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Let $Q$ be a symmetric $n$ by $n$ square matrix, there exists an orthogonal matrix $F$ such that $$F^TQF=\operatorname{diag}(\lambda_1,\ldots,\lambda_n),$$ with $\lambda_1,\ldots,\lambda_n$ being its eigenvalues.

I know little about linear algebra, but I hope somebody could help me prove it because I'm studying differential geometry

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    $\begingroup$ this is in every linear algebra/matrix theory book on earth en.wikipedia.org/wiki/Orthogonal_diagonalization $\endgroup$ – Will Jagy Sep 20 '14 at 21:43
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    $\begingroup$ I'm quite surprised you're studying differential geometry without prior knowledge of linear algebra. $\endgroup$ – egreg Sep 20 '14 at 22:26
  • $\begingroup$ Not know little.it's a little bit actually $\endgroup$ – pxc3110 Sep 22 '14 at 1:23
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This is known as Sylvester's Law of Inertia (more or less). This Wikipedia article refers to Sylvester's original proof, so you could read that. There are, of course, more modern treatments around. :)

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