eigenvalues and eigenvectors of functions I am quite confused about the following question below and how I should understand it.
Let $V=M_{3 \times 3}(\mathbb{R}) = \{$ all $ 3 \times 3 $ matricies with real entries}.
Let function $L: V \rightarrow V$ be defined by: L(X) = $A \cdot X $ - $X \cdot  A$
I intend to compute all eignevalues of L and chose a corresponding eigenvector for each eigenvalue.
Would it be possible to simplify $A \cdot X $ - $X \cdot  A$ into a simple matrix expression? In addition how would it be possible to calculate eigenvalues and eigenvectors of matrices containing variables? Thank you very much and any insights are well appreciated.
 A: You can rewrite this as a single matrix-vector operation by "vectorizing" the $3\times 3$ matrices into vectors in $\mathbb{R}^9$. Stack the columns of $X$ like 
$$
x = \begin{pmatrix}
x_{11} \\
x_{21} \\
\vdots \\
x_{23} \\
x_{33}
\end{pmatrix}.
$$
You can then write this out the operator $L$ as $L = A\otimes I_{3}-I_{3}\otimes A$. This can be written explicitly in block matrix notation as
$$
L = \begin{pmatrix}
a_{11}I_{3} & a_{12}I_{3} & a_{13}I_{3} \\
a_{21}I_{3} & a_{22}I_{3} & a_{23}I_{3} \\
a_{31}I_{3} & a_{32}I_{3} & a_{33}I_{3} 
\end{pmatrix} - 
\begin{pmatrix}
A & 0 &0 \\
0 & A & 0 \\
0 & 0 & A
\end{pmatrix}.
$$
Hopefully this makes the eigenvalue problem a little easier. I haven't worked it out myself, but one can notice that for a general matrix $A$, we have that $I$, $A$, and $A^2$ are all eigen-elements corresponding to the eigenvalue $0$. Higher order polynomials are also eigen-elements corresponding to $0$ but they are not linearly independent from these as a result of the characteristic polynomial of $A$ having degree $3$. This also implies that $A^{-1}$ is contained within this eigenspace.
