We were asked in the exam to find base for $W$ , such that $W$ is the the matrices $B$ of the form $M_{2 \times 2 } (\mathbb{R})$ such that $A B = B A$
Were $A = \begin{bmatrix} 1 && 2\\ 3 && 4 \end{bmatrix}$
I found that $ W= span (\begin{bmatrix} 0 && \frac{2}{3}\\ 1 && 1 \end{bmatrix})$
Is my solution correct ??!
No, your solution is not correct. The space $W$ is spanned by all the matrices of the form $$B=\begin{bmatrix} a & b \\ 3b/2 & a+3b/2 \end{bmatrix}, $$ where $a,b$ are any real number. In particular, the matrix you got is obtained for $a=0$ and $b=2/3$; the identity matrix, that commutes with any other matrix, so it must be in $W$, is obtained for $a=1$ and $b=0$; and the matrix $A$, which obviously commutes with itself, is obtained for $a=1$ and $b=2$. If you prefer, you can say that the 2-dimensional space $W$ is spanned by the matrices $$ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\quad \text{and}\qquad \begin{bmatrix} 0 & 1 \\ 3/2 & 3/2 \end{bmatrix}. $$ You can verify the result in Wolfram Alpha.