The Matrix of a Linear Map

This is a series of notes taken during my review of linear algebra, using Axler's excellent textbook Linear Algera Done Right, which will be heavily referenced.

This is based on book section (3.C).

Suppose $T \in \mathcal{L}(V, W)$ and $v_1,...,v_n$ is a basis of $V$ and $w_1,...,w_m$ is a basis of $W$. The matrix of $T$ with respect to these bases is the $m$-by-$n$ matrix $\mathcal{M}(T)$ whose entries $A_{j,k}$ are defined by

$$T v_k = \sum_{j=1}^m A_{j,k} w_j = A_{1,k} w_1 + ... +A_{m,k} w_m$$

The $k$th column of $\mathcal{M}(T)$ consists of the scalars needed to write $T v_k$ as a linear combination of $w_1,...,w_m$.

Example (3.33): Suppose $T \in \mathcal{L}(\mathbb{F}^2,\mathbb{F}^3)$ is defined by

$$T(x, y) = (x +3y, 2x + 5y, 7x + 9y)$$

Find the matrix of $T$ with respect to the standard bases of $\mathbb{F}^2$ and $\mathbb{F}^3$.

Solution: Because $T(1, 0)=(1,2,7)$ and $T(0,1)=(3,5,9)$, the matrix of $T$ with respect to the standard bases is then

$$ \begin{bmatrix} 1 & 3\\ 2 & 5\\ 7 & 9 \end{bmatrix} $$

Follow-up question: find the matrix of $T$ with respect to the basis $(1,1), (-1,1)$ of $\mathbb{F}^2$ and $(1,3,0),(0,-1,0),(0,0,2)$ of $\mathbb{F}^3$.

Solution: Because $T(1,1) = (4,7,16) = 4(1,3,0) + 5(0,-1,0) + 8(0,0,2)$, and $T(-1,1) = (2,3,2) = 2(1,3,0) -3(0,-1,0) + (0,0,2)$, the matrix with respect to these (non-standard) bases is

$$ \begin{bmatrix} 4 & 2\\ 5 & -3\\ 8 & 1 \end{bmatrix} $$

Key point: the matrices can change with bases, but the underlying operator stays the same! Later on we will see how trace and determinant are really properties of operators, not just their matrix representations.