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! 

Hint:
Left multiplication of a square matrix by  $D=$ first (diagonal) matrix amounts to multiply its rows by the diagonal elements (for this matrix,  the first row is multiplied  by $1$, the second row by $2$, &c.). Right multiplication amounts to multiplying its columns by the diagonal elements. If both results are equal, by identification, you can deduce that $A$ is a diagonal matrix.
Commutativity of multiplication by the second matrix will then let you show, by identification, that all elements on the diagonal are equal.
