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I'm at a loss of how to show this. I know that this is implied by the Spectral Theorem but not sure exactly how to show this in a simple, straightforward proof.

I've tried laying down what I know about symmetric matrices and diagonalizable ones but I'm not too sure what to use when. My guess is I'd have to reason with regards to the eigenvalues but again I'm not sure.

A matrix is symmetric if $A = A^T$. A matrix $A$ is symmetric if it can be expressed in the form $A = QDQ^{T}$. A square matrix $A$ is called diagonalizable if $\exists$ invertible P such that $P^{-1}AP$ is a diagonal matrix.

Would really appreciate any help.

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  • $\begingroup$ Of course being diagonalisable doesn't imply being symmetric. $\endgroup$ – Vim Sep 17 '17 at 1:19
  • $\begingroup$ My mistake, omitted that out. $\endgroup$ – SS' Sep 17 '17 at 1:25
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    $\begingroup$ Those links don't quite have any clear or concise proofs. $\endgroup$ – SS' Sep 17 '17 at 1:37
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    $\begingroup$ @ChristianBlatter perhaps OP wants proof that is so well structured and simplistic that they can imprint it on their mind and build it from scratch whenever requested. I think it's a meaningful question, to ask about simple, sound alternative proofs. (For example, this answer, if fully fledged, will be much more intuitive than most existing proofs.) $\endgroup$ – Vim Sep 20 '17 at 4:45

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