How to show the fact that the set of all orthogonal matrices over $\mathbb C$ is compact

By an orthogonal matrix over $\mathbb C$ I mean a matrix $A$ satisfying $AA^T=I$ and here $A^T=(a_{ji})$ where $A=(a_{ij})$

It is not the same as unitary matrix where in unitary matrix we take transpose and then conjugate or vice versa

I know that set of all orthogonal matrices over $\mathbb R$ is compact.

I think the closedness of the set will follow from the same arguements as in the above case. But the boundedness part not sure

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    $\begingroup$ Do you mean unitary? Because an orthogonal matrix is real by definition. $\endgroup$ Apr 9, 2015 at 4:42
  • $\begingroup$ Possibly a duplicate? This question is for 2x2 matrices... But one of the answers addreses the n-dimensional case.... math.stackexchange.com/questions/137354/… $\endgroup$ Apr 9, 2015 at 5:45
  • $\begingroup$ Who says that @JohnColanduoni $\endgroup$
    – Learnmore
    Apr 9, 2015 at 6:15
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    $\begingroup$ @learnmore Every definition I've ever seen, here for example. $\endgroup$ Apr 9, 2015 at 6:16
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    $\begingroup$ unitary and orthogonal over $\mathbb C$ are not same !so how is this a duplicate $\endgroup$
    – Learnmore
    Apr 9, 2015 at 6:16

1 Answer 1


It is not bounded, unless you are working with $1\times 1$ matrices.

The only complex, $1\times 1$ orthogonal matrix are $(1)$ and $(-1)$.

In $2\times 2$ (and by extension you get $n\times n$) you can consider the family, where $\lambda \geq 1$ is a real parameter, $$ A_\lambda = \begin{pmatrix} \lambda & i \sqrt{\lambda^2 - 1} \\ - i \sqrt{\lambda^2 - 1} & \lambda \end{pmatrix} $$ which you can easily check to be orthogonal. But any matrix norm of $A_\lambda$ would tell you it is of size $\approx \lambda$.

  • $\begingroup$ So except the 1 X 1 case, the subgroup is not compact, right? $\endgroup$
    – user422112
    Feb 6, 2018 at 5:48
  • $\begingroup$ ... yes (for the usual topologies you may want to consider). $\endgroup$ Feb 6, 2018 at 22:18

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