Finding the matrix of a linear endomorphism of $2\times 3$ matrices, given by $\xi \mapsto \psi \xi \varphi$ for $\psi$ and $\varphi$ given matrices I would like to solve a problem in which I'm asked to determine trace and determinant of a linear endomorphism $f:x \mapsto A_1xA_2$ where $$A_1 = \left(\begin{array}{cc}7 & 5\\ 0 & 3\end{array}\right)$$ and $$A_2 = \left(\begin{array}{ccc}1 & 0 & 0\\ -1 & 4 & 0\\2 & -3 & 2\end{array}\right)$$ are both real-valued matrices.
Edit
For those interested, I add my solution below, which I was able to find thanks to @Matthew Leingang's advice! Here it is:
I picked base $\{\left(\begin{array}{cc}1&0\\0&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&1\\0&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\1&0\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&1\\0&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&0\\1&0\end{array}\right),\left(\begin{array}{cc}0&0\\0&0\\0&1\end{array}\right),\} = \{x_1,x_2,x_3,x_4,x_5,x_6\}$
of the space of $3\times 2$- real valued matrices.
Then I computed $f(x_i) = A_1 x_i A_2$, $\forall i \in \{1, ...,6\}$ and reduced the obtained matrices in terms of linear combinations of $x_i$; explicitely I obtained:

*

*$L(x_1) = 7(x_1 - x_3 + 2x_5) + 5(x_2 - x_4 + 2x_6)$


*$L(x_2) = 3(x_2 - x_4 + 2x_6)$


*$L(x_3) = 7(4x_3 - 3x_5) + 5(4x_4 - 2x_6)$


*$L(x_4) = 3(4x_4 - 3x_6)$


*$L(x_5) = 2(7x_5 + 5x_6)$


*$L(x_6) = 6x_6$
Hence I conclude that $f$ can be represented with the matrix:
$$ [f]  = \left(\begin{array}{cccccc}
7 & 0 & 0 & 0 & 0 & 0\\
5 & 3 & 0 & 0 & 0 & 0\\
- 7 & 0 & 28 & 0 & 0 & 0\\
- 5 & - 3 & 20 & 12 & 0 & 0\\
14 & 0 & - 21 & 0 & 14 & 0\\
10 & 6 & - 15 & - 9 & - 10 & 6\\
\end{array}\right)$$
Therefore $tr([f]) = 70$ and $\det([f]) = 592704$. Hope it helps! (and don't hesitate to tell me if you spot any mistake).

First, the map $f$ is a map from $M_{2 \times 3}(\mathbb{R})$ to itself (and not $M_{3 \times 2}(\mathbb{R}))$. Instead of calculating some matrix representing $f$ directly, we can use the following trick:
Note that $A_1$ is upper triangular with distinct diagonal values, hence $A_1$ is diagonalizable with eigenvalues $7,3$. Choose (linearly independent) eigenvectors $v ,w$ such that $A_1 v = 7v, A_1 w = 3w$. Similarly, $A_2^T$ is upper triangular with distinct diagonal values, hence $A_2^T$ is also diagonalizable with eigenvalues $1,4,2$. Choose (linearly independent) eigenvectors $x,y,z$ such that $A_2^T x = x, A_2^T y = 4y, A_2^T z = 2z$.
Now, note that
$$ f \left(  v \cdot x^T \right) = A_1 \cdot v \cdot x^T \cdot A_2 = \left( 7v \right) \cdot \left( A_2^T \cdot x \right)^T = \left( 7v \right) \cdot \left( x \right)^T = 7 v \cdot x^T, \\
f \left( v \cdot y^T \right) = \left( 7 v \right) \cdot \left( 4y \right)^T = 28 v \cdot y^T, \\
f \left( v \cdot z^T \right) = \left( 7v \right) \cdot \left( 2z \right)^T = 14 v \cdot z^T, \\
f \left( w \cdot x^T \right) = \left( 3w \right) \cdot \left( x \right)^T = 3 w \cdot x^T,  \\
f \left( w \cdot y^T \right) = \left( 3w \right) \cdot \left( 4y \right)^T = 12 w \cdot y^T,\\
f \left( w \cdot z^T \right) = \left( 3w \right) \cdot \left( 2z \right)^T = 6 w\cdot z^T.$$
Hence, we found six eigenvectors of $f$ associated to the distinct eigenvalues $7,28,14,3,12,6$. This means that $f$ is diagonalizable and so
$$ \operatorname{trace}(f) = 7 + 28 + 14 + 3 + 12 + 6 = 70, \\
\det(f) = 7 \cdot 28 \cdot 14 \cdot 3 \cdot 12 \cdot 6 = 592704. $$
This argument obviously generalizes.
