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.


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

  • $\begingroup$ 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

Try $$ U=\left(\matrix{1&0\\0&-1} \right) \qquad\mbox{and}\qquad V=\left(\matrix{1/\sqrt{2}&-1/\sqrt{2}\\1/\sqrt{2}&1/\sqrt{2}} \right). $$

  • $\begingroup$ Thanks, they look like two simple ones to try and remember actually! $\endgroup$ – Noble. Mar 10 '13 at 19:13
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.