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Let $A$ and $B$ be matrices such that $$A = \begin{pmatrix}0 & 1\\15 & 2\end{pmatrix},\: B = \begin{pmatrix}0 & 2 & -4 \\ 2 & -3 & -2 \\ -4 & -2 & 0 \end{pmatrix}$$ 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 $6\times6$ 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 $n\times n$ 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.

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  • $\begingroup$ Recall the definition of eigenvalues, eigenvectors for linear transformation. Note that "vectors" are not merely the 1 column matrices. $\endgroup$ – xbh Nov 14 '18 at 4:30
  • $\begingroup$ Okay, you're right. In this case we can call $X$ a vector because it is an element of the vector space $M_{3x2}(\mathbb{R})$, but this is only a technical correction. I still don't know how to solve for $X$ $\endgroup$ – Ryan Greyling Nov 14 '18 at 4:57
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    $\begingroup$ Seems like a "hard" computation… let $X = [x_{j,k}]_{3\times 2}$ and compute $f(X)$ and solve $f(X) = cX$. Or take a basis of $M_{3,2}(\Bbb R)$ then find out the matrix of $f$ in this basis. Now you could apply your knowledge about solving $My = cy$ where $M \in \mathrm M_6 (\Bbb R), y \in \mathrm M_{6,1}(\Bbb R)$. $\endgroup$ – xbh Nov 14 '18 at 5:05
  • $\begingroup$ Oh I see, that'll definitely work. I'll post an answer once I've done the computations. I understand that the eigenvectors for $My = cy$ will have to be converted from 6 dimensional column vectors back to $3$ x $2$ matrices using my chosen basis for $M_{3,2}(\mathbb{R})$. But just to make sure, the eigenvalues will remain the same right? $\endgroup$ – Ryan Greyling Nov 14 '18 at 5:23
  • $\begingroup$ Yeah, just a formal change. This is just an application of linear algebra notions. The eigenvalues are totally determined by the transformations on their own. $\endgroup$ – xbh Nov 14 '18 at 5:28
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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 $C\begin{pmatrix}1 & 0 \\ 0 & 0 \\ 0 & 0 \end{pmatrix}E^{-1}$

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

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