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For a diagonal matrix $D=\mathrm{diag}(d_1,\dots ,d_n)$ over a field, matrix multiplication gives $(BD-DB{)}_{ij}=(d_j-d_i)b_{ij}$. Hence a matrix $B$ commutes with $D$ exactly when $b_{ij}=0$ wherever $d_i\ne d_j$. If all diagonal entries are distinct, the centralizer of a diagonal matrix consists exactly of diagonal matrices. When entries repeat, the allowed entries form unrestricted blocks on the corresponding eigenspaces.

Write $C=(c_{ij})$ and $(d_1,d_2,d_3)=(0,2,4)$. Because $D$ is a diagonal matrix,Thus $CD=DC$ exactly when $(d_j-d_i)c_{ij}=0$ for every $i,j$. The three eigenvalues are distinct, so every off-diagonal entry of $C$ must be zero; diagonal entries are unrestricted. Consequently the centralizer of a diagonal matrix here isEvery displayed matrix indeed commutes with $D$, since diagonal matrices multiply entry by entry on the diagonal.