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

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

• this is in every linear algebra/matrix theory book on earth en.wikipedia.org/wiki/Orthogonal_diagonalization – Will Jagy Sep 20 '14 at 21:43
• I'm quite surprised you're studying differential geometry without prior knowledge of linear algebra. – egreg Sep 20 '14 at 22:26
• Not know little.it's a little bit actually – pxc3110 Sep 22 '14 at 1:23