Problem
Let $V$ be the vector space of all $2 x 2 $ symmetric matrices, with a basis $\beta$ consisting of:
\begin{equation*}
\beta =
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},
\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}
\end{equation*}
We define the linear transformation $L : V \rightarrow V $:
$$L(A)=S^TA S$$
for $A \in V$ and \begin{equation*}S = \begin{bmatrix}
0 & 1 \\
0 & 3
\end{bmatrix} \end{equation*}
Find the matrix representation $B$ of $L$ with the respect to the basis $\beta$
My attempt
I defined A to be:
$$A= \begin{bmatrix}
a & b \\
b & c
\end{bmatrix}$$
I then compute $L(A)$ :
$$L(A) = \begin{bmatrix}
0 & 0\\
0 & a+6b+9c
\end{bmatrix}$$
After this I am not sure what to do, the textbook answer for $B$ is:
$$B =
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
1 & 6 & 9
\end{bmatrix}$$
Question
I am completely stuck and I do not know how to proceed to get the same answer as the book. Any guidance will be much appreciated.
Best Answer
We can rewrite your formula for $L(A)$ as follows: if $\beta = \{M_1,M_2,M_3\}$, then $$ L(a M_1 + bM_2 + cM_3) = 0M_1 + 0M_2 + (a + 6b + 9c) M_3. $$ If we rewrite this in terms of the coordinate vectors (of the input and output) relative to $\beta$, we find that the matrix $B$ that represents your linear transformation must satisfy $$ B \pmatrix{a\\b\\c} = \pmatrix{0\\0\\a + 6b + 9c}. $$ It is clear, then, that the matrix $B$ that we're looking for is the matrix of coefficients in the right-hand-side expression.