# 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?

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.

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.

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.

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.