Endomorphism $Mat(2\times3,\Bbb R) \to Mat(2\times3,\Bbb R)$ Preparing for an exam, while studying old exams I came across this:
Let $ V = Mat(2 \times 3,\mathbb{R}) $ the $\mathbb{R}$-Vectorspace of real matrices of size $2 \times 3$. Let
$$ A = \begin{bmatrix}
-3 & 1 \\
-25 & 7 \\
\end{bmatrix}
$$
The Endomorphism $L_A: Mat(2 \times 3, K) \rightarrow Mat(2 \times 3, K)$ is given by
$L_A := A \cdot B$
Calculate the determinant, Eigenvalues and Eigenvectors of $L_A$
Hint: Left-multiplication with A affect the columns of B separately.
My issue: I don't understand how to get a square matrix from the $2 \times 3$ matrix. I asked an assistant, the only answer I got was that it's supposed to be a $6 \times 6$ matrix. I know how to solve the exercise ones I get the matrix, but I'm clueless.
 A: The roll-up-your-sleeves, explicit way is to find a basis for $M_{2\times 3}(\mathbb{R})$ and calculate the $6\times6$ matrix for $L_A$ defined by $L_A(X)=AX$.
The most obvious choice of six basis vectors is
$$ e_1=\begin{bmatrix}1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad
e_2=\begin{bmatrix}0 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix},$$
$$ e_3=\begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}, \quad
e_4=\begin{bmatrix}0 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix},  $$
$$ e_5=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}, \quad
e_6=\begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix},  $$
What may not be obvious is why I paired them up this way. This is because $L_A$ acts on the three columns of $X\in M_{2\times 3}(\mathbb{R})$ independently. You will see the effect of this on the $6\times 6$ matrix $[L_A]$ when you finish working it out.
Let $A=\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ for full generality.
The first column of a matrix (with respect to a basis) can be read off the coefficients of the basis vectors we get by applying the linear transformation to the first basis vector. We calculate $L_Ae_1$:
$$ \begin{bmatrix} a & b \\ c & d \end{bmatrix}
\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}
=
\begin{bmatrix} a & 0 & 0 \\ c & 0 & 0 \end{bmatrix}, $$
or in other words $L_A(e_1)=ae_1+ce_2$. This tells us the first column of $[L_A]$:
$$ [L_A]=\begin{bmatrix}a & \ast & \ast & \ast & \ast & \ast \\
c & \ast & \ast & \ast & \ast & \ast \\
0 & \ast & \ast & \ast & \ast & \ast \\
0 & \ast & \ast & \ast & \ast & \ast \\
0 & \ast & \ast & \ast & \ast & \ast \\
0 & \ast & \ast & \ast & \ast & \ast \end{bmatrix} $$
(The $\ast$s are unknown at the moment.) That's the first column down. I'll let you find the other five columns. Note the determinant of a block-diagonal matrix is the product of the diagonal blocks' determinants.
