The Matrix of a Linear Map
 Table of Contents

# The Matrix of a Linear Map

We will now look at using matrices to represent linear maps.

 Definition: Let $V$ and $W$ be finite-dimensional $\mathbb{F}$-vector spaces such that $\dim V = n$ and $\dim W = m$ and let $B_V = \{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and $B_W = \{ w_1, w_2, ..., w_m \}$ be a basis of $W$. Let $T \in \mathcal L(V, W)$ such that for each $k = 1, 2, ..., n$ we have $T(v_k) = a_{1,k}w_1 + a_{2,k}w_2 + ... + a_{m,k}w_m$. Then the Matrix of The Linear Map $T$ with Respect to the Bases $B_V$ and $B_W$ is $\mathcal M (T, B_V, B_W ) = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ a_{m,1} & a_{m,2} & \cdots & a_{m,n} \end{bmatrix}$.

By the definition above, for $T \in \mathcal L (V, W)$, given the bases $B_V = \{ v_1, v_2, ..., v_n \}$ for the vector space $V$ and $B_W = \{ w_1, w_2, ..., w_m \}$ for the vector space $W$, then we can construct $\mathcal M (T, B_V, B_W)$ from the coefficients by applying $T$ to the basis vector $v_1$ to get $T(v_1) = a_{1,1}w_1 + a_{2,1}w_2 + ... + a_{m,1}w_m$, then applying $T$ to the basis vector $v_2$ to get $T(v_2) = a_{1,2}w_1 + a_{2,2}w_2 + ... + a_{m,2}w_m$, …, then applying $T$ to the basis vector $v_n$ to get $T(v_n) = a_{1,n}w_1 + a_{2,n}w_2 + ... + a_{m,n}w_m$.

We will now look at some examples of deriving a matrix to represent a linear transformation.

## Example 1

Let $T \in \mathcal L (\wp_4 ( \mathbb{R}), \wp_4 ( \mathbb{R})$ be defined by $T(p(x)) = 2x p'(x)$, and let $B_1 = \{1, x, x^2, x^3, x^4 \}$ be a basis of $\wp_4 ( \mathbb{R})$. Determine $\mathcal M (T, B_1, B_1)$.

We will first calculate the images of the basis vectors in $B_1$ under the linear transformation $T$.

(1)
\begin{align} T(1) = 2x(1)' = 2x(0) = 0(1) + 0(x) + 0(x^2) + 0(x^3) + 0(x^4) \\ T(x) = 2x(x)' = 2x(1) = 0(1) + 2(x) + 0(x^2) + 0(x^3) + 0(x^4) \\ \quad T(x^2) = 2x(x^2)' = 2x(2x) = 0(1) + 0(x) + 4(x^2) + 0(x^3) + 0(x^4) \\ \quad T(x^3) = 2x(x^3)' = 2x(3x^2) = 0(1) + 0(x) + 0(x^2) + 6(x^3) + 0(x^4) \\ \quad T(x^4) = 2x(x^4)' = 2x(4x^3) = 0(1) + 0(x) + 0(x^2) + 0(x^3) + 8(x^4) \end{align}

Therefore we can now construct our matrix as follows. The first column of $\mathcal M (T, B_1, B_1)$ corresponds to the coefficients determined by $T(1)$. The second column corresponds to the coefficients determined by $T(x)$, and so forth, and so:

(2)
\begin{align} \mathcal M (T, B_1, B_1) = \begin{bmatrix} 0 & 0 & 0 & 0 & 0\\ 0 & 2 & 0 & 0 & 0\\ 0 & 0 & 4 & 0 & 0\\ 0 & 0 & 0 & 6 & 0\\ 0 & 0 & 0 & 0 & 8\end{bmatrix} \end{align}

## Example 2

Let $T \in \mathcal L (M_{22}, M_{22})$ be defined by $T \left ( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right ) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ for $a, b, c, d \in \mathbb{R}$, and let $B_1 = \left \{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right \}$ be a basis of $M_{22}$. Determine $\mathcal M (T, B_1, B_1)$.

We will first calculate the images of our basis vectors under $T$.

(3)
\begin{align} \quad T \left ( \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \right ) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = 0\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + 1\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \\ \quad T \left ( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right ) = \begin{bmatrix} 0 & -1 \\ 0 & 0 \end{bmatrix} = 0\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} -1\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \\ \quad T \left ( \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \right ) = \begin{bmatrix} 0 & 0 \\ -1 & 0 \end{bmatrix} = 0\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} -1\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \\ \quad T \left ( \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right ) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = 1\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \\ \end{align}

Therefore we can construct $\mathcal M (T, B_1, B_1)$ as follows:

(4)
\begin{align} \mathcal M (T, B_1, B_1) = \begin{bmatrix}0 & 0 & 0 & 1\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 1 & 0 & 0 & 0\end{bmatrix} \end{align}
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