Consider the vector space $V= \{A \in \text{Mat}(2,2;\mathbb{R}) | A^t = A \} \subset \text{Mat}(2,2;\mathbb{R})$ and the matrix $$L = \begin{pmatrix}1 & 2 \\0 & -2 \\ \end{pmatrix}\in\text{Mat}(2,2;\mathbb{R}). $$
How can I show that for $A \in V$ also $L^t \cdot A \cdot L \in V$ and how do I calculate the determinant of the endomorphism
$f :V \rightarrow V, A \mapsto L^t \cdot A \cdot L$?
I know that obviously $$L^t= \begin{pmatrix}1 & 0 \\2 & -2 \\ \end{pmatrix}$$ but I don't know how to advance from there.
HINT For your first question, whenever we have matrices $A,B$ of compatible dimension, $$ (AB)^t=B^tA^t. $$ For your second question, observe that $L$ is isomorphic to $\mathbb{R}^3$, under the isomorphism $$ A=\begin{pmatrix}a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix}\mapsto \begin{pmatrix} a_{11} \\ a_{12} \\ a_{22} \end{pmatrix}. $$ Using this isomorphism, we can write $f$ as a matrix, whose determinant can be readily computed.