<s>[INST] Finding a basis for a set of matrices in a vector space. I am trying to find a basis for the following vector space: 
V = {2x2 matrices A | $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$A = A$\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$}
So far, I have augmented $\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)$ with the identity matrix to find that A = $\bigl( \begin{smallmatrix} -3 & 2 \\ 2 & -1 \end{smallmatrix} \bigr)$. Is the identity matrix the only basis for this question? 
 [/INST] No, it is not. For example, also $\;A\;$ and all its powers commute with A, and also for example
$$\begin{pmatrix}-1&1\\1&0\end{pmatrix}\;\ldots$$
To solve this, write 
$$B:=\begin{pmatrix}a&b\\c&d\end{pmatrix}\;,\;\;\text{so that}\;\;AB=BA\iff \begin{pmatrix}1&2\\2&3\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&2\\2&3\end{pmatrix}\iff$$
$$\iff\begin{cases}a+2c=a+2b\\b+2d=2a+3b\\2a+3c=c+2d\\2b+3d=2c+3d\end{cases}$$
Now solve this system....for example, from the first eq. we get $\;b=c\;$ already. Play around a little with this. The set of all the matrices commuting with $\;A\;$ is a subspace of dimension $\;2\;$ , and thus you'll need two linearly independent such matrices to have a basis.
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