# Are there infinitely many unitary matrices?

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).

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

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

• (+1) Thank you for the example and explanation. Jul 20, 2019 at 1:46

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.

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