LINEAR-ALGEBRA

Columns Tell the Story: How Any Linear Transformation Becomes a Matrix

IKey Formulas

Formula	Name	Description
$\text{Column } j \text{ of } A = [T(v_j)]_w$	Column rule	Apply $T$ to input basis vector $v_j$ , write in output basis $w$
$A \mathbf{x} = [T(\mathbf{x})]_w$	Matrix-transformation link	The matrix reproduces $T$ on coordinates
$\text{Matrix}(S \circ T) = B A$	Composition rule	Composing transformations = multiplying matrices
$P^2 = P$	Projection property	Any projection matrix satisfies this

Shape rule: A matrix representing $T : \mathbb{R}^n \to \mathbb{R}^m$ has $m$ rows (output dim) and $n$ columns (input dim).

IIVisualization 1 — Columns Are the Images of the Basis Vectors

A $2\times 2$ matrix is just the two places $T$ sends $e_1$ and $e_2$ , stacked as columns.

IIIVisualization 2 — One Projection, Two Matrices

Projecting onto a line is a single operation, yet its matrix depends on the basis you write it in.

Line angle 45°

💡 The eigenvector-basis matrix stays $\text{diag}(1, 0)$ at every angle — that fixed, diagonal form is exactly what diagonalization buys you.

IVVisualization 3 — Differentiation Is a Matrix

Differentiation is linear, so on the basis $\{1, x, x^2\}$ it acts as a fixed matrix on coefficient vectors.

a (const) 5 b (x) 3 c (x²) 4

💡 The columns of $D$ are $\tfrac{d}{dx}$ applied to each basis vector: $1 \to 0$ , $x \to 1$ , $x^2 \to 2x$ .

VQuiz Questions

Test your understanding of the column rule for building matrices.

Question 1

Suppose a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ satisfies $T(e_1) = \begin{bmatrix} 3 \\ -2 \end{bmatrix}, \quad T(e_2) = \begin{bmatrix} 1 \\ 4 \end{bmatrix}.$

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

Select the matrix (entries shown row-by-row):

✅ Correct! The output vectors become the columns of the matrix.

❌ Not quite. Remember: $T(e_1)$ and $T(e_2)$ become the columns of the matrix, not the rows.

Show solution

Solution:

The column rule says: column $j$ of the matrix is $T(e_j)$ written in the output basis. So:

Column 1 is $T(e_1) = (3, -2)^T$
Column 2 is $T(e_2) = (1, 4)^T$

Stacking these as columns gives: $A = \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix}$

Reading row-by-row: $[[3, 1], [-2, 4]]$ .

The common mistake $[[3, -2], [1, 4]]$ puts the output vectors as rows instead of columns.

Question 2

Consider the projection $T$ onto the $x$ -axis (the horizontal line $y=0$ ) in $\mathbb{R}^2$ .

Which matrix represents $T$ in the standard basis $\{e_1, e_2\}$ ?

Select the matrix:

✅ Correct! $T(e_1) = e_1$ (unchanged) and $T(e_2) = 0$ (crushed), giving columns $(1,0)$ and $(0,0)$ .

❌ Not quite. Apply $T$ to each standard basis vector: where does $e_1=(1,0)$ go under projection onto the x-axis? Where does $e_2=(0,1)$ go?

Show solution

Solution:

Apply the column rule. Projection onto the $x$ -axis sends a point $(x, y)$ to $(x, 0)$ .

$T(e_1) = T(1, 0) = (1, 0)$ — already on the $x$ -axis, unchanged. Column 1 is $(1, 0)^T$ .
$T(e_2) = T(0, 1) = (0, 0)$ — gets crushed to the origin. Column 2 is $(0, 0)^T$ .

So the matrix is: $P = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

Notice this happens to be the same as the projection onto the 45° line in the eigenvector basis — because the $x$ -axis IS the eigenvector basis for projecting onto the $x$ -axis. The standard basis already lines up with the line.

$[[0,0],[0,1]]$ would project onto the $y$ -axis.
$[[1,0],[0,1]]$ is the identity (no projection).
$[[1/2, 1/2], [1/2, 1/2]]$ is the projection onto the 45° line in the standard basis.

Question 3

True or False: If two matrices look different (have different entries), they must represent different linear transformations.

True or False?

✅ Correct! Different matrices can represent the same transformation when expressed in different bases.

❌ Not quite. Recall the video's opening: two different-looking matrices both represented the same projection onto the 45° line.

Show solution

Solution: False.

A matrix is a transformation expressed in a chosen basis. Change the basis, and the matrix entries change — but the underlying transformation stays the same.

The video's central example: projection onto the 45° line is represented by $\begin{bmatrix} 1/2 & 1/2 \\ 1/2 & 1/2 \end{bmatrix} \quad \text{(standard basis)}$ but also by $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \quad \text{(eigenvector basis)}.$

Both matrices describe the same geometric projection. They are related by a change-of-basis transformation.

Key takeaway: Matrix = transformation + choice of basis. Neither matrix is more "correct" than the other.

Question 4

Let $D$ be the derivative as a linear map from polynomials of degree $\leq 2$ to polynomials of degree $\leq 1$ , using bases $\{1, x, x^2\}$ and $\{1, x\}$ . So $D = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}.$

Apply $D$ to the polynomial $p(x) = 7 + 2x - 3x^2$ . The coordinate vector of $p$ is $(7, 2, -3)^T$ . What is $p'(x)$ ?

Select the derivative:

✅ Correct! Matrix multiplication gives $(2, -6)$ , which translates back to $2 - 6x$ .

❌ Not quite. Compute $D \cdot (7, 2, -3)^T$ row by row, then translate the output coordinates back to a polynomial in basis $\{1, x\}$ .

Show solution

Solution:

Multiply the matrix by the coordinate vector: $D \begin{bmatrix} 7 \\ 2 \\ -3 \end{bmatrix} = \begin{bmatrix} 0 \cdot 7 + 1 \cdot 2 + 0 \cdot (-3) \\ 0 \cdot 7 + 0 \cdot 2 + 2 \cdot (-3) \end{bmatrix} = \begin{bmatrix} 2 \\ -6 \end{bmatrix}.$

Output coordinates $(2, -6)$ in the basis $\{1, x\}$ mean: $p'(x) = 2 \cdot 1 + (-6) \cdot x = 2 - 6x.$

Direct check: $\frac{d}{dx}(7 + 2x - 3x^2) = 2 - 6x$ . They agree.

$p'(x) = -6 + 2x$ has the right coefficients but in the wrong order (you mixed up the constant and linear terms).
$p'(x) = 2 - 3x$ forgot the factor of 2 from differentiating $x^2$ .

VISummary

One rule, three lessons:

Matrix = transformation + basis choice. A matrix is what a transformation looks like once you pick coordinates. Change the basis, change the matrix.
Eigenvector basis → diagonal matrix. The cleanest representation puts eigenvalues on the diagonal and reveals what the transformation truly does.
Calculus is linear algebra in disguise. Differentiation acts linearly on polynomials, so it has a matrix — once you know its action on the basis, you know everything.

The mantra: Apply $T$ to the basis vectors. Write the results in output coordinates. Stack them as columns.

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