# 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.

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Nope, not even when $C$ is invertible. For counterexample, just add $I$ to $A,B,C$ in wasting_chalk's answer. – achille hui Feb 25 '13 at 17:12
And if $[A,B] = AB - BA$ is invertible? – Learning is a mess Feb 25 '13 at 17:13
What I have in mind is a bit more precise, if you look at a certain 3*3 matrix C, which commutes with any element of SO(3) it needs to be proportional to the identity. – Learning is a mess Feb 25 '13 at 17:16
So you need to make another edit. You need $A^{\top}A = B^{\top}B = I$ and $\det(A) = \det(B) = 1$. – Fly by Night Feb 25 '13 at 17:46
What is true for sure is: if $C$ commutes with every matrix, then it is scalar. This can be strengthened a bit: if $C$ commutes with every invertible matrix of determinant $1$, then it is scalar. I don't know how much further we can go this way, I admit. – 1015 Feb 25 '13 at 18:08

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.

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+1: Your counter example works, I've tried it. $AB \neq BA$ while $AC=CA$ and $BC=CB$, as required. – Fly by Night Feb 25 '13 at 16:57
Yes you're right I should have added that A and B are invertible. – Learning is a mess Feb 25 '13 at 17:05

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 $AB-BA=\begin{pmatrix}0&\sqrt{2}&0\\ \sqrt{2}&0&0\\0&0&0\end{pmatrix}\not=0.$

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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.

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