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How do we prove that the algebraic multiplicity is equal to the geometric multiplicity in a symmetric matrix?

Rama
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  • I have answered the same question at here without using spectral theorem. https://math.stackexchange.com/a/4402827/128942 – bfhaha Mar 16 '22 at 07:44

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Generally, this is true for all diagonalizable matrices. Symmetric matrices being diagonalizable by the spectral theorem, the result follows for them.

Ulysse Mizrahi
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  • Yes, you can cite the spectral theorem, and the fact that diagonalizability is equivalent to geometric and algebraic multiplicity coinciding, but that doesn't really answer "why" in a substantive way (although it does say what to google for). A good answer should give some intuition about why the spectral theorem is true. – Aaron Oct 26 '16 at 10:02
  • There really is no "intuition" for the spectral theorem. Really there barely is any intuitive interpretation for symmetric matrices (or self-adjoint operators) so finding intuition for the spectral theorem is pretty much hopeless.The question asks to "prove" something, the answer proves it. By stackexchange standards, it is a good answer. – Ulysse Mizrahi Oct 26 '16 at 10:12
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    it only proves it in the weakest sense of the word. It mostly just shifts the burden by using basic facts to rephrase the question. It would be like asking how to show that a number is a product of prime factors and just citing the fundamental theorem of arithmetic. You have given an answer, but devoid of content of illumination. And I disagree that there is no intuition to be found, every proof I've seen of the spectral theorem relies on having some useful way to think about what symmetric matrices do/represent, or how they act. E.g., bilinear forms. – Aaron Oct 26 '16 at 10:35
  • Then by all means post a better answer – Ulysse Mizrahi Oct 26 '16 at 11:55