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Will there be infinitely many unitary matrices of the form of $U_{n \times n}$ with complex coefficients?

How about unitary matrices of the same form but with only real coefficients?


(Not a homework question, just a curious thought).

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    $\begingroup$ There are infinitely many orthonormal basis. $\endgroup$
    – lhf
    Jul 18, 2019 at 21:32
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    $\begingroup$ It is enough to prove for $n=2$. Then you can complete it to a block matrix with $I_{n-2}$ as one block. $\endgroup$
    – lhf
    Jul 18, 2019 at 21:34
  • $\begingroup$ If you try to explain enough conditions instead of "the same form" maybe some not obvious results can be concluded. $\endgroup$ Jul 18, 2019 at 22:23

2 Answers 2

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Any rotation matrix is unitary, and there is an uncountable infinity even of $2 \times 2$ kind.

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  • $\begingroup$ (+1) Thank you for the example and explanation. $\endgroup$ Jul 20, 2019 at 1:46
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The unitary matrices with real coefficients are the orthogonal matrices. That is, $O(n)\subset U (n)$.

In dimension two, there are uncountably many orthogonal matrices, the rotations: $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}$, for $\theta\in [0,2\pi)$, forming $SO(2)\subset O(2)$.

$SO(2)$ is thus isomorphic to the circle group, sometimes denoted $U(1)$.

You can use induction to prove it for $n\gt2$, as suggested in the comments by @lhf.

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  • $\begingroup$ Thank you for the answer! This is what I wanted! $\endgroup$ Jul 20, 2019 at 1:46

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