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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!

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$A$ 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 $c_1=c_2=c_3=c_4$.

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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.

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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

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My answer can be treated as the supplement to the Loup Blanc's answer, I would like to express similar result in a more elementary way.

Denote both mentioned matrices as $D$ and $P$.

It is easy to check that if $A$ commutes with matrices $D$ and $P$ then it commutes also with any power of matrices $D$ and $P$, any polynomial of $D$ and $P$ and generally with any product or linear combination of these matrices.

For powers of $D$ we have results

$D=\text{diag} ( 1 \ \ 2 \ \ 3 \ \ 4) , \\ D^2= \text{diag} ( 1 \ \ 2^2 \ \ 3^2 \ \ 4 ^2) , \\ D^3=\text{diag} ( 1 \ \ 2^3 \ \ 3^3 \ \ 4^3 ) , \\D^4=\text{diag} ( 1 \ \ 2^4 \ \ 3^4 \ \ 4^4) $

Four vectors formed from diagonal entries are linearly independent (if they are columns of $4 \times 4$ matrix then they form Vandermonde matrix) so linear combination of them can generate any diagonal matrix, denote it generally as ${D_i}$.

On the other hand $P= \begin{pmatrix} 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \end{pmatrix}$ is a permutation matrix with its powers

$P^2= \begin{pmatrix} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{pmatrix}, P^3= \begin{pmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 \end{pmatrix}, P^4=I$.

It is visible that with the expression $D_0+P D_1 + P^2D_2+P^3D_3$ we can generate any $4 \times 4$ matrix (assuming first we generate appropriate diagonal matrices $D_i$) and hence the matrix $A$ has to commute with all possible $4 \times 4$ matrices.

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