Find all $2\times 2$ matrices that commute with $AX = XA$? Find all $2\times 2$ matrices that commute with  $$ A = \left( \begin{array}{cc} 1 & 1 \\ 0 & 0 \end{array} \right)$$ where $AX = XA$.
Solution and ask:
$$ X = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$$
$$AX = \left( \begin{array}{cc} 1\cdot a + 1\cdot c  & 1\cdot b + 1\cdot d  \\ 0\cdot a + 0\cdot c  & 0\cdot b + 0\cdot d  \end{array} \right) = \left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right)$$
$$XA = \left( \begin{array}{cc} a\cdot 1 + b\cdot 0  & a\cdot 1 + b\cdot 0  \\ c\cdot 1+d\cdot 0  & c\cdot 1+d\cdot 0  \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
Since $AX = XA$, we obtain
$$\left( \begin{array}{cc} a+c & b+d \\ 0 & 0 \end{array} \right) = \left( \begin{array}{cc} a & a \\ c & c \end{array} \right)$$
so that $a+c = a$, $b+d = a$, $0 = c$ and $0 = c$.
Is the calculation correct?
What is the value of $a,b,c,d$ ? 
$c = 0, a = ?, b = ?, d = ?$

@Servaes answer is exactly right, but let me add one more thing: the set of all such matrices $X$ forms a subspace of the space of $2 \times 2$ matrices. 
One way to see this is to add two of them and see that the resulting matrix has the same form; the other is a little more abstract: suppose that $AX = XA$ and $AY = YA$. Then $A(X+Y) = AX + AY = XA + YA = (X+Y) A$, so the set of such matrices is closed under addition. It's also closed under scalar multiplication (which you can probably show for yourself). 
When you come across a subspace like this, it's not a bad idea to ask yourself, "What's the dimension of the subspace? What's a basis for it?" 
You can then choose a different matrix $A$, and ask yourself whether the dimension changes when $A$ changes, and if so, what's the relationship? 
I know you're not likely to pursue this if this happens to just be a homework problem and you've got others to do as well, but I offer it as a possible general approach as you advance in mathematics. 
