june-5

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 9 — June 5, 2026

Administrative

• Tutorial section changes: contact Irene Sun (address on the course website) — no longer handled by the previous administrator.
• Reflection 2: will be posted tonight. Deadline will be set on Wednesday.
• Pre-midterm tutorial (week of June 9): problems posted Monday by ~3 PM. Expect ~15 questions: a short computation component, a conceptual component, and a proof component (no open-ended investigative questions).

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1. Recap: columns of a matrix are images of basis vectors

Before moving to new material, let’s re-examine the recipe for building the matrix of a linear map (standard bases throughout). Let be the standard basis of and the matrix of . Then:

the -th column of . So : the -th column of is the image of the -th standard basis vector. To find the matrix, apply to each basis vector and stack the results as columns. The full formula then follows by decomposing in the standard basis and using linearity.

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2. A concrete example: differentiation on polynomials

Last lecture ended with the abstract statement that a linear map between finite-dimensional vector spaces is given by a matrix once bases are chosen. The following example makes this concrete.

Setup. Let denote polynomials of degree at most 3, and polynomials of degree at most 2. The differentiation map

is linear (we verified this in Lecture 8 for , and polynomials are a subspace of it).

Choosing bases. To associate a matrix to , we must choose a basis for each space.

For the domain, take — a basis for listed in decreasing degree. A polynomial then has coordinate vector

For the codomain, take — the analogous basis for .

Computing the matrix. Apply to each basis vector of and express the result in basis :

Stacking these as columns gives the matrix

Verification. Take . Its coordinate vector is . The matrix gives

And indeed .

The commutative diagram. What we have constructed is:

The two paths from to agree: differentiate then take coordinates, or take coordinates then multiply by . In symbols:

This is the general formula for any linear map with bases for and for :

The abstract geometric map and its matrix say the same thing — one in the language of vectors, the other in the language of coordinates.

Why the basis order matters

The matrix computed here depends on the choice of basis and the ordering of the basis vectors. Reversing the order of to gives a different matrix for the same map . Neither matrix is more “correct” — they both encode , just in different coordinate systems. This is a key source of confusion when comparing linear algebra problems across different books.

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3. Kernel and image

With linear transformations in hand, two subspaces naturally arise.

Definition

Let be a linear transformation. The kernel (or null space) of is

The image (or range) of is

In words: the kernel is what “kills” (sends to ), and the image is what “reaches” (the set of all outputs). Both are subspaces of their respective spaces — and .

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4. Examples

Projection onto the -axis. Define by . Matrix: .

• — the -axis; dimension 1.
• — the -axis; dimension 1.

Check: .

Zero map. for all . Matrix: all zeros.

• (everything is killed); dimension .
• ; dimension 0.

Check: .

Differentiation on polynomials. .

• constants ; dimension 1. (Any polynomial with is constant.)
• ; dimension 3. (Every quadratic polynomial is the derivative of , e.g. maps to .)

Check: .

Note also: the pre-image of any quadratic polynomial is not unique. If , then as well, for any constant — and the freedom is exactly the kernel (the space of constants). The kernel measures the non-uniqueness of pre-images.

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5. The rank-nullity theorem

The three examples above all satisfy the same pattern. This is not a coincidence.

Theorem (Rank-nullity)

Let be a linear transformation with finite-dimensional. Then

The two quantities have standard names: is the nullity of , and is the rank of .

Intuition. Think of as the total number of “degrees of freedom” in the input. The kernel “absorbs” of them — those directions disappear in the output. The remaining directions survive and fill the image. The theorem says this accounting is exact.

Proof strategy (details next week). Take a basis of . Extend it to a basis of (using the basis extension proposition). Then show that is a basis for . This gives with and .

Connection to RREF

For the map : the kernel is the solution set of (the homogeneous system), so = number of free variables in the RREF. The image is the column space of , so = number of pivot columns = rank of . The rank-nullity theorem says pivots + free variables = total variables. This is exactly what RREF has been telling us all semester — the theorem is the abstract version of that fact.

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6. Lemma: linear maps send to

To close the lecture, a proof that can be written in two and a half minutes — a model of a short axiomatic argument.

Lemma

Let be a linear transformation. Then .

Proof

Using condition 1 of linearity with :

Add to both sides:

Alternatively: condition 2 with gives in one line. Both arguments are valid. A proof via matrices (” for any matrix ”) also works but only covers the finite-dimensional case; the axiomatic version handles all vector spaces uniformly.

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Looking ahead

Next week: the full proof of rank-nullity, and more examples working with specific linear maps and their kernels and images. This material is not on the midterm; it will be developed after June 13.

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Textbook references

Topic	Axler (2nd ed.)	Hefferon (4th ed.)	Treil
Matrix of w.r.t. chosen bases	Ch. 3, §3C	Ch. 3, §IV.2	Ch. 3, §2
Polynomial differentiation example	Ch. 3 (exercises)	Ch. 3, §IV.2	Ch. 3
Kernel (null space)	Ch. 3, 3.16	Ch. 3, §IV.2	Ch. 3, §3
Image (range)	Ch. 3, 3.19	Ch. 3, §IV.2	Ch. 3, §3
	Ch. 3, 3.11	Ch. 3, §IV.1	Ch. 3, §1
Rank-nullity theorem	Ch. 3, 3.22	Ch. 3, §IV.2	Ch. 3, §3