june-26

These lecture notes were generated by Claude inside a Cowork environment with knowledge of the syllabus and course materials, and generated from the audio transcript of the lecture. The lecture notes are not a transcript, and skip some parts while adding small details with sources/clarifications. These notes have been checked by me to be correct. I have added and edited the file for correctness and alignment with necessary material. All right-margin aside notes are additional to the course, and are not part of the learning objectives.

Lecture 14 — June 26, 2026

Administrative

• Days off: Monday June 29, Tuesday June 30, and Wednesday July 1 (Canada Day long weekend). No course activity on any of these days.
• TA office hours: see the Google calendar linked on the Schedule and appointments page.
• Make-up midterm: Saturday, July 4, 9–11 AM (room announced separately). Format: 12 multiple-choice questions + 2 long-answer questions. Covers all material through today (including determinants and isomorphisms). The make-up is not graded for extra credit — it is a practice opportunity in the same format as the final. For students who cannot attend in person: the make-up PDF will be posted at 12pm on July 4th; set a 2-hour timer and attempt it under closed-book conditions.
• Solutions to the first midterm: being posted shortly. Slight delay because the rubric is being embedded into the solutions for each question.

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1. Why areas must be signed

In Lecture 13 we motivated the determinant as a function that measures volume scaling. The first two properties we wanted were normalisation and multilinearity (linearity in each argument). But multilinearity forces the function to take signed (possibly negative) values.

The problem with unsigned area. Suppose Area denotes the (non-negative) area of the parallelogram spanned by and . By multilinearity in the first argument, we would need:

i.e., , giving . If area is always non-negative, this forces every area to be zero. Nonsense.

The fix: allow negative values. The sign encodes orientation.

Oriented area in . In the plane, the two vectors are either arranged counter-clockwise or clockwise. We assign positive area to the counter-clockwise arrangement and negative to the clockwise:

The blue-green picture. With oriented area, the multilinearity rule now works even when and point in opposite directions. When the two parallelograms sit “side by side” (same orientation), their areas add — this is the blue + green = red case. When one reverses orientation relative to the other, its contribution subtracts — the blue green red case. Both are captured cleanly by the signed framework.

Orientation in higher dimensions

In we can think of positive orientation as “looking from above” and negative as “looking from below.” In , orientation is even more tangible: a right-handed coordinate system has positive orientation, a left-handed one has negative. A linear transformation with negative determinant swaps the handedness of the space — it involves a reflection.

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2. The determinant theorem — statement

Theorem (Existence and uniqueness of the determinant)

Let . There is exactly one function satisfying all three of:

(1) Normalisation: .

(2) Multilinearity: for every position and every ,

(3) Skew-symmetry: swapping any two arguments negates the value:

This unique function is called the determinant, denoted or where is the matrix with columns .

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3. Proof of uniqueness (supplementary — not covered in lecture)

Note on lecture coverage

This section was not presented in lecture. The lecture stated the uniqueness theorem and then worked the case explicitly (§4), which already contains all the essential ideas. The general proof below requires the language of permutations and their signatures — a non-trivial combinatorial detour — and was deliberately omitted. §4 is the content you are responsible for; this section is here for completeness.

The strategy is to show that the three properties force the value of on any input — hence at most one such function can exist. (Existence can be shown by verifying that the explicit formula we derive actually satisfies all three properties.)

Step 1: expand each vector in the standard basis. Write

where is the -th component of .

Step 2: apply multilinearity repeatedly. Begin with the first argument:

Apply again in the second argument of each resulting term, then the third, and so on. After expanding all arguments, we have terms, each of the form

for some sequence of indices .

Step 3: most terms vanish.

Lemma

If any two arguments of are equal, then .

Proof

Suppose position and position both hold the vector . Swap them: by skew-symmetry, . But the input is unchanged by swapping two equal arguments. So , giving .

Therefore any term in the expansion with a repeated index ( for some ) contributes zero. The only surviving terms are those where is a permutation of — all indices distinct. There are such permutations.

Step 4: evaluate the surviving terms. For a permutation (where ):

To determine the sign, we use property (A) repeatedly: any permutation can be returned to the identity by a sequence of transpositions — each transposition swaps exactly two positions and leaves the rest fixed. Each transposition, by (A), introduces a factor of . So if requires transpositions to reach the identity:

Definition (Signature of a permutation)

A transposition is a permutation that swaps exactly two elements and fixes all others. The signature (or sign) of a permutation , written , is if can be decomposed into an even number of transpositions, and if into an odd number:

The critical — and non-obvious — fact is that the parity of does not depend on which decomposition you choose. The same permutation can be written as a product of transpositions in many different ways, but they will all use either an even number or an odd number of transpositions; never both. This makes well-defined.

Why the parity is well-defined

One elegant way to see this: the determinant itself witnesses it. Suppose can be decomposed into transpositions and also into transpositions. Applying each decomposition to the standard basis gives and also . Since the function has a single value, , so and have the same parity. The argument is circular in a satisfying way: the uniqueness of the determinant implies the well-definedness of the signature, while the signature is needed to write down the determinant formula.

An alternative proof — not using the determinant — goes via the polynomial . Applying permutes the variables; the product either stays the same or flips sign. Which one happens depends only on , not on how is decomposed.

Conclusion. For any :

This formula is entirely determined by the three properties — it could not be anything else. Hence the function is unique.

Existence

To show the formula actually works (existence), one verifies that the right-hand side satisfies all three properties. Multilinearity is immediate from the product structure. Skew-symmetry takes slightly more work — swapping and amounts to composing every permutation with a transposition, which flips the signature of every term. Normalisation is checked directly: the only non-zero term for the standard basis is the identity permutation, giving .

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4. The determinant derived

Let us carry out the proof explicitly in dimension , with vectors and .

Expand in the first argument:

Expand in the second argument of each term:

Apply the Lemma ( since two arguments coincide) and skew-symmetry ():

In matrix notation, where are the columns of the matrix:

Written with the more familiar row-based notation for a matrix:

Definition ( determinant)

The formula was not postulated — it was the only possible answer consistent with the three axioms.

Why this formula detects linear dependence

The columns and are linearly dependent iff one is a scalar multiple of the other: (assuming ), which is equivalent to , i.e., . So iff the columns are dependent iff is not invertible.

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5. Geometric interpretation of the determinant

Volume scaling. is the factor by which scales -dimensional volumes. The unit -cube maps to a parallelotope of volume .

Sign and orientation.

• : the transformation preserves orientation.
• : the transformation reverses orientation (it includes a reflection).
• : the transformation collapses volume — something is sent to , so :

This gives a computationally accessible criterion for invertibility: no row reduction needed — just compute a number.

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6. The determinant and cofactor expansion

For there are permutations, giving six terms in the formula (three positive, three negative).

Definition ( determinant)

Cofactor expansion along the first row. A useful recursive formula:

Each minor is the submatrix obtained by deleting the row and column of the entry being expanded. The signs alternate: along the first row.

Why cofactor expansion works. Suppose we plug in for the first vector. We get a function of the remaining vectors, which still satisfies all three properties — it is therefore the determinant. This is the mechanism behind cofactor expansion, and it generalises: the determinant of any matrix can be computed by expanding along any row or column (with appropriate alternating signs).

Sarrus' rule (mnemonic for )

Write the matrix and repeat the first two columns to the right. The three diagonals going top-left to bottom-right give the positive terms (, , ); the three diagonals going top-right to bottom-left give the negative terms (, , ). This mnemonic does not generalise beyond .

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Textbook references

Topic	Axler	Hefferon	Treil
Oriented area and signed volume	Ch. 4, §4A	Ch. 4, §I.1	Ch. 4, §1
Determinant theorem (axioms)	Ch. 4, 4.1–4.3	Ch. 4, §I.1	Ch. 4, §1
Proof of uniqueness	Ch. 4, 4.5	Ch. 4, §I.2	Ch. 4, §1
and formulas	Ch. 4, 4.6	Ch. 4, §I.2	Ch. 4, §1
Cofactor expansion	Ch. 4, 4.8	Ch. 4, §II.2	Ch. 4, §2
Determinant and invertibility	Ch. 4, 4.10	Ch. 4, §I.3	Ch. 4, §1