Csdrhrt

Let $A$ and $B$ be matrices such that $$A = begin{pmatrix}0 & 1\15 & 2end{pmatrix},: B = begin{pmatrix}0 & 2 & -4 \ 2 & -3 & -2 \ -4 & -2 & 0 end{pmatrix}$$ Let $f$ and $g$ be linear transformations such that $$f, g: M_{3x2}(mathbb{R})
rightarrow M_{3x2}(mathbb{R})$$ where $f(X) = BX - XA$ and $g(X) = BXA$ for every $3$ x $2$ real matrix $X$. I want to find the eigenvalues and bases for the eigenspaces of these functions.

EDIT: While @xbh 's method is correct, it involves computing the eigenvalues of a $6times6$ matrix. I've recently learned of a method that can solve this problem with less computation, but I only know half of it.

This method relies on the observation that $$begin{pmatrix}lambda_1 & 0 & 0 \ 0 & lambda_2 & 0 \ 0 & 0 & lambda_3 end{pmatrix}begin{pmatrix}a_{11} & a_{12}\ a_{21} & a_{22} \ a_{31} & a_{32}end{pmatrix}= begin{pmatrix}lambda_1a_{11} & lambda_1a_{12} \ lambda_2a_{21} & lambda_2a_{22} \ lambda_3a_{31} & lambda_3a_{32} end{pmatrix}$$ and$$begin{pmatrix}a_{11} & a_{12} \ a_{21} & a_{22} \ a_{31} & a_{32} end{pmatrix}begin{pmatrix}mu_1 & 0 \ 0 & mu_2 end{pmatrix} = begin{pmatrix} mu_1a_{11} & mu_2a_{12} \ mu_1a_{21} & mu_2a_{22} \ mu_1a_{31} & mu_2a_{32} end{pmatrix} $$
Also, it is easy enough to see that the eigenvalues for $A$ are $mu = -3,5$ with associated eigenvectors $v_1 = (1, -3) : v_2 = (1, 5)$. For $B$, $lambda = -4$ has two-dimensional eigenspace given by $t_1 = (1,0,1)$ and $t_2 = (0,2,1)$ and for $lambda = 5, : t_3 = (-2,-1,2)$.

Since $A$ and $B$ are $ntimes n$ matrices with $n$ linearly independent eigenvectors. They are diagonizable. Thus for $f$,$$A = ED_aE^{-1},: B = CD_bC^{-1}$$ and then $$CD_bC^{-1}X - XED_aE^{-1} = delta X$$ $$Longrightarrow D_b(C^{-1}XE) - (C^{-1}XE)D_a = delta C^{-1}XE$$ From here I'm supposed to deduce that the eigenvalues of $f$ are every combination $lambda - mu$; however, I don't understand how to reach this conclusion since substituting any combination $lambda - mu$ for $delta$ cannot make $$(lambda_i-mu_j)begin{pmatrix}a_{11} & a_{12}\ a_{21} & a_{22} \ a_{31} & a_{32}end{pmatrix} = begin{pmatrix}(lambda_1-mu_1)a_{11} & (lambda_1-mu_2)a_{12} \ (lambda_2-mu_1)a_{21} & (lambda_2-mu_2)a_{22} \ (lambda_3-mu_1)a_{31} & (lambda_3-mu_2)a_{32} end{pmatrix}$$ where $a_{ij}$ denotes the scalars of $C^{-1}XE$.

Also I don't know how to find the eigenvectors with this method.

Using the methods described in the edited question, we end up with 6 equations. $$(lambda_1-mu_1)a_{11}=delta a_{11}$$ $$(lambda_1-mu_2)a_{12}=delta a_{12}$$ $$(lambda_2-mu_1)a_{21}=delta a_{21}$$ $$(lambda_2-mu_2)a_{22}=delta a_{22}$$ $$(lambda_3-mu_1)a_{31}=delta a_{31}$$ $$(lambda_3-mu_2)a_{32}=delta a_{32}$$ where $a_{ij}$ denotes the entries of $C^{-1}XE$ and $delta$ are the eigenvalues of $f$ we are trying to find. From this list of equations it is simple to see that the eigenvalues $delta$ are every combination $lambda-mu$. We can then find the eigenvectors by solving for $a_{ij}$ and then solving $begin{pmatrix}a_{11} & a_{12} \ a_{21} & a_{22} \ a_{31} & a_{32} end{pmatrix} = C^{-1}XE$ for $X$

For example, with the given eigenvalues/vectors of $A$ and $B$ consider $delta=-1$ $$-a_{11}=-a_{11}$$ $$-9a_{12}=-a_{12}$$ $$-a_{21}=-a_{21}$$ $$-9a_{22}=-a_{22}$$ $$8a_{31}=-a_{31}$$ $$0*a_{32}=-a_{32}$$ Every $a_{ij}$ other than $a_{11}$ and $a_{21}$ are $0$, so the eigen vector for $delta=-1$ is given by $begin{pmatrix}a_{11} & 0 \ a_{21} & 0 \ 0 & 0 end{pmatrix}$ which can be expressed by the basis ${begin{pmatrix}1 & 0 \ 0 & 0 \ 0 & 0 end{pmatrix},begin{pmatrix} 0 & 0 \ 1 & 0 \ 0 & 0 end{pmatrix} }$

For example, to solve $begin{pmatrix}1 & 0 \ 0 & 0 \ 0 & 0 end{pmatrix}=C^{-1}XE :$ simply compute $Cbegin{pmatrix}1 & 0 \ 0 & 0 \ 0 & 0 end{pmatrix}E^{-1}$

As for $g$, the eigenvalues will be every combination $lambda*mu$ and the process for finding the eigenvectors is very similar.