Looking for a quick way to prove a matrix identity It's quite simple, it takes two seconds to understand it, but I can't find a quick demonstration. Let's say I have three matrices $A$, $B$ and $C$. The matrices $A$ and $B$ do not commute with one other, but $C$ commutes with both.
Does this imply that $C$ must be proportional to the identity?
EDIT 
The matrices $A$ and $B$ are supposed invertible.
 A: I think the claim is not true. Let $$A=\begin{bmatrix}1&1 &0&0\\0 &1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},$$
$$B=\begin{bmatrix}1&0 &0&0\\1 &1&0&0\\0&0&0&0\\0&0&0&0\end{bmatrix},$$
$$C=\begin{bmatrix}0&0 &0&0\\0 &0&0&0\\0&0&1&1\\0&0&1&1\end{bmatrix}.$$
C is not a multiple of the identity matrix.
A: The OP asks for an example with $A,B\in SO(3)$ in the comments. Let $r=1/\sqrt{2}$,
$$
A=\begin{pmatrix}1\\&-1\\&&-1\end{pmatrix},
\ B=\begin{pmatrix}r&r&0\\-r&r&0\\0&0&1\end{pmatrix},
\ C=\begin{pmatrix}1\\&1\\&&0\end{pmatrix}.
$$
Then $A,B\in SO(3)$ and $C$ is not a multiple of $I$. One can verify that $AC-CA=BC-CB=0$ but $AB-BA=\begin{pmatrix}0&\sqrt{2}&0\\ \sqrt{2}&0&0\\0&0&0\end{pmatrix}\not=0.$
A: I recently came across a problem similar to yours. 
What you want to prove can be proved using Schur's Lemma.
 If A is a complex matrix of order n that commutes with all matrices from G then A is a scalar matrix. If G is not irreducible, then this is not true.
We can see that all the counter examples are wrong.
First of all we can see that one of them is not irreducible and the one that "proves" it using SO(3) elements forgot that that you have to prove that C commutes with ALL SO(3) matrices, which is not true.
Hope this helps.
