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I know I could show this by counter-example, finding two unitary square matrices of size $2 \times 2$ at least, and conclude $U_n$ is non-abelian. The problem with this, is I think it's somewhat time consuming trying to work out two unitary matrices and showing they don't commute, so I'm hoping there's a more concise, and clever, way of doing it.

I've tried looking at contradiction, assuming for two unitary matrices $A$ and $B$ we have

$AB = BA$

$A^{-1}ABB^{-1} = A^{-1}BAB^{-1}$

$I_n = \bar{A}^T BA \bar{B}^T$

Then maybe trying to show

$(I_n)_{11} = 1 = \left(\bar{A}^T BA \bar{B}^T \right)_{11}$

Doesn't hold for all unitary matrices $A$ and $B$, but short of actually finding $A$ and $B$ to disprove this I'm unsure what could be done.

Any ideas greatly appreciated.

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Or take $U=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $V=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. We have $UV=-VU\neq0$.

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  • $\begingroup$ Why do I always make it slightly more complicated than it is? I'm afraid I answer too quickly...+1. $\endgroup$ – Julien Mar 10 '13 at 19:20
  • $\begingroup$ @user1551 Thanks, this one is even simpler! $\endgroup$ – Noble. Mar 10 '13 at 20:42
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Try $$ U=\left(\matrix{1&0\\0&-1} \right) \qquad\mbox{and}\qquad V=\left(\matrix{1/\sqrt{2}&-1/\sqrt{2}\\1/\sqrt{2}&1/\sqrt{2}} \right). $$

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  • $\begingroup$ Thanks, they look like two simple ones to try and remember actually! $\endgroup$ – Noble. Mar 10 '13 at 19:13
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    $\begingroup$ Since $U = \pmatrix{1 & 0\cr 0 & -1\cr}$ is diagonal with distinct eigenvalues, every matrix that commutes with it is diagonal (i.e. if $V$ commutes with $U$ and $U v = \lambda v$, then also $UVv = \lambda Vv$). So it suffices to note that there are $2 \times 2$ unitary matrices that are not diagonal. $\endgroup$ – Robert Israel Mar 10 '13 at 19:17
  • $\begingroup$ @RobertIsrael Right. Thanks for the note. I'm sure the OP will appreciate it. $\endgroup$ – Julien Mar 10 '13 at 19:19
  • $\begingroup$ @RobertIsrael Thanks Robert, that is very useful to note. $\endgroup$ – Noble. Mar 10 '13 at 20:41
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Since this question has been answered explicitly, let me suggest a general fact that it useful here (and elsewhere), of which julien's answer is one case. If we take an $n \times n$ matrix (unitary or not), which is diagonal, with $n$ distinct entries on its main diagonal, it will only commute with other diagonal matrices. This is easy to check, and I omit the details. Therefore when $n >1,$ any time you can dream up two unitary $n \times n$ matrices, one diagonal with distinct entries on its diagonal, and the other not diagonal, they will not commute.

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  • $\begingroup$ I see user1551 came up with another example of the same kind in the meantime! $\endgroup$ – Geoff Robinson Mar 10 '13 at 19:24
  • $\begingroup$ This is indeed an important fact, +1. $\endgroup$ – Julien Mar 10 '13 at 19:24
  • $\begingroup$ And Robert Israel's comment notes the same fact. $\endgroup$ – Geoff Robinson Mar 10 '13 at 19:25
  • $\begingroup$ @GeoffRobinson Thanks a lot Geoff! $\endgroup$ – Noble. Mar 10 '13 at 20:42

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