Two $n × n$ matrices $A$ and $B$ are said to be simultaneously diagonalizable if there exists an invertible $n × n$ matrix $S$ such that $S^{−1}AS$ and $S^{−1}BS$ are both diagonal.
Let $D$ be a diagonal $n × n$ matrix with distinct entries on the diagonal. How would I find all $n × n$ matrices $C$ such that $CD = DC$?
Isn't it just the $0$ matrix, because there exist no such matrices?
Hint First, note that the answer clearly depends on the matrix $D$. If $D$ is a multiple $\lambda I_n$ of the identity matrix, every $n \times n$ matrix $C$ satisfies $CD = DC$ (we say that $C$ commutes with $D$). On the other hand, one can show that the multiples $\lambda I_n$ are the only matrices that commute will all matrices.
We can approach the problem naively and write out the equation $CD = DC$ in entries: Taking the $(i, j)$ entry of each side gives $$\sum_{k = 1}^n c_{ik} d_{kj} = \sum_{k = 1}^n d_{ik} c_{kj} .$$ This is a rather complicated quadratic relation, and, indeed if $D$ were a general matrix, this wouldn't be a very efficient approach---but since in our case $D$ is diagonal, so $d_{kj}$ is zero unless $k = j$. If we substitute these relations into the equation, it simplifies to $$c_{ij} d_{jj} = d_{ii} c_{ij} .$$ So, if $d_{ii} = d_{jj}$, then $c_{ij}$ can assume any value, but if $d_{ii} \neq d_{jj}$, we must have $c_{ij} = 0$.
Hint: Let $C$ be a matrix with entries $(C)_{ij} = c_{ij}$, and let $d_i = (D)_{ii}$ denote the diagonal entries of $D$. Then we have $$ (DC)_{ij} = d_ic_{ij} \qquad (CD)_{ij} = d_jc_{ij} $$ What conditions on $c_{ij}$ cause these two to be equal for every $i$ and $j$?
It is useful to remember (and easy to show) that if $D$ is diagonal and $A$ is any matrix of the same size, then $DA$ is obtained from $A$ by multiplying every entry by the diagonal entry of$~D$ corresponding to its row, while $AD$ is obtained from $A$ by multiplying every entry by the diagonal entry of$~D$ corresponding to its column. (More concisely, left-multiplication by a diagonal matrix multiplies each row by a scalar, while right-multiplication does so with every column). Now since the diagonal entries of $D$ are assumed to be all different, it follows that as soon as $A$ has any off-diagonal entry that is nonzero, then $DA$ and $AD$ will differ in that position. On the other hand if all off-diagonal entries of $A$ are zero (that is, if $A$ is a diagonal matrix), then clearly $DA=AD$.
So the answer to your question is precisely the set of diagonal matrices.
