# 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 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. Sep 12, 2022 at 19:33
• 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 $\mathrm{Mat}(2 \times 3, \mathbb{K})$. Sep 12, 2022 at 19:35
• 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. Sep 12, 2022 at 19:35

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.

• 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: $$e_2 = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$
– josh
Sep 12, 2022 at 20:02
• @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.
– anon
Sep 12, 2022 at 20:09
• Perfect, that's what I figured, tested it with some variation :) Thanks very much!
– josh
Sep 12, 2022 at 20:18