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? 
 A: 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.
A: The element-by-element comparison of $A\bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 1 & 2 \\ 2 & 3 \end{smallmatrix} \bigr)A$ yields two linearly independent relationship between the four elements of matrix $A$.  That leaves only two linearly independent basis of the vector space.
A: You've misunderstood the question: it asks for all matrices A that commute with the given matrix.  Instead, you just found the inverse of the given matrix: that certainly commutes with it, but it might very well not be the only matrix that has that property.
A: Start with the definitions
$$
  \mathbf{A} = 
\left[
\begin{array}{cc}
 1 & 2 \\
 2 & 3 \\
\end{array}
\right],
\qquad
  \mathbf{B} = 
\left[
\begin{array}{cc}
 a & b \\
 c & d \\
\end{array}
\right].
$$
Following the logic of @DonAntonio, compute the commutator:
$$
 \left[ \mathbf{A}, \mathbf{B} \right] = 
   \mathbf{A} \mathbf{B} - 
   \mathbf{B} \mathbf{A} = 
   2\left[ \begin{array}{cc}
      -b + c & -a - b + d \\
     a+c-d  & b -c
    \end{array} \right]
= \mathbf{0}.
$$
From the diagonal terms we see that $b = c$. We now have matrix.
$$
  \mathbf{B} = 
\left[
\begin{array}{cc}
 a & b \\
 b & c \\
\end{array}
\right], \qquad
\left[ \mathbf{A}, \mathbf{B} \right] = 
   2\left[ \begin{array}{cc}
      0 & -a - b + d \\
     a+b-d  & 0
    \end{array} \right]
= \mathbf{0}.
$$
From this, $d = a+b$. 
The set of matrices which commute with $\mathbf{A}$ are
$$
\mathbf{B} = 
\left[
\begin{array}{cc}
 a & b \\
 b & a + b \\
\end{array}
\right], \quad \alpha, \beta \in \mathbb{C}.
$$
Notice the original matrix $\mathbf{A}$ has this form.
