LINEAR-ALGEBRA

The Right Basis Changes Everything

IKey Formulas

Formula	Name	Description
$T(v_j) = a_{1j}v_1 + a_{2j}v_2 + \cdots + a_{nj}v_n$	Column recipe	Column $j$ of $A$ tells how $T(v_j)$ decomposes in the basis
$B = M^{-1} A M$	Change of basis	Relates two matrix representations of the same $T$
$T(v_i) = \lambda_i v_i$	Eigenvector condition	If basis is eigenvectors, $A = \text{diag}(\lambda_1, \dots, \lambda_n)$
$D^n = \text{diag}(\lambda_1^n, \dots, \lambda_n^n)$	Power of diagonal	Eigenbasis makes iteration trivial

IIVisualization 1 — Same Transformation, Two Matrices

One transformation $T$ wears a tangled matrix in the standard basis and a clean diagonal one in its eigenbasis.

λ₁ 2 λ₂ 0.5 eigenbasis angle θ 30°

💡 Challenge: set $\theta = 0\degree$ or $\lambda_1 = \lambda_2$ and watch $A$ 's off-diagonal entries vanish — both matrices coincide.

IIIVisualization 2 — Iterating T: Why Diagonal Wins

Applying $T$ $n$ times means $A^n$ in the standard basis, but only $D^n=\operatorname{diag}(\lambda_1^n,\lambda_2^n)$ in the eigenbasis.

power n 3

💡 The eigenbasis is the dream, but finding it costs about $O(n^3)$ — which is why large-scale work leans on the cheaper, nearly-diagonalizing Fourier and wavelet bases.

IVVisualization 3 — Columns Are Where Basis Vectors Land

Each column of a transformation's matrix is simply the coordinates of where a basis vector lands.

transformation T

VQuiz Questions

Question 1

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation. Using the standard basis $(e_1, e_2)$ , suppose

$T(e_1) = 3e_1 + 5e_2, \quad T(e_2) = -2e_1 + 4e_2.$

What is the matrix of $T$ in the standard basis?

✅ Correct! Each $T(e_j)$ becomes column $j$ , not row $j$ .

❌ Not quite. You wrote the coefficients as rows instead of columns. The vector $T(e_1) = (3, 5)$ goes down column 1, not across row 1.

❌ Not quite. Remember: column $j$ of $A$ is exactly the coefficient vector of $T(e_j)$ in the chosen basis.

Show solution

Solution:

Column $j$ of the matrix is the coefficients of $T(e_j)$ in the basis.

$T(e_1) = 3e_1 + 5e_2$ → column 1 is $(3, 5)$
$T(e_2) = -2e_1 + 4e_2$ → column 2 is $(-2, 4)$

Stacking these as columns: $A = \begin{bmatrix} 3 & -2 \\ 5 & 4 \end{bmatrix}$

The answer is [[3, -2], [5, 4]].

Question 2

Suppose $A$ and $B$ are similar matrices related by $B = M^{-1} A M$ for some invertible $M$ .

True or False: $A$ and $B$ represent different linear transformations.

✅ Correct! Similar matrices describe the same $T$ in different bases — the matrix is just a costume.

❌ Not quite. Similar matrices describe the same transformation. The formula $B = M^{-1}AM$ is exactly how the same $T$ looks under a change of basis.

❌ Not quite. Try again — the hints above can help.

Show solution

Solution:

The statement is False.

Similar matrices represent the same linear transformation $T$ , just expressed in different bases. The matrix $M$ is the change-of-basis matrix that translates coordinates from one basis to the other:

$B = M^{-1} A M$

Read right to left: start with a vector in the $B$ -basis, apply $M$ to translate to the $A$ -basis, apply $A$ (the transformation), then apply $M^{-1}$ to translate back. Same transformation $T$ — different coordinate language.

This is the central insight of the video: a transformation is one thing, but its matrix depends on the basis.

Question 3

A linear transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ has three eigenvectors $v_1, v_2, v_3$ forming a basis, with eigenvalues $\lambda_1 = 4$ , $\lambda_2 = -1$ , $\lambda_3 = 2$ .

What is the matrix of $T$ in the eigenvector basis $(v_1, v_2, v_3)$ ?

✅ Correct! In the eigenvector basis, $T$ is diagonal with the eigenvalues on the diagonal.

❌ Not quite. Those would be eigenvalues of $T^{-1}$ , not $T$ . Since $T(v_i) = \lambda_i v_i$ , column $i$ has $\lambda_i$ (not $1/\lambda_i$ ) on the diagonal.

❌ Not quite. Using the column recipe with the eigenvector condition $T(v_i) = \lambda_i v_i$ produces a diagonal matrix.

Show solution

Solution:

Apply the column recipe in the eigenvector basis:

$T(v_1) = \lambda_1 v_1 = 4 v_1 = 4v_1 + 0 v_2 + 0 v_3$ → column 1 is $(4, 0, 0)$
$T(v_2) = \lambda_2 v_2 = -v_2 = 0 v_1 + (-1) v_2 + 0 v_3$ → column 2 is $(0, -1, 0)$
$T(v_3) = \lambda_3 v_3 = 2 v_3 = 0 v_1 + 0 v_2 + 2 v_3$ → column 3 is $(0, 0, 2)$

So the matrix is:

$D = \begin{bmatrix} 4 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \text{diag}(4, -1, 2)$

This is the central magic of the eigenvector basis: $T$ becomes diagonal, and every direction is independent.

Question 4

For an image with about $n = 10^6$ pixels, we'd love to diagonalize transformations like blurring or compression using the eigenvector basis. In practice, engineers use the Fourier basis or wavelet basis instead.

What is the main reason for this choice?

✅ Correct! The eigenvector basis is the ideal but unreachable choice; Fourier and wavelet bases are the practical near-diagonal alternatives.

❌ Not quite. The eigenvector basis diagonalizes $T$ exactly, but its cost ( $O(n^3)$ ) is prohibitive at scale. Fourier and wavelets diagonalize approximately — that's the trade.

Show solution

Solution:

The eigenvector basis would be the ideal choice — it diagonalizes $T$ perfectly. But finding eigenvectors of an $n \times n$ matrix costs roughly $O(n^3)$ operations, and at $n = 10^6$ this is around $10^{18}$ operations — completely impractical, and numerically unstable too.

The engineering compromise:

Fourier basis ( $O(n \log n)$ via FFT) — nearly diagonalizes shift-invariant operations. Used in JPEG, MP3.
Wavelet basis ( $O(n)$ ) — nearly diagonalizes localized operations. Used in JPEG 2000, MRI reconstruction.

These bases aren't the true eigenvectors, but they're close enough that the resulting matrix is approximately diagonal, while being cheap to compute. That trade — perfect diagonalization for computational feasibility — is the heart of modern signal processing.

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