cell; the finger holds no number of its own.

Simple, fails. The squared length of a column is NOT the diagonal entry for a general grid: $Q p_{1} = (2; 1)$ , $‖ Q p_{1} ‖^{2} = 4 + 1 = 5 \neq 2 = Q_{11}$ .

Simple, works. On the projector $Z$ of §16, $Z p_{2}$ is column $2$ , $\frac{1}{6} (2; 2; 2) = (\frac{1}{3}; \frac{1}{3}; \frac{1}{3})$ , and $p_{2^{⊤}}$ of that is $\frac{1}{3} = Z_{22}$ . Here the squared length DOES agree: $‖ Z p_{2} ‖^{2} = 3 \cdot \frac{1}{9} = \frac{1}{3} = Z_{22}$ .

Parallel worked example. For a symmetric idempotent $Z$ ,

Z_{i i} = p_{i^{⊤}} Z p_{i} = p_{i^{⊤}} Z Z p_{i} = p_{i^{⊤}} Z^{⊤} Z p_{i}

= (Z p_{i})^{⊤} (Z p_{i}) = ‖ Z p_{i} ‖^{2} \geq 0

idempotent writes $Z$ as $Z Z$ , symmetric flips the first $Z$ , §8 regroups. Numbers, $i = 1$ : $Z p_{1} = \frac{1}{6} (5; 2; - 1)$ , $‖ Z p_{1} ‖^{2} = \frac{25 + 4 + 1}{36} = \frac{5}{6} = Z_{11}$ . In your homework's letters these two lines are the bound of Exercise 1.6a - yours to write.

The complement half, and the bound. Run the same three lines on $I - Z$ , which §16 showed is also symmetric and idempotent: $(I - Z)_{i i} = ‖ (I - Z) p_{i} ‖^{2}$ . And $(I - Z)_{i i} = 1 - Z_{i i}$ , because $I$ has a $1$ in every diagonal slot. So $1 - Z_{i i}$ is a squared length too. A squared length is never negative, so $Z_{i i} \geq 0$ and $1 - Z_{i i} \geq 0$ - that is $0 \leq Z_{i i} \leq 1$ , every diagonal entry of a projector caught between zero and one. On the section's own $Z$ : $Z_{11} = \frac{5}{6}$ and $1 - Z_{11} = \frac{1}{6}$ , both in range.

Fund read. Probing the thirty-by-thirty hat grid with the $i$ -th probe returns name $i$ 's leverage.

Write this down: $p_{i^{⊤}} Z p_{i} = Z_{i i}$ ; for symmetric idempotent $Z$ : $Z_{i i} = ‖ Z p_{i} ‖^{2}$ and $1 - Z_{i i} = ‖ (I - Z) p_{i} ‖^{2}$ , so $0 \leq Z_{i i} \leq 1$ .

18. The trace and its cyclicity - light: Lecture 1, slide 23 carries it

The trace of a square grid is the sum of its diagonal entries. A product's trace is unchanged when the first factor moves to the back, even when the two products differ in size.

Symbols, defined before use. $tr (Q)$ is $Q_{11} + Q_{22} + \dots$ , the sum $\sum_{i = 1}^{d} Q_{i i}$ (§4). Cyclic means rotate: an $n \times d$ against a $d \times n$ gives $n \times n$ one way and $d \times d$ the other, and the two traces agree. $K = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ and $L = (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ are §3's grids; $R = diag (1, 2)$ ; $I_{2}$ is the $2 \times 2$ identity.

Formal (Lecture 1, slide 23).

tr (Q) = \sum_{i = 1}^{d} Q_{i i}, tr (K L) = tr (L K)

Read: the trace of a product is unchanged when the first factor moves to the end.

Middle-school version. A shared dinner bill totals the same whether you add it person by person or dish by dish.

Simple, works. $K L = (\begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix})$ , trace $2 + 3 = 5$ ; $L K = (\begin{matrix} 3 & 4 \\ 1 & 2 \end{matrix})$ , trace $3 + 2 = 5$ .

Harder, fails. Cyclic means rotate, not shuffle. $K L R = (\begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix}) (\begin{matrix} 1 & 0 \\ 0 & 2 \end{matrix}) = (\begin{matrix} 2 & 2 \\ 4 & 6 \end{matrix})$ , trace $8$ . Swap the last two: $K R L = (\begin{matrix} 1 & 4 \\ 3 & 8 \end{matrix}) (\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}) = (\begin{matrix} 4 & 1 \\ 8 & 3 \end{matrix})$ , trace $7$ . A swap changed it; a rotation does not: $L R K = (\begin{matrix} 0 & 2 \\ 1 & 0 \end{matrix}) (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = (\begin{matrix} 6 & 8 \\ 1 & 2 \end{matrix})$ , trace $8$ .

HOW TO CHECK

Add the diagonal entries to get the trace.
Rotate the first factor to the back, never swap two factors.
Both orders must give a square product.

The one the homework wants. Put the projector of §16 through cyclicity: in $tr (F (F^{⊤} F)^{- 1} F^{⊤})$ move the first factor $F$ to the back, which the rule above allows, and the trace becomes $tr ((F^{⊤} F)^{- 1} F^{⊤} F)$ . The inverse now meets the grid it inverts, leaving $tr (I_{d}) = d$ - the identity of size $d$ has $d$ ones down its diagonal. Check it on §16's numbers: that $F$ has $d = 2$ columns, and its $Z$ has diagonal $\frac{5}{6} + \frac{2}{6} + \frac{5}{6} = \frac{12}{6} = 2$ . The trace of a projector counts the columns it projects onto.

Fund read. The board's hat grid has trace five, so the thirty leverages average $\frac{5}{30} = \frac{1}{6}$ .

Write this down: tr = sum of the diagonal; $tr (K L) = tr (L K)$ , rotate only, never swap; $tr (F (F^{⊤} F)^{- 1} F^{⊤}) = tr (I_{d}) = d$ .

19. Linearity of expectation with a fixed grid - light: §6 has the one-number version

Averaging and applying a fixed grid can be done in either order. Take a random column that is one-on-zero half the time and three-on-four the other half; its average is two-on-two, and pushing each outcome through the grid one-two-three-four and averaging, or pushing the average through the grid, both give $(6; 14)$ .

Symbols, defined before use. $u$ (read "u") is a random column - its entries depend on the outcome. $K$ (read "K") is a fixed grid: the same in every outcome, never built from the random thing. $𝔼 [u]$ (read "the expectation of u") is the column of the coordinate averages (§6). $η$ (read "eta") is the noise column used from here to §24: each entry $η_{i}$ is a coin showing $2$ or $- 2$ with equal odds, the coins independent - the face of one tells you nothing about the odds of another, so every combination of faces is equally likely (§22) - and $𝔼 [η] = 0$ , the column of zeros.

Formal (Lecture 3, slide 18 uses it to pull a fixed vector out of an average over the noise). For a fixed $K$ and a random column $u$ ,

𝔼 [K u] = K 𝔼 [u]

Read: the average of K times u is K times the average of u.

Middle-school version. If rent is a fixed quarter of a varying paycheck, average rent is a quarter of average pay; a share that changes with the paycheck breaks that.

Simple, fails. The rule is for fixed multiply-and-add, not for squaring: $𝔼 [η_{1^{2}}] = \frac{4 + 4}{2} = 4$ , but $(𝔼 [η_{1}])^{2} = 0^{2} = 0$ .

Harder, works. $u = (1; 0)$ or $(3; 4)$ , half each, and $K = (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix})$ . $K u$ is $(1 + 0; 3 + 0) = (1; 3)$ or $(3 + 8; 9 + 16) = (11; 25)$ ; average $(\frac{1 + 11}{2}; \frac{3 + 25}{2}) = (6; 14)$ . Other order: $𝔼 [u] = (2; 2)$ , and $K (2; 2) = (2 + 4; 6 + 8) = (6; 14)$ . Same column.

Fund read. Under the noise model the average fitted score of every name is its true score, and a rule re-fitted after seeing the day's outcome is a grid that is not fixed.

Write this down: $𝔼 [K u] = K 𝔼 [u]$ for a FIXED $K$ ; not through squares ( $𝔼 [η_{1^{2}}] = 4 \neq 0$ ); fitted minus true $= Z η$ , which averages to zero.

20. Covariance tau-squared I: what it says - light

Flip two fair coins and score each as plus two or minus two. Each score averages zero, each score squared averages four, and the product of the two scores averages zero, because the four combinations cancel two by two: $\frac{4 - 4 - 4 + 4}{4} = 0$ .

Symbols, defined before use. $η = (η_{1}; η_{2})$ is the noise column of §19, two coins; your homework uses a different Greek letter. $Var (η_{1})$ (read "variance of eta one") is $𝔼 [η_{1^{2}}] - (𝔼 [η_{1}])^{2}$ (§6). $τ^{2}$ (read "tau squared") is the common variance, here $4$ ; your homework's letter for it is another Greek letter. $η η^{⊤}$ (read "eta eta transpose") is the $2 \times 2$ grid of products $η_{i} η_{j}$ , the outer product of §11. $Cov (η)$ (read "cov of eta") is the covariance grid; $I_{2}$ the $2 \times 2$ identity.

Formal. For a zero-mean noise column,

Cov (η) = 𝔼 [η η^{⊤}] = (\begin{matrix} 𝔼 [η_{1^{2}}] & 𝔼 [η_{1} η_{2}] \\ 𝔼 [η_{2} η_{1}] & 𝔼 [η_{2^{2}}] \end{matrix})

and the assumption your 1.6 makes is $Cov (η) = τ^{2} I$ : every diagonal entry $τ^{2}$ , every off-diagonal entry $0$ . Read: on the diagonal, each coordinate's variance; off it, how a pair moves together. Tau-squared-I says: same variance everywhere, no pair linked.

Middle-school version. Two dice at one table, equally jumpy and neither knowing what the other showed; a die that copied its neighbour would put a number where a zero belongs.

Simple, fails. Equal jumpiness but copying: $η_{2} = η_{1}$ always. Outcomes $(2; 2)$ and $(- 2; - 2)$ , half each. $𝔼 [η_{1} η_{2}] = \frac{4 + 4}{2} = 4$ , so $Cov (η) = (\begin{matrix} 4 & 4 \\ 4 & 4 \end{matrix}) \neq 4 I_{2}$ .

Simple, works. Two independent coins: outcomes $(2; 2)$ , $(2; - 2)$ , $(- 2; 2)$ , $(- 2; - 2)$ , a quarter each. $𝔼 [η_{1}] = \frac{2 + 2 - 2 - 2}{4} = 0$ ; $𝔼 [η_{1^{2}}] = 4$ ; $𝔼 [η_{1} η_{2}] = \frac{4 - 4 - 4 + 4}{4} = 0$ . $Cov (η) = (\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}) = 4 I_{2}$ , so $τ^{2} = 4$ .

Fund read. On the board this says every name's surprise is the same size and no two are linked - and a sector move that puts one surprise into eight names at once is an off-diagonal $4$ where the model has a $0$ .

Write this down: $Cov (η) = 𝔼 [η η^{⊤}]$ : diagonal = variances, off-diagonal = how pairs move together; $τ^{2} I$ = same variance everywhere, no pair linked.

21. Variance of a linear form

Weigh two independent coin-noises, one times the first plus two times the second. The swing needs no outcome list: the common variance, four, times the sum of the squared weights, one plus four, gives twenty.

Symbols, defined before use. $a$ is a fixed weight column. $a^{⊤} η$ (read "a transpose eta") is the weighted sum $\sum_{i} a_{i} η_{i}$ , one random number. $Var (a^{⊤} η)$ is its average squared distance from its mean. $‖ a ‖^{2} = \sum_{i} a_{i^{2}}$ . $Cov (η) = τ^{2} I$ with $τ^{2} = 4$ is §20's grid.

Formal. For a fixed $a$ and a zero-mean $η$ with $Cov (η) = τ^{2} I$ ,

Var (a^{⊤} η) = a^{⊤} Cov (η) a = τ^{2} ‖ a ‖^{2}

Read: the variance of a fixed weighted sum is $a$ sandwiched around the covariance grid, which for a tau-squared identity grid is tau squared times the squared length of $a$ . The first equals sign holds for any grid; the second needs the zeros.

Middle-school version. Three people flip coins for you, your stakes one dollar, two and two, heads win and tails lose. Your total's swing depends on the squares of the stakes, and since the coins do not talk the shares simply add.

Simple, works. Independent coins, $a = (1; 2)$ : values $6, - 2, 2, - 6$ , squares $36, 4, 4, 36$ , average $20 = 4 \cdot 5$ .

Harder, fails. Correlated noise: $η_{2}$ copies $η_{1}$ three times in four and flips it once in four, so $𝔼 [η_{1} η_{2}] = 4 \cdot \frac{3}{4} - 4 \cdot \frac{1}{4} = 2$ and $Cov (η) = (\begin{matrix} 4 & 2 \\ 2 & 4 \end{matrix})$ . With $a = (1; 1)$ the sandwich is $4 + 2 + 2 + 4 = 12$ . The sandwich holds; the $τ^{2} ‖ a ‖^{2} = 8$ shortcut does not.

Harder, works. Three independent coins, $a = (1; 2; 2)$ , $‖ a ‖^{2} = 9$ : the formula says $36$ . The values of $η_{1} + 2 η_{2} + 2 η_{3}$ are $10, 2, 2, - 6, 6, - 2, - 2, - 10$ ; their squares $100, 4, 4, 36, 36, 4, 4, 100$ add to $288$ , and $\frac{288}{8} = 36$ .

Parallel worked example (different letters and sizes from your homework). Numbers ( $a = (1; 2)$ , independent coins). $(a^{⊤} η)^{2} = (η_{1} + 2 η_{2})^{2} = η_{1^{2}} + 4 η_{1} η_{2} + 4 η_{2^{2}}$ . Average term by term (§19, coefficients fixed): $4 + 4 \cdot 0 + 4 \cdot 4 = 20$ . As a sandwich: $a^{⊤} (\begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix}) a = (1, 2) \cdot (4; 8) = 4 + 16 = 20$ .

Letters. $Var (a^{⊤} η) = 𝔼 [(a^{⊤} η)^{2}]$ , the mean being $a^{⊤} 𝔼 [η] = 0$ (§19). Flip the second copy: $(a^{⊤} η)^{2} = (a^{⊤} η) (a^{⊤} η) = (a^{⊤} η) (η^{⊤} a)$ - a $1 \times 1$ is its own transpose (§1). Regroup: $= a^{⊤} (η η^{⊤}) a$ . §19 pulls the fixed outer factors out:

𝔼 [a^{⊤} (η η^{⊤}) a] = a^{⊤} 𝔼 [η η^{⊤}] a = a^{⊤} (τ^{2} I) a

= τ^{2} a^{⊤} a = τ^{2} ‖ a ‖^{2}

Its use in 1.6b, on the projector $Z$ of §16: entry $2$ of $Z η$ is $a^{⊤} η$ with $a$ row $2$ of $Z$ , $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ ; for symmetric $Z$ that row upright is column $2$ , which is $Z p_{2}$ (§17). So $Var ((Z η)_{2}) = 4 \cdot ‖ Z p_{2} ‖^{2} = 4 \cdot \frac{1}{3} = \frac{4}{3} = τ^{2} Z_{22}$ . In your homework's letters this is the variance line of Exercise 1.6b - yours to write.

HOW TO CHECK: is my variance of a weighted sum right?

Weight column fixed, noise zero-mean? Then $Var = a^{⊤} Cov (η) a$ .
With $Cov = τ^{2} I$ it collapses to $τ^{2} \sum_{i} a_{i^{2}}$ ; without the zeros, use the full sandwich.
If $a$ is a row of a symmetric idempotent $Z$ , $‖ a ‖^{2}$ is the diagonal entry (§17).
Sanity: $a = (1; 2)$ , $τ^{2} = 4$ : $20$ by listing, $20$ by formula.

Fund read. A name's fitted score has variance tau squared times its leverage, so edge-of-cloud names get the noisiest scores.

Write this down: $Var (a^{⊤} η) = a^{⊤} Cov (η) a = τ^{2} ‖ a ‖^{2}$ when $Cov = τ^{2} I$ ; a row of $Z$ : $Var ((Z η)_{2}) = 4 Z_{22} = \frac{4}{3}$ .

22. THE TRAP: when a cross term dies and when it does not - the difference between 1.6b and 1.6c

The average of a product is the product of the averages only when the two factors have nothing to do with each other.

THE TRAP, in one sentence. If you kill the training residual's cross term the way the fresh miss's cross term dies, you get the SAME value for both - $4 + \frac{4}{3} = \frac{16}{3}$ for the training residual as well as for the fresh miss - and the optimism gap between them comes out $\frac{16}{3} - \frac{16}{3} = 0$ . A zero optimism gap is the visible sign that this trap has been sprung, and it destroys 1.6c, whose whole content is that the gap is not zero.

Symbols, defined before use. Two random quantities are independent when the outcome of one tells you nothing about the odds of the other: every pair of outcomes has probability the product of the two separate probabilities. $δ$ (read "delta") is a fresh noise column, independent of $η$ , same distribution - three new coins; your homework has its own letter for it. A cross term is the middle term of an expanded square. $γ$ (read "gamma") is $(Z η)_{2} = \frac{η_{1} + η_{2} + η_{3}}{3}$ , the §19 random number built from the coins of $η$ , with $𝔼 [γ^{2}] = \frac{4}{3}$ (§21); $β$ (read "beta") is any second random number.

Formal. If two random numbers $β$ and $γ$ are independent, then

𝔼 [β γ] = 𝔼 [β] 𝔼 [γ]

so if either has mean zero the product averages zero. Read: for independent things, the average of the product is the product of the averages.

Middle-school version. Orlando's weather and Tokyo's tell you nothing about each other; Orlando's and a town ten miles away nearly everything.

Harder, fails. Two zero-mean things built from the SAME coins, $γ$ and $η_{2}$ : $𝔼 [η_{2} γ] = \frac{𝔼 [η_{2} η_{1}] + 𝔼 [η_{2^{2}}] + 𝔼 [η_{2} η_{3}]}{3} = \frac{0 + 4 + 0}{3} = \frac{4}{3} \neq 0$ . Both factors average zero; the product does not, because $η_{2}$ is inside $γ$ .

Harder, works. The same $γ$ against a fresh coin $δ_{2}$ , independent of all three coins in $η$ : $𝔼 [δ_{2} γ] = 𝔼 [δ_{2}] 𝔼 [γ] = 0 \cdot 0 = 0$ .

In your homework's letters, with the one-half its loss carries, the fresh-response line of Exercise 1.6b and the training line of 1.6c are yours to write.

HOW TO CHECK.

Ask whether the two factors share a coin.
Independent, either one mean zero: the cross term is zero.
Shared coins: the cross term is a covariance, compute it.

Fund read. The backtest's loss on the thirty names that FIT the weights is the training case and a fresh day is the delta case - that gap is the optimism the ledger measures when $\hat{ℛ}$ beats $ℛ$ .

Where the two differences come from - the truth cancels, twice. Neither $δ_{2} - γ$ nor $η_{2} - γ$ is handed to you: each is one subtraction, and each time the truth drops out. Set it up on §16's grid $F = (\begin{matrix} 1 & - 1 \\ 1 & 0 \\ 1 & 1 \end{matrix})$ with §23's true weights $r_{0} = (1; 2)$ . The outcome is the truth plus noise, $t = F r_{0} + η$ , and the truth behind it is $F r_{0} = (1 - 2; 1 + 0; 1 + 2) = (- 1; 1; 3)$ , so at name 2 the truth is $(F r_{0})_{2} = 1$ . The fitted values are the projector applied to the outcome, and $Z F = F$ (§16) carries that truth through untouched:

Z t = Z F r_{0} + Z η = F r_{0} + Z η

so fitted value 2 is $(Z t)_{2} = (F r_{0})_{2} + (Z η)_{2} = 1 + γ$ - the truth at name 2, plus the fitted error $γ$ .

The fresh miss. A FRESH outcome at name 2 carries the same truth and brand-new noise: $1 + δ_{2}$ . Subtract the fitted value:

(1 + δ_{2}) - (1 + γ) = δ_{2} - γ

The $1$ cancels. What is left is the fresh noise minus the fitted error - pure noise, no truth in it.

The training residual. The TRAINING outcome at name 2 is $t_{2} = 1 + η_{2}$ . Subtract the same fitted value:

(1 + η_{2}) - (1 + γ) = η_{2} - γ

The $1$ cancels for the same reason: the projector fixes the truth, so the truth sits inside the fitted value exactly as it sits inside the outcome. What is left is the training noise minus the fitted error. The two differences look alike, and the ONLY difference between them is that $η_{2}$ is one of the three coins inside $γ$ while $δ_{2}$ is not.

The two numbers 1.6c turns on, printed. The fresh miss uses the independent copy: $𝔼 [(δ_{2} - γ)^{2}] = 𝔼 [δ_{2^{2}}] - 2 𝔼 [δ_{2} γ] + 𝔼 [γ^{2}]$ , which is $4 - 0 + \frac{4}{3} = 4 + \frac{4}{3}$ , the cross term dying because $δ$ and $γ$ are built from different coins. The training residual uses the same coins: $𝔼 [(η_{2} - γ)^{2}] = 𝔼 [η_{2^{2}}] - 2 𝔼 [η_{2} γ] + 𝔼 [γ^{2}]$ , which is $4 - 2 \cdot \frac{4}{3} + \frac{4}{3} = 4 - \frac{4}{3}$ - the cross term surviving at $\frac{4}{3}$ . The gap between them is twice that cross term.

THE SAFE ROUTE for the training residual - the one to use under exam pressure. There is a second way to the training number that never writes a cross term at all, because there is none to write. Since $(Z η)_{2} = γ$ , the residual at name 2 is one entry of the complementary projector applied to the noise column:

η_{2} - γ = η_{2} - (Z η)_{2} = ((I - Z) η)_{2}

Row 2 of $Z$ is $\frac{1}{6} (2, 2, 2)$ from §16, so row 2 of $I - Z$ is $(0 - \frac{2}{6}, 1 - \frac{2}{6}, 0 - \frac{2}{6}) = (- \frac{2}{6}, \frac{4}{6}, - \frac{2}{6}) = (- \frac{1}{3}, \frac{2}{3}, - \frac{1}{3})$ - the middle row of the grid $I - Z = \frac{1}{6} (\begin{matrix} 1 & - 2 & 1 \\ - 2 & 4 & - 2 \\ 1 & - 2 & 1 \end{matrix})$ that §16 already printed. On this instance the residual entry is therefore ONE fixed weighted sum of the three coins:

((I - Z) η)_{2} = \frac{- 2 η_{1} + 4 η_{2} - 2 η_{3}}{6} = \frac{- η_{1} + 2 η_{2} - η_{3}}{3}

Its coefficient column, in §21's letters, is $a = (- \frac{1}{3}; \frac{2}{3}; - \frac{1}{3})$ - fixed numbers, with no randomness anywhere in it. Its squared length:

‖ a ‖^{2} = \frac{1 + 4 + 1}{9} = \frac{6}{9} = \frac{2}{3}

and that squared length is exactly the matching diagonal entry of the complementary projector: $(I - Z)_{22} = 1 - Z_{22} = 1 - \frac{1}{3} = \frac{2}{3}$ . Same number, as §17 says it must be. The residual is now a fixed weighted sum of zero-mean coins, so its own mean is $0$ (§19) and its mean square IS its variance - and §21's variance-of-a-linear-form rule finishes it in one line, with $τ^{2} = 4$ :

𝔼 [(η_{2} - γ)^{2}] = τ^{2} ‖ a ‖^{2} = 4 \cdot \frac{2}{3} = \frac{8}{3}

The cross-term route above gave $4 - \frac{4}{3}$ , and $4 - \frac{4}{3} = \frac{12 - 4}{3} = \frac{8}{3}$ . The two routes land on the same number. The fresh miss, read the same way, is $4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{16}{3}$ , and the gap is $\frac{16}{3} - \frac{8}{3} = \frac{8}{3}$ - twice the cross term, exactly as the line above said. This second route is the SAFE one, and the reason is stateable in one clause: no cross term ever appears in it, because $I - Z$ absorbed the cross term before any squaring began. Under exam pressure, take this route - there is no middle term to forget, and so no way to spring the trap. Keep the expansion for the FRESH line, where the cross term genuinely does die.

Write this down: independent: $𝔼 [β γ] = 𝔼 [β] 𝔼 [γ]$ ; same coins: a covariance, compute it; fresh miss $4 + \frac{4}{3} = \frac{16}{3}$ , training residual $4 - \frac{4}{3} = \frac{8}{3}$ ; that cross term is NOT zero, and a gap of zero means you killed it. SAFE ROUTE: $η_{2} - γ = ((I - Z) η)_{2} = \frac{- η_{1} + 2 η_{2} - η_{3}}{3}$ , weights $a = (- \frac{1}{3}; \frac{2}{3}; - \frac{1}{3})$ , $‖ a ‖^{2} = \frac{2}{3} = (I - Z)_{22}$ , mean square $4 \cdot \frac{2}{3} = \frac{8}{3}$ - no cross term to mishandle.

23. Unbiasedness - light: the homework defines it

An estimate is unbiased when it is right on average - its long-run average is the number it aims at, even if every reading is off. A two-or-minus-two coin read once is never zero, yet such readings average zero.

Symbols, defined before use. An estimator $\hat{r}$ (read "r hat") is a rule computed from data; the hat means "estimated from the sample" (§6). $r_{0}$ (read "r zero") is the fixed unknown truth it aims at - your homework writes the truth with a star on its own letter. Bias is $𝔼 [\hat{r}] - r_{0}$ ; pointwise means coordinate by coordinate.

Formal. $\hat{r}$ is an unbiased estimator of $r_{0}$ when

𝔼 [\hat{r}] = r_{0}

Read: the average of the estimate, over the noise, is the truth.

Middle-school version. A scale that reads two pounds heavy one morning and two light the next is honest on average; one that always reads three heavy is precise and wrong.

Simple, works. Estimate the coins' mean (which is $0$ ) by the average of two readings, $\frac{η_{1} + η_{2}}{2}$ : its average is $\frac{0 + 0}{2} = 0$ .

Harder, fails. Estimate the variance $τ^{2} = 4$ from two coins by the average squared distance from the sample mean. With the centre $\frac{η_{1} + η_{2}}{2}$ , each reading sits $\frac{η_{1} - η_{2}}{2}$ from it, so the estimate is $\frac{1}{2} (\frac{(η_{1} - η_{2})^{2}}{4} + \frac{(η_{1} - η_{2})^{2}}{4}) = \frac{(η_{1} - η_{2})^{2}}{4}$ , whose average is $\frac{𝔼 [η_{1^{2}}] - 2 𝔼 [η_{1} η_{2}] + 𝔼 [η_{2^{2}}]}{4} = \frac{4 - 0 + 4}{4} = 2$ . Half the truth: biased.

Parallel worked example (different letters from your homework). Set it up on §16's grid $F = (\begin{matrix} 1 & - 1 \\ 1 & 0 \\ 1 & 1 \end{matrix})$ with true weights $r_{0} = (1; 2)$ , so the truth behind the outcome is $F r_{0} = (1 - 2; 1 + 0; 1 + 2) = (- 1; 1; 3)$ , and the outcome is $t = F r_{0} + η$ . Push it through the projector, using $Z F = F$ from §16: $Z t = Z F r_{0} + Z η = F r_{0} + Z η$ , so $𝔼 [Z t] = F r_{0} + Z 𝔼 [η] = F r_{0} = (- 1; 1; 3)$ - unbiased, because every coordinate is a fixed weighted sum of zero-mean coins. In your homework's letters this is the unbiasedness deduction of Exercise 1.6b - yours to write.

HOW TO CHECK

Average the estimator over the noise.
Compare it to $r_{0}$ , coordinate by coordinate.
A centre fitted from the same data makes the spread come out low.

Fund read. Fitted scores are unbiased under the noise model while the backtest loss on the fitting board is biased low.

Write this down: unbiased = the average of the estimate is the truth, coordinate by coordinate; $Z t = F r_{0} + Z η$ averages to $F r_{0}$ ; spread around a fit from the same data is biased low.

24. Conditional expectation and the tower property - light: §6 has the add-and-subtract move

An average can be taken in two stages: the average at each value of the input, then the average of those averages, weighted by how often each input shows up. Suppose the input is one half the time and two the other half, with outcome averages two and six: the two-stage average is $\frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 6 = 4$ , and the one-stage average over the four equally likely outcomes one, three, four, eight is $\frac{16}{4} = 4$ .

Symbols, defined before use. $s$ (read "s") is the input and $t$ (read "t") the outcome - your homework's pair has different letters. $ℙ (s = 1)$ (read "the probability that s equals one"). $𝔼 [t | s = 1]$ (read "the expectation of t given s equals one") is the average of $t$ over cases where $s$ is $1$ ; $𝔼 [t | s]$ is the function of $s$ whose value at each input is that average.

Formal (Lecture 2, slides 9 and 10 - the tower property, the law of iterated expectations).

𝔼 [t] = 𝔼 [𝔼 [t | s]] = \sum_{s} ℙ (s) 𝔼 [t | s]

Read: the overall average is the weighted average of the conditional averages. The same holds for any loss, which is why 1.1 minimises at a fixed input.

Middle-school version. A season batting average blends the average against lefties and against righties by how many of each you faced, and hitting your best against each kind separately is the best season you can have.

Simple, fails. Unequal input frequencies, averaged without weights: $s = 1, 2, 3$ with probabilities $\frac{1}{2}, \frac{1}{4}, \frac{1}{4}$ and conditional averages $2$ , $6$ , $10$ . Unweighted: $\frac{2 + 6 + 10}{3} = 6$ . Wrong: $𝔼 [t] = \frac{1}{2} \cdot 2 + \frac{1}{4} \cdot 6 + \frac{1}{4} \cdot 10 = 1 + \frac{3}{2} + \frac{5}{2} = 5$ .

Simple, works. The opener, weighted by its halves: $\frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 6 = 4 = \frac{1 + 3 + 4 + 8}{4}$ .

Fund read. The conditional risk at a fixed feature row is the average loss on names that look exactly like this one, and the tower says that minimising it row by row is also best for the board.

Write this down: tower: $𝔼 [t] = \sum_{s} ℙ (s) 𝔼 [t | s]$ , weighted ( $5$ , not $6$ ); a rule's risk is the weighted average of its conditional risks, so minimise at each $s$ .

25. What to write at home - the list, in order

Copy these by hand, in this order, doing every piece of arithmetic they name yourself - each one is printed in the section named, so you can check every number against the page. Nothing else from this guide needs to reach paper before the homework. The three sentences at the end are the teach-back the exam asked for and you did not write.

(§2) shapes under every symbol, walk left to right, inner sizes match, outer survive · $a^{⊤} K c$ : $(1 \times 2) (2 \times 3) (3 \times 1) = 1 \times 1$ , checked $= 7$ on numbers · $θ^{⊤} X^{⊤} X θ$ : $(1 \times 5) (5 \times 30) (30 \times 5) (5 \times 1) = 1 \times 1$ · the other three chains: $30 \times 30$ , $30 \times 1$ , $5 \times 5$ - none a number · forced sizes: left neighbour forces rows, right neighbour forces columns; a lone letter added to numbers is a number · the four-row table for $h (a, c)$ : $K$ $2 \times 3$ , $H$ $3 \times 3$ , $g$ $2 \times 1$ , $k$ $1 \times 1$ .
(§3) $(K L)^{⊤} = L^{⊤} K^{⊤}$ , checked on $K = (1, 2; 3, 4)$ , $L = (1, 0; 1, 1)$ : both sides $(3, 7; 2, 4)$ ; the same order gives $(1, 4; 2, 6)$ · for non-square grids the same order is not even legal: $G^{⊤} H^{⊤}$ is $(2 \times 3) (2 \times 2)$ · $(A B C)^{⊤} = C^{⊤} B^{⊤} A^{⊤}$ · $X^{⊤} X$ is symmetric for every $X$ · $(X θ - y)^{⊤} = θ^{⊤} X^{⊤} - y^{⊤}$ .
(§4) $\sum$ = a for-loop; a double sum = a loop inside a loop, one term per $(i, j)$ pair · $\sum_{i} \sum_{j} u_{i} S_{i j} u_{j} = u^{⊤} S u$ , checked: $18$ both ways for $S = (2, 1; 1, 3)$ , $u = (1; 2)$ · quadratic form = two copies of $u$ with a matrix between them = one number · $(S u)_{i} = \sum_{j} S_{i j} u_{j}$ is the bridge · $x^{⊤} x$ = the squares ADDED, one number ( $1 + 4 + 9 = 14$ ; never a column of squares)
(§5) partial derivative = the slope from nudging ONE knob · gradient = the column of all partial slopes, same shape as $u$ , never a row · $\nabla (c^{⊤} u) = c$ (checked: $(5, - 3)$ and $(2, - 1, 4)$ ) · $\nabla (u^{⊤} S u) = 2 S u$ only for symmetric $S$ ; the 2 is the two cross terms (checked: slope 8 by nudge, $S u$ said 4) · for non-symmetric $N$ it is $(N + N^{⊤}) u$ · nudge test: $0.01$ in, rise $\approx$ slope $\times 0.01$ out.
(§6) hat = estimated from the sample, never direction · $\hat{ℛ} = \frac{2}{5}$ from five games is printable, $ℛ$ is not · the Bayes predictor = the best announcement at each $x$ ; its risk is the floor · at a fixed $x$ , minimising average squared loss over $z$ is a one-variable problem · add and subtract the average $m$ : the cross term averages to zero, only $(m - z)^{2}$ depends on $z$ (checked on $y = 1, 3$ ).
(§7) invertible = undoable = every output has one input = $K v = 0$ only for $v = 0$ · $I$ changes nothing; $K^{- 1} K = I$ · $2 \times 2$ test: $a d - b c \neq 0$ (checked: $(2, 0; 0, 3)$ yes, $(2, 4; 1, 2)$ no) · a column that is a combination of the others → singular · least squares = make $‖ X θ - y ‖^{2}$ smallest, gradient to zero gives $X^{⊤} X θ = X^{⊤} y$ · proportional feature columns → $X^{⊤} X$ singular → many $θ$ , one backtest · full column rank → $X^{⊤} X$ invertible → $\hat{θ} = (X^{⊤} X)^{- 1} X^{⊤} y$ .
(§8) $v^{⊤} (G^{⊤} G) v = ‖ G v ‖^{2}$ ; from $G^{⊤} G v = 0$ , left-multiply by $v^{⊤}$ , get $G v = 0$ ; then $v = 0$ needs full column rank.
(§9) positive definite: $v^{⊤} Q v > 0$ for EVERY nonzero $v$ (semidefinite: $\geq 0$ ); one negative column kills it, positives prove nothing; prove it by completing the square.
(§10) $Q v = λ v$ with $v \neq 0$ : $v$ an eigenvector, $λ$ its eigenvalue; at an eigenvector the sandwich is $λ ‖ v ‖^{2}$ , the same sign as $λ$ .
(§11) orthogonal: $W^{⊤} W = I$ , so $W^{- 1} = W^{⊤}$ ; spectral theorem: symmetric $Q = W Λ W^{⊤}$ , columns of $W$ unit eigenvectors, $Λ$ the eigenvalues on a diagonal.
(§12) $Λ^{1 / 2}$ = root each diagonal entry (needs every $λ_{k} > 0$ ); $S = Λ^{1 / 2} W^{⊤}$ gives $S^{⊤} S = W Λ W^{⊤}$ , and $S$ is invertible.
(§13) (a) iff (b) iff (c) is proved by the ring (a) $\to$ (b) $\to$ (c) $\to$ (a); each leg "Assume [tail] ... So [head]"; say why the ring suffices.
(§14) full column rank = only the zero column in the null space = rank $d$ ; from $G^{⊤} G v = 0$ get $‖ G v ‖^{2} = 0$ , $v = 0$ , so $G^{⊤} G$ is invertible.
(§15) zero gradient = flat, not bottom; around the candidate $ℓ (\hat{r} + q) = ℓ (\hat{r}) + ‖ G q ‖^{2}$ because the normal equation kills the cross term; unique under full column rank.
(§16) $Z = F (F^{⊤} F)^{- 1} F^{⊤}$ : $Z^{⊤} = Z$ , $Z^{2} = Z$ , $Z F = F$ ; $I - Z$ symmetric, idempotent, $(I - Z) F = 0$ ; every column is $Z v + (I - Z) v$ ; on the numbers here, $Z$ leaves both columns of $F$ where they are, $Z (1; 1; 1) = (1; 1; 1)$ and $Z (- 1; 0; 1) = (- 1; 0; 1)$ , while $I - Z = \frac{1}{6} (1, - 2, 1; - 2, 4, - 2; 1, - 2, 1)$ sends both of them to the zero column.
(§17) $p_{i^{⊤}} Z p_{i} = Z_{i i}$ ; for symmetric idempotent $Z$ : $Z_{i i} = ‖ Z p_{i} ‖^{2}$ and $1 - Z_{i i} = ‖ (I - Z) p_{i} ‖^{2}$ , so $0 \leq Z_{i i} \leq 1$ .
(§18) tr = sum of the diagonal; $tr (K L) = tr (L K)$ , rotate only, never swap; $tr (F (F^{⊤} F)^{- 1} F^{⊤}) = tr (I_{d}) = d$ .
(§19) $𝔼 [K u] = K 𝔼 [u]$ for a FIXED $K$ ; not through squares ( $𝔼 [η_{1^{2}}] = 4 \neq 0$ ); fitted minus true $= Z η$ , which averages to zero.
(§20) $Cov (η) = 𝔼 [η η^{⊤}]$ : diagonal = variances, off-diagonal = how pairs move together; $τ^{2} I$ = same variance everywhere, no pair linked.
(§21) $Var (a^{⊤} η) = a^{⊤} Cov (η) a = τ^{2} ‖ a ‖^{2}$ when $Cov = τ^{2} I$ ; a row of $Z$ : $Var ((Z η)_{2}) = 4 Z_{22} = \frac{4}{3}$ .
(§22) independent: $𝔼 [β γ] = 𝔼 [β] 𝔼 [γ]$ ; same coins: a covariance, compute it; fresh miss $4 + \frac{4}{3} = \frac{16}{3}$ , training residual $4 - \frac{4}{3} = \frac{8}{3}$ ; that cross term is NOT zero, and a gap of zero means you killed it. SAFE ROUTE: $η_{2} - γ = ((I - Z) η)_{2} = \frac{- η_{1} + 2 η_{2} - η_{3}}{3}$ , weights $a = (- \frac{1}{3}; \frac{2}{3}; - \frac{1}{3})$ , $‖ a ‖^{2} = \frac{2}{3} = (I - Z)_{22}$ , mean square $4 \cdot \frac{2}{3} = \frac{8}{3}$ - no cross term to mishandle.
(§23) unbiased = the average of the estimate is the truth, coordinate by coordinate; $Z t = F r_{0} + Z η$ averages to $F r_{0}$ ; spread around a fit from the same data is biased low.
(§24) tower: $𝔼 [t] = \sum_{s} ℙ (s) 𝔼 [t | s]$ , weighted ( $5$ , not $6$ ); a rule's risk is the weighted average of its conditional risks, so minimise at each $s$ .
Your own line, in your words, for the teach-back - the one the cycle-2 exam asked for and you left blank. Three sentences, nothing open in front of you, and they must contain these three things and no more: (i) what shape a column of $n$ is and what shape a row of $n$ is; (ii) why two columns of the same length cannot be multiplied as they stand - name the sizes that fail to meet; (iii) what changes if you tip the FIRST one over versus the SECOND one, and which of the two gives a single number. Write it in your words, not mine.

26. Coverage contract - every row of the demand list

The points, once, so the two halves add up. 1.1 = 2 · 1.2 = 2 · 1.3a = 0.5 · 1.3b = 0.5 · 1.4a = 1 · 1.4b = 1 · 1.4c = 1 · 1.5 = 3 · 1.6a = 1.3 · 1.6b = 1.3 · 1.6c = 1.4. Total 15. Part A carries 1.1, 1.3a, 1.3b, 1.4a, 1.4b, 1.4c = 6; Part B carries 1.2, 1.5, 1.6a, 1.6b, 1.6c = 9.

#	Demand (from `demands.md`)	Needed by	Where this guide teaches it
1	Shape bookkeeping: inner-dimension rule, columns as $d \times 1$ , $1 \times 1$ = a number	1.3-1.6	owned at 10 on the cycle-2 exam (§1, one line); the chain through three and four factors - the exam's gap - rebuilt in §2 with the forced-sizes method worked twice; the sizes in the homework's own letters deliberately not given
2	Transpose rules: reversal, symmetric = own transpose, $1 \times 1$ = own transpose	1.2, 1.4c, 1.5, 1.6	§3, on numbers - $(K L)^{⊤}$ entry by entry, the same order failing, the non-square case, three factors, $X^{⊤} X$ symmetric; the transpose of an inverse in §16
3	Quadratic form $w^{⊤} Q w$ as double sum, both directions	1.3, 1.4, 1.5, 1.6a	§4, from zero (carried whole from Guide #2); the statement in the homework's letters deliberately not given
4	Expanding a squared norm, and its matrix-weighted version	1.1, 1.4c, 1.5, 1.6	§3 and §5 (the merge justified in §3; the four-term expansion merged to three on numbers in §5) and §15 (the same expansion around a candidate). The matrix-weighted version of Lecture 4 slide 5 is not taught - no part of this homework needs it
5	Gradient as stacked partials: un-compress, differentiate, re-compress	1.4, 1.5	§5, from zero (carried whole)
6	The two gradient facts	1.4, 1.5	§5, derived on numbers (Facts A and B, each with a failing case); the general- $d$ statements in the homework's letters are deliberately not written - they ARE 1.4a and 1.4b
7	Definition of positive definite / semidefinite	1.2, 1.5, 1.6a	§9 - the sandwich test, two failing grids, completing the square, the $G^{⊤} G$ route
8	Linearity of expectation	1.1, 1.6	§6 (scalar, on two numbers) and §19 (with a fixed grid in front, on a column of coins)
9	Left-multiply by $v^{⊤}$ to get a squared norm	1.2, 1.5, 1.6a	§8 - the one trick, with the CONCLUSION shown failing when the tall grid sends a nonzero column to zero, which is the separate hypothesis the last step needs
10	Projector algebra	1.6	§16 - symmetric, idempotent, $Z F = F$ , the complementary projector, on a three-name grid
11	Independence of random vectors; expectation of a product of independent zero-mean terms	1.6b, 1.6c	§22, THE TRAP - when a cross term dies and when it does not, on coins, plus the SAFE route that avoids the cross term entirely
12	Covariance $σ^{2} I$	1.6	§20 ( $τ^{2} I$ on other letters: what it says entry by entry)
13	Variance of a linear form	1.6	§21 - $Var (a^{⊤} η)$ on coins, then the grid version
14	Standard basis vectors and $e_{i}^{⊤} M e_{i} = M_{i i}$	1.6	§17 - the probe columns pull out one diagonal entry
15	Trace and its cyclicity	1.6a, 1.6c	§18 - the trace on Guide #2's $K$ and $L$ , cyclic, and the trace of a projector
16	Full column rank, independence, null space	1.5, 1.6	§7 (the $3 \times 3$ null-space example, the duplicated-signal board) and §14 (the definitions on three tall grids)
17	Rank-nullity theorem	1.5	§14 - counted on the tall grids; nullity $0$ is full column rank
18	Invertibility of $X^{⊤} X$ under full column rank	1.5, 1.6	§7 as a statement on a $3 \times 2$ board both ways; §14 the proof route on a parallel instance
19	Eigenvalues and eigenvectors	1.2	§10 - stretch factors on $Q$ , found and checked
20	Spectral theorem, orthogonal matrices	1.2	§11 - $Q = W Λ W^{⊤}$ on numbers, what orthogonal buys, and the unscaled eigenvector grid that fails $W^{⊤} W = I$
21	Matrix square root of a diagonal	1.2	§12 - $Λ^{1 / 2}$ and the factor $S$ with $S^{⊤} S = H$
22	Proof structure for a three-way equivalence	1.2	§13 - the ring, on a toy first, then the three statements about $Q$
23	Uniqueness of a minimiser beyond zero gradient	1.5	§15 - completing the square around the candidate, on one knob then on a column
24	Conditional expectation, tower property, conditional risk	1.1	§6 (the floor and the four announcements) and §24 (conditional expectation and the tower property on a two-input table)
25	Bayes predictor as pointwise minimiser of conditional risk	1.1	§6 - the floor in the strikeout story, the four-announcement ladder, and the $argmin$ notation with the min-versus-argmin distinction 1.1 opens by asking for
26	Minimiser of expected squared error is the conditional mean - the proof	1.1	§6 and §24: the technique (add and subtract the mean, the cross term averages to zero) on two numbers, then on a conditional table; the proof in general is deliberately not given - it is what 1.1 asks
27	Reading a normal distribution's parameters	1.1	§6 ( $N (m, s^{2})$ : average $m$ , variance $s^{2}$ )
28	Unbiasedness of an estimator	1.6b	§23 - the fitted VALUES average to the true values, on the three-name grid with coin noise; carrying that back to the weights is the step 1.6b asks for
29	The squared-loss convention and the $\frac{1}{2}$ / $\frac{1}{2 n}$ scaling	1.1, 1.4c, 1.5, 1.6	§6 (the $\frac{1}{2}$ in the two-number computation) and §5 (the 1.4c scaling warning)
30	Overfitting - definition (diagnostic item 19)	(fund band)	owned - taught light in §22: the training residual comes out smaller than the fresh miss on the same coins ( $4 - \frac{4}{3}$ against $4 + \frac{4}{3}$ ), and that gap IS overfitting, measured. Owned material is taught lightly, never dropped

The order of work at home, after the write-at-home list - cheapest points first: 1.3a (0.5 pt, §2's table method) → 1.3b (0.5 pt, §4: count the copies of the vector; inner sum first) → 1.4a (1 pt, §5, Fact A, three lines, the answer is a column) → 1.4b (1 pt, §5, Fact B, both families, say where symmetric is used) → 1.1 (2 pt, §6 and §24: write the conditional risk for its loss, add and subtract the conditional mean, the cross term averages to zero) → 1.4c (1 pt, stretch, §5's four-term expansion merged to three; Fact B on the first term, Fact A on the second) → 1.5 (3 pt, the largest single prize, §5 and §7 first for the gradient and where the normal equation comes from, then §8, §14, §15: left-multiply, full column rank, complete the square) → 1.2 (2 pt, §9-§13: the ring, one leg at a time, §8's trick for the sign) → 1.6a (1.3 pt, §16-§18) → 1.6b (1.3 pt, §19-§21, §22 for the fresh-outcome cross term, §23) → 1.6c (1.4 pt, §22 - read it twice, it is the trap - plus §16 and §17 for the complementary projector and §18 for the averaging). That order banks 6 points before you meet anything marked hard.

Sources cited: your notes of 2026-09-07 on Guide #2 §0-§3 (15 lines); the cycle-2 exam of the same evening; Lecture 1 slides 21-23; Lecture 2 slides 4-5, 9-10, 12-13, 15; Lecture 3 slides 12-17; Lecture 4 slides 3-6; Bach, pages 47-51, 63-64; Grosse, chapter 1; demands.md master_demand_list; Study Guide #2 §4-§7 (carried); the example bank build/cycle3_example_bank.md and its two checker passes. assembled: 28660 words, sections 0..26, 24 write lines