On eigenvalues of complex-orthogonal matrices Suppose that $A$ is a complex matrix satisfying $A^TA = I$ (so $A$ is the entrywise transpose, not the conjugate transpose).  What can be said about the eigenvalues of $A$, if $A$ is "complex-orthogonal" in this sense?  
Of course, for any eigenpair $(\lambda,x)$, we have
$$
x^Tx = x^TA^TAx = (Ax)^TAx = \lambda^2 (x^Tx)
$$
which allows us to conclude that $\lambda^2 = 1$... so long as $x^Tx \neq 0$.  Can anything else be said?  Does the case in which $A$ has real entries allow us to conclude that $|\lambda| = 1$?
 A: $$A=\pmatrix{\frac{a+a^{-1}}2&i\frac{a-a^{-1}}2\\
-i\frac{a-a^{-1}}2&\frac{a+a^{-1}}2}$$
has $A^tA=I$ and has $a$ and $a^{-1}$ as eigenvalues.
As an example $a=2$ gives
$$A=\pmatrix{\frac54&\frac34i\\-\frac34i&\frac54}$$
and
$$A^tA=\pmatrix{\frac54&-\frac34i\\\frac34i&\frac54}
\pmatrix{\frac54&\frac34i\\-\frac34i&\frac54}=\pmatrix{1&0\\0&1}.$$
A: At the risk of stating the obvious, as it hasn't been explicitly stated here, I add that:

*

*Each eigenvalue is either (a) one of a pair of eigenvalues $(\lambda, \lambda^{-1})$, or (b) $\lambda = \pm 1$.

*The product of all the eigenvalues satisfies $\prod_i \lambda_i = \pm 1$.

Both of these claim are straightforward to show:

*

*Let $A x = \lambda x$ define the left eigenvector $x$, then by taking the transpose it follows that $x^T A = x^T \lambda^{-1}$, i.e. $x^T$ is a right eigenvector with eigenvalue $\lambda^{-1}$. If (a) $x^T x = 0$ then these are distinct eigenvalues and $\lambda$ may take any value, if (b) $x^T x \neq 0$ then they are the same eigenvalue, and we require $\lambda = \lambda^{-1}$, i.e. $\lambda = \pm 1$.

*Note $1 = \det(I) = \det(A A^T) = \det(A)\det(A^T)= \det(A)^2$. This implies $\det(A) = \prod_i \lambda_i = \pm 1$.

A: This has been thoroughly studied in the paper "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices" by Horn and Merino (1999) and also in Olga Ruff's master thesis "The Jordan Canonical Forms of complex orthogonal and skew-symmetric matrices: characterisation and examples" (2007). In particular, theorem 1.2.3 (pp. 31-32) of Ruff's thesis states that

An $n\times n$ complex matrix is similar to a complex orthogonal matrix if and only if its Jordan Canonical Form can be expressed as a direct sum of matrices of only the following three types:
(a) $J_k(\lambda)\oplus J_k(\lambda^{-1})$ for $\lambda\in\mathbb C\setminus\{0\}$ and any $k$,
(b) $J_k(1)$ for any odd $k$ and
(c) $J_k(-1)$ for any odd $k$.

In particular, every nonzero complex number is an eigenvalue of some complex orthogonal matrix, and for each complex orthogonal matrix, all eigenvalues $\ne\pm1$ must occur in reciprocal pairs.
A: I have some simple ideas:
Let $A$ be an orthogonal matrix $i.e.$ $A^TA=AA^T=I$, from the very definition of the orthogonal matrix, $A$ and $A^T$ are both inverses of each other. So the definition says that their eigenvalues are reciprocals. $i.e.$ if $a$ $\in \mathbb{C}$  is an eigenvalue of $A$ then $\frac{1}{a}$ is an eigenvalue of $A^T$.  But we know that the eigenvalues of  $A$ and $A^T$ are same.  So $a=\frac{1}{a}$ implies $a^2=1$. If we take $a=a_1+ia_2$ then $a_1^2-a_2^2=1$, $a_1a_2=0$ which implies modulus of $a$ is $1$.
A: To clarify the result from the existing answer: we can get a complex-orthogonal matrix with eigenvalues $a,a^{-1}$ with
$$
A = \frac 12 \pmatrix{a+a^{-1} & i(a - a^{-1})\\ -i(a - a^{-1}) & a + a^{-1}} = \pmatrix{i&-i\\1&1} \pmatrix{a\\&a^{-1}}\pmatrix{i&-i\\1&1}^{-1}
$$
