Recipe

Find the standard matrix of a linear transformation

Use when: you have a linear transformation's formula T:\mathbb R^n\to\mathbb R^m and need the matrix A with T(\mathbf x)=A\mathbf x

Goal: the m\times n standard matrix A, built column by column with nothing guessed

Taught in The standard matrix of a linear transformation

1. 1

   A is m\times n: n, the number of columns, is the number of entries T's inputs have; m, the number of rows, is the number of entries T's outputs have. Write this down before computing a single entry.

   Example

   T(x_1,x_2)=(2x_1-x_2,\ x_1+x_2,\ 3x_2) takes 2 numbers in and returns 3, so A will be 3\times 2 — two columns, three rows.

   Why: the shape isn't a convention to memorize, it's forced by T's own domain and codomain.

2. 2

   Plug \mathbf e_1=(1,0,\dots,0), then \mathbf e_2=(0,1,0,\dots,0), and so on through \mathbf e_n, into T's formula one at a time — each evaluation has a single 1 and the rest 0s.

   Example

   T(\mathbf e_1)=T(1,0)=(2(1)-0,\ 1+0,\ 3(0))=(2,1,0), and T(\mathbf e_2)=T(0,1)=(2(0)-1,\ 0+1,\ 3(1))=(-1,1,3).

   Why: linearity forces T(\mathbf x)=x_1T(\mathbf e_1)+\cdots+x_nT(\mathbf e_n) for every input, so these n vectors already contain everything T ever does.

3. 3

   T(\mathbf e_1) becomes column 1 of A, T(\mathbf e_2) becomes column 2, and so on — never a row, and never out of order.

   Example

   A=\begin{pmatrix} 2 & -1 \\ 1 & 1 \\ 0 & 3\end{pmatrix}

   The first column is T(\mathbf e_1)=(2,1,0); the second is T(\mathbf e_2)=(-1,1,3).

   Why: this is exactly the column picture of matrix-vector multiplication — A\mathbf x built from columns is, by definition, the same combination linearity already forced T(\mathbf x) to equal.

4. 4

   Pick an \mathbf x you haven't used yet, compute T(\mathbf x) straight from the original formula, then compute A\mathbf x, and confirm the two agree.

   Example

   T(2,-1) = (2(2)-(-1),\ 2+(-1),\ 3(-1)) = (5,1,-3), and A\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}2(2)+(-1)(-1)\\1(2)+1(-1)\\0(2)+3(-1)\end{pmatrix}=\begin{pmatrix}5\\1\\-3\end{pmatrix} — the same vector.

   Why: A is unique, so any mismatch here means an arithmetic slip in an earlier step, not a real exception to the theorem.