Matrix represenation of linear operator

Question

Observe the matrix $$ A =\begin{pmatrix} 6 & 2 \\ 2 & 3\\ \end{pmatrix}\in Mat_2(\mathbb{R})$$ and let $L = L_A: \mathbb{R^2}\to\mathbb{R^2}$ be the corresponding linear operator. Let $V=(v_1,v_2)$ be a basis for $\mathbb{R^2}$ where: $$ v_1 =\begin{pmatrix} 1 \\ -2 \\ \end{pmatrix}, v_2 =\begin{pmatrix} 2 \\ 1 \\ \end{pmatrix}$$ Show that $L(v_1) = 2v_1$ and $L(v_2) = 7v_2$ and find the matrix representation $B =\ _v[L]_v$ for L with respect to $V$.

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I have shown the first part:

$$L(v_1)=\begin{pmatrix} 6 & 2 \\ 2 & 3\\ \end{pmatrix}\cdot\begin{pmatrix} 1 \\ -2 \\ \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ \end{pmatrix}=2v_1$$ $$L(v_2)=\begin{pmatrix} 6 & 2 \\ 2 & 3\\ \end{pmatrix}\cdot\begin{pmatrix} 2 \\ 1 \\ \end{pmatrix} = \begin{pmatrix} 14 \\ 7 \\ \end{pmatrix}=7v_2$$

This is how far I have got with finding the matrix representation

$$_v[L]_v= ([L(v_1)]_v \ [L(v_2)]_v)=([2v_1]_v \ [7v_2]_v)$$ I am a bit unsure how to evaluate the last part, my instructor says that the matrix representation should be: $$ \begin{pmatrix} 2 & 0\\ 0 & 7 \\ \end{pmatrix}$$

Answer 1

You almost had it:

$$_v[L]_v= ([L(v_1)]_v \ [L(v_2)]_v)=([2v_1]_v \ [7v_2]_v)$$

Note $[2v_1]_V=\begin{bmatrix}2\\0\end{bmatrix}$ and $[7v_2]_V=\begin{bmatrix}0\\7\end{bmatrix}$ so

$([2v_1]_v \ [7v_2]_v)=\begin{bmatrix}2 & 0 \\ 0 & 7 \end{bmatrix}$

Note by definition given an ordered basis $\mathcal{B}=\{b_1,b_2,\ldots,b_m\}$ for a vector space $V$ then for each $v\in V$ $$ v=x_1b_1+\cdots+x_mb_m $$ for a unique scalar values $x_1,x_2,\ldots,x_m$ (use linear independence of $\mathcal{B}$ to show uniqueness and that $\mathcal{B}$ spans $V$ to show they exist).

Then the definition of $[v]_\mathcal{B}$ is

$$[v]_\mathcal{B}=\begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m\end{bmatrix}$$.

That is the bracket is an invertible linear transformation $[\cdot]_\mathcal{B}:V \to \mathbb{R}^m$

Answer 2

That's the meaning of matrix of linear map $T$ with respect to a basis $(b_1,b_2)$: if $T(b_1)=\alpha b_1+\beta b_2$ and $T(b_2)=\gamma b_1+\delta b_2$, then the matrix of $T$ with respect to that basis will be $\left(\begin{smallmatrix}\alpha&\gamma\\\beta&\delta\end{smallmatrix}\right)$. In your case, $\alpha=2$, $\beta=\gamma=0$, and $\delta=7$ and so…