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

Let $U,V$ be the $2$ given matrices and $(e_i)_i$ be the canonical basis of $K^4$.
The invariant proper subspaces of $U$ are the $span(\mathcal{B})$ where $\mathcal{B}$ is any proper subset of $(e_i)_i$. For every such $\mathcal{B}$, $span(\mathcal{B})$ is not $V$-stable. then $U,V$ have no common proper invariant subspaces. According to the Burnside's theorem (*), the algebra generated by $U,V$ is whole $M_4(K)$; the required result follows.  $\square$
(*) cf. https://core.ac.uk/download/pdf/82680953.pdf
