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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 $|\lambda| = 1$?

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  • $\begingroup$ For the first question, what about $x^*x=x^*A^*Ax=(Ax)^*Ax=\overline{\lambda}\lambda x^*x=|\lambda|^2x^*x$, so $|\lambda|^2=1$? $\endgroup$ May 21 '17 at 12:25
  • $\begingroup$ @GerryMyerson $A^*$ is the adjoint of $A$ relative to the Hermitian inner product, i.e. $$ \langle x, y \rangle = y^*x = \sum_{j=1}^n x_j \overline{y_j} $$ the point is to avoid this. $\endgroup$ May 21 '17 at 12:46
  • $\begingroup$ For any of you interested in the (now gone) first part of this question, I've asked it separately over here. $\endgroup$ May 21 '17 at 13:38
  • $\begingroup$ $A^*$ is the matrix you get from $A$ by flipping it across the main diagonal and taking the complex conjugate. You can do both those actions without ever having heard of inner products. $\endgroup$ May 21 '17 at 23:08
  • $\begingroup$ @GerryMyerson sure, I guess you could. It would a pretty unintuitive thing to do, though. Why, for instance, should we expect that $(AB)^* = B^*A^*$? $\endgroup$ May 22 '17 at 0:41
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$$A=\pmatrix{\frac{a+a^{-1}}2&i\frac{a-a^{-1}}2\\ -i\frac{a-a^{-1}}2&\frac{a+a^{-1}}2}$$ 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 $$A^tA=\pmatrix{\frac54&-\frac34i\\\frac34i&\frac54} \pmatrix{\frac54&\frac34i\\-\frac34i&\frac54}=\pmatrix{1&0\\0&1}.$$

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  • $\begingroup$ That answers part of the second half of my question, so thanks for that. Still, I'm looking for something more comprehensive. $\endgroup$ May 21 '17 at 13:01
  • $\begingroup$ @Omnomnomnom For even $n$ the quadratic form $z_1^2+\cdots+z_n^2$ is equivalent to $w_1w_2+w_3w_4+\cdots$. Is this a useful step towards your "comprehensive" vision? $\endgroup$ May 21 '17 at 13:07
  • $\begingroup$ About your other comment, I suppose that depends on what exactly an "equivalence" of quadratic forms refers to in this case $\endgroup$ May 21 '17 at 13:15
  • $\begingroup$ @Omnomnomnom For equivalence of quadratic forms see Wikipedia en.wikipedia.org/wiki/Quadratic_form $\endgroup$ May 21 '17 at 13:30
  • $\begingroup$ You're right about the matrix... it was quick to check by hand after all. I'm surprised that W|A made a mistake there; I don't think I have a typo. $\endgroup$ May 21 '17 at 13:33
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At the risk of stating the obvious, as it hasn't been explicitly stated here, I add that:

  1. Each eigenvalue is either (a) one of a pair of eigenvalues $(\lambda, \lambda^{-1})$, or (b) $\lambda = \pm 1$.
  2. The product of all the eigenvalues satisfies $\prod_i \lambda_i = \pm 1$.

Both of these claim are straightforward to show:

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

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  • $\begingroup$ Interestingly Ruff and Horn and Merino seem inequivalent. Horn and Merino further explicitly exclude the possibility of $J_k(\pm1) \oplus J_k(\pm 1)$ for odd $k$ (theorem 1), whereas Ruff (quoted here) does not. $\endgroup$ Mar 3 at 15:30
  • $\begingroup$ @ComptonScattering $J_k(1)\oplus J_k(1)$ for odd $k$ is allowed in Horn and Merino. It is the direct sum of two matrices of type (d). The authors just state the same theorem differently. It might be clearer to say that even-sized Jordan blocks for $\lambda=\pm1$ must occur an even number of times but odd-sized Jordan blocks for $\lambda=\pm1$ can occur any number of times. $\endgroup$
    – user1551
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  • $\begingroup$ Yes you are correct. $\endgroup$ Mar 3 at 15:48
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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 $1$.

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    $\begingroup$ Noting that $A$ has the same eigenvalues as $A^{-1}$ is not enough to deduce that any eigenvalues satisfy $a = (1/a)$. For instance, note that $$ A = \pmatrix{2&0\\0&1/2} $$ has the same eigenvalues as its inverse. $\endgroup$ May 21 '17 at 13:25
  • $\begingroup$ ohh... I see. I made a mistake doing $a=\frac{1}{a}$. thank you $\endgroup$ May 21 '17 at 13:29
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To clarify the result from the existing answer: we can get a complex-orthogonal matrix with eigenvalues $a,a^{-1}$ with $$ A = \frac 12 \pmatrix{a+a^{-1} & i(a - a^{-1})\\ -i(a - a^{-1}) & a + a^{-1}} = \pmatrix{i&-i\\1&1} \pmatrix{a\\&a^{-1}}\pmatrix{i&-i\\1&1}^{-1} $$

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