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I have (informally) encountered the following theorem:

Since Hessians are symmetric, it holds that Hessians have all real eigenvalues.

Based on this theorem, I'm presuming that the more general theorem would be as follows:

Any symmetric matrix has all real eigenvalues.

Am I correct in thinking this?

If so, I would greatly appreciate it if someone could please explain and prove why/that this is true.

The Pointer
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  • That statement is not completely true like this, consider $\begin{pmatrix} i & 0 \ 0 & i \end{pmatrix}$. So what's true is, that any real valued symmetric Matrix only has real eigenvalues (depending on your definition of symmetric though). Did you check probably any book on linear algebra for a proof? Or typed it into google? ^^ – hal4math Sep 26 '19 at 14:09
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    @hal4math Thanks for the clarification. I checked my linear algebra textbook , but I didn't find anything like this. I did also Google it, but, again, I didn't find anything. I found this http://pi.math.cornell.edu/~jerison/math2940/real-eigenvalues.pdf for "any real valued symmetric matrix only has real eigenvalues". – The Pointer Sep 26 '19 at 14:12
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    I typed in "symmetric matrix real eigenvalues proof" and found for example llink. Does that help? – hal4math Sep 26 '19 at 14:16
  • Looks like a nice proof for it. – hal4math Sep 26 '19 at 14:20
  • @hal4math In Lepidopterist's proof, do you know how it is that $\langle Av,Av\rangle = v^A^Av$? I'm presuming that the $*$ represents conjugate transpose, but I'm not familiar with the result. – The Pointer Sep 26 '19 at 16:01
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    There you use, that you can represent the scalar product with Matrix matrix multiplication. So in general (for the complex scalar product use conjugation), we have $\langle v,w\rangle = v^* w$. So with a linear map in there this become $\langle Av , w \rangle = (Av)^* w = v^A^ w$. – hal4math Sep 26 '19 at 16:10
  • @hal4math I don't think any of those are good answers to my question: 1. The top-rated answer by Lepidopterist is, according to his answer, not a proof, but rather an explanation of why that result does not prove the necessary result; 2. None of them provide any explanations of their proofs, and instead just copy-and-paste, which is not informative. – The Pointer Sep 26 '19 at 16:52
  • I disagree, the main idea to get to the desired result is stated clearly for me. Maybe articulate what is unclear in these proofs? – hal4math Sep 26 '19 at 17:05
  • @hal4math Are you saying that Lepidopterist's answer is a proof of the result? He states in his answer that it is not -- he was just showing that the stated method does not prove the result. Note that the author did not accept any of the answers, so it is likely that they also found the answers to be unsatisfactory. – The Pointer Sep 26 '19 at 17:07
  • Of course it couldn't work, what he is doing since then he would have proofed the statement above, for which I gave you a counterexample. So you have to put in one more thing you know, that is also mentioned (in your pdf). – hal4math Sep 26 '19 at 17:11
  • @hal4math I don't understand what your point is. As you say, your link does not prove my desired result -- at least, not in a clear and informative way. – The Pointer Sep 26 '19 at 17:22
  • But there is more than one answer, isn't it? Do you get the idea how to proof it yet? I can of course write something up, I just want to be sure I know which points I have to address in more detail. – hal4math Sep 26 '19 at 17:29
  • @hal4math I'm going through the pdf http://pi.math.cornell.edu/~jerison/math2940/real-eigenvalues.pdf now. It seems to be clearer and better explained than any of the answers to that question. – The Pointer Sep 26 '19 at 17:32
  • Still not sure what mean by a copy and paste solution. Sometimes all you need is a computation. Now, of course it is fine to not immediately see why that proofs it, but that is what you could ask about, isn't? But your last comments were just complaints about the quality of the answers. If you have another mathematical question I am more then happy to try to help. Otherwise I will just delete my comments, since they seem no to help you with your question. Sorry, about that. – hal4math Sep 26 '19 at 18:03

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