If A commutes with both of these matrices, then A must be a scalar multiple of the identity matrix

Question

I am working on the following problem:

Let $A$ be a $4 \times 4$ matrix with entries in a field of characteristic zero. Suppose that $A$ commutes with both $\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 4 \end{pmatrix}$ and $\begin{pmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}$. Prove that $A$ is a scalar multiple of the identity matrix.

I know that $A$ is a scalar multiple of the identity matrix if and only if $AB = BA$ for all other possible $4 \times 4$ matrices $B$ with entries in a field of characteristic $0$. However, I'm struggling with deducing here that $A$ commuting with these specific matrices forces $A$ to be a scalar multiple of the identity matrix. Does commuting with these specific matrices force $A$ to commute with all $4 \times 4$ matrices with entries in a field of characteristic $0$? If so, how can I deduce this?

Thanks!

Answer 1

$A$ commutes with the first matrix implies that $A$ preserves its eigenspaces. This implies that $A(e_i)=c_ie_i,i=1,2,3,4$.

$A$ commutes with the second matrix $C$ implies that $AC(e_1)=A(e_2)=c_2e_2=C(A(e_1)=C(c_1e_1)=c_1e_2$ implies $c_1=c_2$,...

since $AC(e_2)=CA(e_2), AC(e_3)=CA(e_3)$ deduce that $c_1=c_2=c_3=c_4$.