Answer
See answer below
Work Step by Step
Let $S$ be the set set of $2 \times 2$ real matrices that commute with the matrix $\begin{bmatrix}
1 & 2 \\
0 & 2
\end{bmatrix}$
We can notice that $A=\begin{bmatrix}
0& 0\\
0 & 0
\end{bmatrix} \times \begin{bmatrix}
1& 2\\
0 & 2
\end{bmatrix}= \begin{bmatrix}
0& 0\\
0 & 0
\end{bmatrix} \in S$. Thus $S$ is non-empty.
Assume that $A,B \in S \rightarrow AM=MA, BM=MB \\ \rightarrow (A+B)M=AM+BM=MA+MB=M(A+B)\\
\rightarrow A+B \in S$
Given a scalar $k$ we have: $AM=MA \\ \rightarrow (kA)M=k(AM)=k(MA)=M(kA) \\
\rightarrow kA \in S$
Hence $S$ is a non-empty subset of $M_2R$ and therefore $S$ is a subspace of $M_2R$. Since then $S$ do form a vector space over $R$
