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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 $|\lambda|^2=1$? $\endgroup$ – Gerry Myerson 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. $\endgroup$ – Omnomnomnom 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. $\endgroup$ – Omnomnomnom 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. $\endgroup$ – Gerry Myerson 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 $(AB)^* = B^*A^*$? $\endgroup$ – Omnomnomnom 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. $\endgroup$ – Omnomnomnom 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? $\endgroup$ – Lord Shark the Unknown 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 $\endgroup$ – Omnomnomnom May 21 '17 at 13:15
  • $\begingroup$ @Omnomnomnom For equivalence of quadratic forms see Wikipedia en.wikipedia.org/wiki/Quadratic_form $\endgroup$ – Lord Shark the Unknown 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. $\endgroup$ – Omnomnomnom May 21 '17 at 13:33
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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. $\endgroup$ – Omnomnomnom May 21 '17 at 13:25
  • $\begingroup$ ohh... I see. I made a mistake doing $a=\frac{1}{a}$. thank you $\endgroup$ – Hirakjyoti Das 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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