A matrix $M$ that commutes with any matrix is of the form $M=\alpha I$ I feel like this is probably a simple proof but I can't quite come up with it in an elegant way nor could I find it here.
Prove that if a matrix $M$ commutes with any matrix then $M$ is of the form $M=\alpha I$.
Proving the contrapositive seems like the natural way to go where we can logically transform $\lnot \forall A(MA = AM)$ into $\exists A (MA \neq AM)$ but assuming that $M \neq \alpha I$ immediately becomes messy. Is there a nice way out of this or is it inevitably going to get messy?

Here's somewhat of an overkill answer for what it is worth.
A normal matrix is a matrix that is unitarily similar to a diagonal matrix. Another characterization is that a matrix $M$ is normal iff $M^* M = M M^*$. 
If $M$ commutes with all matrices then it is clear it is normal. From this we have $M = UDU^*$ for some unitary matrix $U$. Write $MU = UDU^* U = UD$ and then use commutativity to write $UM = UD$. Finally cancel the $U$'s to conclude that $M$ must be diagonal.
Then if $P$ is the elementary matrix that swaps rows $i$ and $j$ when applied to $M$ as $PM$, since $MP$ swaps columns and is the same as $MP$, we conclude each diagonal element is the same.
