Find Basis and Dimension of specific subspace of $M_{2 \times 2}$. Let $V=M_{2 \times 2}(\mathbb{R})$ be the set of all $2 \times 2$ real-valued matrices, and let the field $K = \mathbb{R}$.  Then $V$ is a vector space under matrix addition and Scaler multiplication.
Let $$A = \begin{bmatrix}1&0\\1&2\end{bmatrix}$$
If $C(A) = \{B \mid B \in V$, such that $AB=BA$} find the dimension of the subspace $C(A)$ and determine a basis.
First of all I believe I would have to prove this is a subspace.  So I would have to show that the subspace is closed under vector addition and scaler multiplication.  Then I would have to construct a basis.  The number of vectors in the basis is the dimension of the subspace.
It is the condition that is tripping me up.  How do show all this with the condition that these $2 \times 2$ matrices are commutative?

Let’s take $B\in C(A)$ to be 
$$\begin{bmatrix} a & b \\ c & d\end{bmatrix},$$
and then compute $AB$ and $BA$. What constraints are there on $a,b,c,$ and $d$? 
To show closure, let’s say that we have $B_1, B_2\in C(A)$. Then for all $x\in\mathbb{R}$ we want:
$$ (xB_1)A = A(xB_1); \quad (B_1 + B_2)A = A(B_1 + B_2).$$
Can you take it from here? 
