The Matrix of a Linear Map Examples 1

# The Matrix of a Linear Map Examples 1

Recall from The Matrix of a Linear Map page that if $V$ and $W$ are both finite-dimensional vector spaces such that $\mathrm{dim} (V) = n$ and $\mathrm{dim} (W) = m$ and where $B_V = \{ v_1, v_2, ..., v_n \}$ is a basis of $V$ and $B_W = \{ w_1, w_2, ..., w_m \}$ is a basis of $W$, then for $T \in \mathcal L (V, W)$ being such that $T(v_k) = a_{1,k}w_1 + a_{2,k}w_2 + ... + a_{m,k}w_m$ for each $k = 1, 2, ..., n$ then the matrix of $T$ with respect to the bases $B_V$ and $B_W$ is:

(1)
\begin{align} \quad \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} \end{align}

We will now look at some examples regarding matrices representing linear maps.

## Example 1

Consider the linear map $T \in \mathcal L (\wp_2 (\mathbb{R}), \mathbb{R}^2)$ defined by $T(p(x)) = \left (2p(1) + 3p'(1), \int_0^1 p(x) \: dx \right )$. Let $B_V \{ 1, x, x^2 \}$ be a basis of $\wp_2 (\mathbb{R})$ and let $B_W = \{(1, 0), (0, 2) \}$ be a basis of $W$. Determine the matrix $\mathcal M (T, B_V, B_W)$.

To find the matrix $\mathcal M (T, B_V, B_W)$ we must apply $T$ to vector in the basis of $V$. We will have that:

(2)
\begin{align} \quad T(1) = \left ( 2(1) + 3(0), \int_0^1 1 \: dx \right ) = \left ( 2, 1 \right ) \end{align}
(3)
\begin{align} \quad T(x) = \left ( 2(1) + 3(1), \int_0^1 x \: dx \right ) = \left ( 5, \frac{1}{2} \right ) \end{align}
(4)
\begin{align} \quad T(x^2) = \left ( 2(1) + 3(2), \int_0^1 x^2 \: dx \right ) = \left ( 8, \frac{1}{3} \right ) \end{align}

We now need to write these images as linear combinations of the basis vectors in $B_W \{ (1, 0), (0, 2) \}$:

(5)
\begin{align} \quad T(1) = (2, 1) = 2(1, 0) + \frac{1}{2} (0, 2) \end{align}
(6)
\begin{align} \quad T(x) = \left ( 5, \frac{1}{2} \right ) = 5(1, 0) + \frac{1}{4} (0, 2) \end{align}
(7)
\begin{align} \quad T(x^2) = \left ( 8, \frac{1}{3} \right ) = 8(1, 0) + \frac{1}{6} (0, 2) \end{align}

The coefficients in the linear combinations above form our matrix. We have that:

(8)
\begin{align} \quad \mathcal M (T, B_V, B_W) = \begin{bmatrix} 2 & 5 & 8 \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{6} \end{bmatrix} \end{align}

## Example 2

Let $V$ and $W$ both be finite-dimensional vector spaces and let $T \in \mathcal L (V, W)$. Prove that for any choice of basis $B_V$ of $V$ and any choice of basis $B_W$ of $W$ that the matrix $\mathcal M (T, B_V, B_W)$ has at least $\mathrm{dim} (\mathrm{range} (T))$ nonzero entries.

Let $\mathrm{dim} (V) = n$ and let $\mathrm{dim} (W) = m$. Let $B_V = \{ v_1, v_2, ..., v_n \}$ be a basis of $V$ and let $B_W = \{ w_1, w_2, ..., w_m \}$ be a basis of $W$. Also let $T \in \mathcal L (V, W)$.

Let's instead assume that $\mathcal M (T, B_V, B_W)$ has at most $\mathrm{dim} (\mathrm{range} (T)) - 1$ nonzero entries. Then there are at most $\mathrm{dim} ( \mathrm{range} (T)) - 1$ nonzero vectors amongst:

(9)
\begin{align} \quad \{ T(v_1), T(v_2), ..., T(v_n) \} \end{align}

Therefore $\mathrm{dim} (\mathrm{span} (T(v_1), T(v_2), ..., T(v_n)) ≤ \mathrm{dim} ( \mathrm{range} (T)) - 1$. However, we have that $\mathrm{range} (T) = \mathrm{span} (T(v_1), T(v_2), ..., T(v_n))$ and so:

(10)
\begin{align} \quad \mathrm{dim} ( \mathrm{range} (T)) ≤ \mathrm{dim} (\mathrm{range} (T)) - 1 \end{align}

This is a contradiction though, so our assumption that $\mathcal M (T, B_V, B_W)$ has at most $\mathrm{dim}(\mathrm{range} (T)) - 1$ nonzetro entries is false.

Therefore $\mathcal M (T, B_V, B_W)$ has at least $\mathrm{dim} (\mathrm{range} (T))$ nonzero entries.