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Is it possible to consider complex eigenvalues without a Hermitian (i.e. sesquilinear) inner product over a complex vector space?


For instance: let $A$ be a real orthogonal matrix (so $A^TA = I$). Without referencing a Hermitian inner product, is it possible to show that the complex eigenvalues of $A$ have magnitude $1$?

The usual proof of this fact is as follows: if $x,\lambda$ is an eigenpair of $A$, then we have $$ \|x\|^2 = x^*x = x^*(A^*A)x = (Ax)^*(Ax) = \lambda\overline{\lambda} (x^*x) = |\lambda|^2 \|x\|^2 $$ from which it follows that $|\lambda| = 1$. Note: this proof required the use of the sesquilinear inner product $\langle x,y \rangle = y^*x$.


A rephrasing of the original question: consider $\Bbb C^n$ with the bilinear from $$ \langle x,y \rangle = y^Tx $$ note that this bilinear form is not an inner product. The complex-orthogonal matrices are those matrices $A$ that satisfy $A^TA = I$, where $T$ is the entrywise transpose. Notably, the complex-orthogonal matrices preserve the above bilinear form. How can we show that if $A$ is complex-orthogonal with real entries, then the eigenvalues of $A$ have magnitude $1$?

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  • $\begingroup$ Might be a silly question, but are you allowed to reference to a symmetric bilinear real inner product on $\Bbb R^n$? Or are you looking for something that flat-out uses only the structure of real vector space + $\overline {\Bbb R}=\Bbb C$? $\endgroup$ – user228113 May 21 '17 at 13:48
  • $\begingroup$ @G.Sassatelli sure, can't see why not. The question I'm avoiding is how this bilinear form should extend to the complexification of the vector space. $\endgroup$ – Omnomnomnom May 21 '17 at 13:50
  • $\begingroup$ Do you mean the conjugate transpose by $A^T$ ? $\endgroup$ – Astyx May 21 '17 at 13:50
  • $\begingroup$ @Astyx well in this context, $A$ is a real matrix, so of course the two are the same. $\endgroup$ – Omnomnomnom May 21 '17 at 13:52
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    $\begingroup$ @Veliko "I think you would agree that the property of the eigenvalues norm actually follows from the matrix being orthogonal and surely can be proven without explicitly talking about complex inner product". I'm not so sure, hence the question. The transpose property of orthogonal matrices is ultimately a reference to a real inner product, hence the issue. $\endgroup$ – Omnomnomnom May 21 '17 at 14:31
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Well, in a way, yes. The real Jordan normal form theorem yields that every real matrix has an invariant subspace $W$ (i.e. $AW\subseteq W$) such that $1\le\dim W\le 2$. Since $\dim W^\perp +\dim W=n$ and, if $A$ is orthogonal, the orthogonal complement of an $A$-invariant subspace is $A$-invariant, there is an orthonormal basis $B=(b^1,\cdots, b^n)$ such that $B^{-1}AB=B^TAB=\begin{pmatrix}U &0\\ 0&U'\end{pmatrix}$, for some $U\in O(2)$ and $U'\in O(n-2)$ - or, repsectively, $O(1)$ and $O(n-1)$. Now, the eigenvalues of $A$ are either eigenvalues of $U$ or eigenvalues of $U'$. The form and eigenvalues of a $2\times 2$ orthogonal matrix can be calculated explicitly, and the rest can be done by induction.

That being said, as far as I know the real Jordan normal form theorem is proved by complexifying the endomorphism of $A$, using the machinery of Jordan normal form in $\Bbb C^n$, and then bringing it back to $\Bbb R^n$. So one could argue that this is just hiding the issue under the carpet.

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  • $\begingroup$ Interesting approach! I agree that the real Jordan form trick is sweeping something under the rug... but I think all that can be framed in terms of polynomials, perhaps. $\endgroup$ – Omnomnomnom May 21 '17 at 14:11
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    $\begingroup$ Seems legit to me. The usual machinery involved in the Jordan theorem does not use anything Hermitian. $\endgroup$ – Amitai Yuval May 21 '17 at 14:15
  • $\begingroup$ @Omnomnomnom I've just seen the part where you state the actual problem. I don't know how to tackle it on top of my head and, for now, I don't know if this approach may work effectively: last time I tackled a similar problem, I thought exactly the thing I wrote and took a grizzling blunder. $\endgroup$ – user228113 May 21 '17 at 14:19
  • $\begingroup$ Uh oh. Still, it's the closest thing that I'll get to an answer quickly. If I really try to tackle it, maybe I'll ask a follow up. $\endgroup$ – Omnomnomnom May 21 '17 at 14:27
  • $\begingroup$ @Omnomnomnom I mean, it can certainly be used to prove that $A^TA=I$ and $A$ real implies that the eigenvalues of $A$ are roots of unity. However, I suggest caution with generalizations: I once tried to craft a proof of $A\in O(n,m)\implies A\text{ diagonalizable in }\Bbb C$ and I ended up using it incorrectly (for the obvious reason that it was a false theorem). $\endgroup$ – user228113 May 21 '17 at 14:58
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Here is an argument without using the sesquilinear inner product on $\mathbb C^n$, but the usual bilinear inner product $b$ on $\mathbb R^n$ is still used. Essentially:

  1. By using the normality of $Q$ with respect to the real inner product $b$, it can be shown that, up to a change of orthonormal basis on $\mathbb R^n$, we may assume that $Q=(-I_s)\oplus R$ for some real orthogonal matrix $R$ that doesn't possess $-1$ in its spectrum.
  2. By using Rayleigh quotients on $\mathbb R^n$ (and so we are still using the real inner product $b$), it can be shown, without stepping into the complex field, that every real symmetric matrix has a real orthonormal eigenbasis. It follows that every real positive semidefinite matrix has a complete and nonnegative spectrum.
  3. By Cayley transform, $R=(I-K)(I+K)^{-1}$ for some real skew-symmetric matrix $K$. As $-K^2=K^TK$ is positive semidefinite, all complex eigenvalues of $K$ are purely imaginary (this is straightforward if we can use the sesquilinear inner product on $\mathbb C^n$; since we cannot use it here, we need item 2). Since $|(1-z)/(1+z)|=1$ for every $z\in i\mathbb R$, we conclude that all eigenvalues of $R$ lie on the unit circle.
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  • $\begingroup$ Very nice. The first two steps seem natural but using the Cayley transform is clever. $\endgroup$ – Omnomnomnom May 21 '17 at 17:14
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As long as you have a notion of complex scaling (which you have by definition of "vector space over $\Bbb C$"), you can define eigenvalues and eigenvectors. You cannot, without a norm or inner product, say anything about normalised eigenvectors, or whether the eigenvectors are orthogonal, or anything like that, but the eigenvalues are just complex numbers. As such they behave as complex numbers always do, which specifically means that they have an absolute value / norm of their own.

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  • $\begingroup$ Thank you for your answer. Just so you know, I've narrowed the question down quite a bit. $\endgroup$ – Omnomnomnom May 21 '17 at 14:08

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