Similarly Commuting Matrices I was wondering, if $D$ and $A$ are similar matrices, over $\mathbb{R,C}$, that is $D=S^{-1}AS$ and $DC=CD$, for some $C$, must $A$ commute with $C$?
For some reason, this one is slipping from me. I want to say yes, and that it is a common fact, but I cannot seem to remember it. If so, could someone leave an example/explanation? 
 A: Don't see why. However, let $C = S^{-1} F S,$ which we can do by defining
$F = S C S^{-1}.$ Your $DC=CD$ becomes
$$  S^{-1} A S S^{-1} F S = S^{-1} F S S^{-1} A S,  $$ 
$$  S^{-1} A  F S = S^{-1} F A S,  $$
$$   A  F  =  F A.  $$
Your problem of about eleven hours ago deals with a diagonalizable matrix where the resulting diagonal has $n$ distinct entries on the diagonal, matrices, $n$ by $n.$ Please do these calculations:
$$
\left(
\begin{array}{rr}
5 & 0 \\
0 & 7
\end{array}
\right)
\left(
\begin{array}{rr}
p & q \\
r & s
\end{array}
\right) = ?
$$ 
$$
\left(
\begin{array}{rr}
p & q \\
r & s
\end{array}
\right)
\left(
\begin{array}{rr}
5 & 0 \\
0 & 7
\end{array}
\right) = ??
$$
So: every matrix commutes with the identity matrix. BUT, what kind of matrices commute with a diagonal matrix that has all diagonal elements different? 
If the first pair was not enough, do
$$
\left(
\begin{array}{rrr}
5 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 11
\end{array}
\right)
\left(
\begin{array}{rrr}
r & s & t \\
u & v & w \\
x & y & z
\end{array}
\right) = ?
$$ 
$$
\left(
\begin{array}{rrr}
r & s & t \\
u & v & w \\
x & y & z
\end{array}
\right)
\left(
\begin{array}{rrr}
5 & 0 & 0 \\
0 & 7 & 0 \\
0 & 0 & 11
\end{array}
\right) = ??
$$
