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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.

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    $\begingroup$ A 2-by-2 times a 2-by-3 is another 2-by-3. 6-by-6 matrices will happen only once you start expressing things in a basis, which is what I guess the TA is getting at. $\endgroup$
    – Randall
    Sep 12, 2022 at 19:33
  • $\begingroup$ Do you know how to represent a linear map by a matrix ? If so, just write the matrix of $L_A$ (which is the application $B \mapsto AB$) w.r.t. the canonical basis of $\mathrm{Mat}(2 \times 3, \mathbb{K})$. $\endgroup$ Sep 12, 2022 at 19:35
  • $\begingroup$ I agree with Randall. I think you need to consider your $2\times 3$ matrix as being multiplied on the left by your $2\times 2$ matrix. $\endgroup$ Sep 12, 2022 at 19:35

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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.

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  • $\begingroup$ Thank you so much, this helps a lot. Here's the matrix I got $$ A = \begin{bmatrix} a & b & 0 & 0 & 0 & 0 \\ c & d & 0 & 0 & 0 & 0 \\ 0 & 0 & a & b & 0 & 0 \\ 0 & 0 & c & d & 0 & 0 \\ 0 & 0 & 0 & 0 & a & b \\ 0 & 0 & 0 & 0 & c & d \\ \end{bmatrix} $$ Quick follow-up. How is the order of base-matrices given, as in, how do I know where the 1 is in any given base-matrix. For example why is this not correct: $$ e_2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ $\endgroup$
    – josh
    Sep 12, 2022 at 20:02
  • $\begingroup$ @josh You can pick whatever basis vectors you want; it's a matter of choice. I picked this one so the $6\times 6$ would come out block-diagonal, following the hint given in the problem. What you got is correct. $\endgroup$
    – anon
    Sep 12, 2022 at 20:09
  • $\begingroup$ Perfect, that's what I figured, tested it with some variation :) Thanks very much! $\endgroup$
    – josh
    Sep 12, 2022 at 20:18

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