Find the matrix representation of the operator $A\in\mathcal L(G)$ in the basis $f$.

Question

In the very beginning, I'm going to refer to my previous question where I applied the same method in a bit different vector space.

Let $G\leqslant M_2(\Bbb R)$ be the subspace of the upper-triangular matrices of the order $2$ and let's define a linear operator $A\in\mathcal L(G)$ with: $$A\left(\begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=\begin{bmatrix}4a+3b-3c&3a-2b-3c\\0&-a+b+2c\end{bmatrix}$$ and let $f=\left\{\begin{bmatrix}1&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix},\begin{bmatrix}1&0\\0&1\end{bmatrix}\right\}$ be a basis for $G$.

Find the matrix representation of the operator $A$ in the basis $f$.

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My attempt:

First, I computed the transformation matrix in the standard canonical basis $e=\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix},\begin{bmatrix}0&1\\0&0\end{bmatrix},\begin{bmatrix}0&0\\0&1\end{bmatrix}\right\}$. $$\begin{aligned}A\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}4&3\\0&-1\end{bmatrix}&=&&\color{red}{4}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}+\color{red}{3}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}\color{red}{-1}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\\A\left(\begin{bmatrix}0&1\\0&0\end{bmatrix}\right)&=\begin{bmatrix}3&-2\\0&1\end{bmatrix}&=&&\color{red}{3}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}\color{red}{-2}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}+\color{red}{1}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\\A\left(\begin{bmatrix}0&0\\0&1\end{bmatrix}\right)&=\begin{bmatrix}-3&-3\\0&2\end{bmatrix}&=&\ \color{red}{-}&\color{red}{3}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}\color{red}{-3}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}+\color{red}{2}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\end{aligned}$$ $$[A]_e=\begin{bmatrix}4&3&-3\\3&-2&-3\\-1&1&2\end{bmatrix}$$ $$\begin{aligned}\begin{bmatrix}1&1\\0&0\end{bmatrix}&=\color{red}{1}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}+\color{red}{1}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}+\color{red}{0}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\\\begin{bmatrix}0&0\\0&1\end{bmatrix}&=\color{red}{0}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}+\color{red}{0}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}+\color{red}{1}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\\\begin{bmatrix}1&0\\0&1\end{bmatrix}&=\color{red}{1}\cdot\begin{bmatrix}1&0\\0&0\end{bmatrix}+\color{red}{0}\cdot\begin{bmatrix}0&1\\0&0\end{bmatrix}+\color{red}{1}\cdot\begin{bmatrix}0&0\\0&1\end{bmatrix}\end{aligned}$$

$T=I^{-1}F=F=\begin{bmatrix}1&0&1\\1&0&0\\0&1&1\end{bmatrix}$ will be the transition matrix representing the change of a standard canonical basis $e$ into $f$, so $$[A]_f=F^{-1}[A]_eF$$ I got $F^{-1}=\begin{bmatrix}0&1&0\\-1&1&1\\1&-1&0\end{bmatrix}$, and then: $$\begin{aligned}[A]_f=F^{-1}[A]_eF&=\begin{bmatrix}0&1&0\\-1&1&1\\1&-1&0\end{bmatrix}\cdot\begin{bmatrix}4&3&-3\\3&-2&-3\\-1&1&2\end{bmatrix}\cdot\begin{bmatrix}1&0&1\\1&0&0\\0&1&1\end{bmatrix}\\&=\begin{bmatrix}3&-2&-3\\-2&-4&2\\1&5&0\end{bmatrix}\cdot\begin{bmatrix}1&0&1\\1&0&0\\0&1&1\end{bmatrix}\\&=\begin{bmatrix}1&-3&0\\-6&2&0\\6&0&1\end{bmatrix}\end{aligned}$$

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Is this correct? If so, how can I improve my answer?

Thank you in advance!

Answer 1

In this particular example, it is a lot easier to work directly with the basis $f$. Just by looking at it, \begin{align} Af_1&=f_1-6f_2+6f_3\\ Af_2&=-3f_1+2f_2\\ Af_3&=f_3 \end{align} If it's not obvious, note that the $1,2$ coordinate can only be determined by $f_1$, so that gives you its coefficient right away. Then you use $f_3$ to adjust the $1,1$ coordinate, and then $f_2$ to adjust the $2,2$.

Now you can read directly that $$ [A]_f=\begin{bmatrix} 1&-3&0\\-6&2&0\\6&0&1\end{bmatrix}. $$

Answer 2

Your solution is correct but long-winded. In order to find the matrix representation of $A$ with respect to a given basis, we compute how $A$ acts on the basis vectors and re-write it in terms of the basis (as you have done in the first part): $$ A\Bigg( \begin{bmatrix}1 & 1 \\ 0 & 0 \end{bmatrix}\Bigg) = \begin{bmatrix} 7 & 1 \\ 0 & 0 \end{bmatrix} = 1 \begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix} -6\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} + 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, $$ $$ A \Bigg( \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \Bigg) = \begin{bmatrix} -3 & -3 \\ 0 & 2 \end{bmatrix} = -3 \begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix} + 2\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} + 0\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, $$ $$ A \Bigg( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \Bigg) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = 0 \begin{bmatrix} 1 & 1 \\ 0 & 0\end{bmatrix} + 0\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} + 1\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ The coefficients of the linear combination of basis vectors used to express each $A(f_j)$ for $f_j \in f$ become the columns of the matrix of $A$. Which is to say: $$ \mathcal{M}_f(A) = \begin{bmatrix} 1 & -3 & 0 \\ -6 & 2 & 0 \\ 6 & 0 & 1 \end{bmatrix}. $$