TITLE: Eigenvectors of real symmetric matrices are orthogonal
QUESTION [230 upvotes]: Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.

REPLY [2 votes]: Note a real symmetric matrix is a linear operator on Euclidean space with respect standard basis (orthonormal). So the fact that it equals to its conjugate transpose implies it is self-adjoint. 
For two distinct eigenvalues $\lambda_1, \lambda_2$ and corresponding eigenvectors $v_2, v_2$, $$(\lambda_1-\lambda_2)<v_1, v_2>=<Tv_1,v_2>-<v_1,\bar{\lambda_2} v_2>=<Tv_1,v_2>-<v_1,T^* v_2>=0$$ where the 2nd last equality follows from properties of self-adjoint (thus normal) linear operator (Lemma below). 
Lemma: Assume $T$ is normal. If $(\lambda, v)$ is eigenvalue and eigenvector of $T$, $(\bar{\lambda}, v)$ is eigenvalue and eigenvector of the adjoint $T^*$. (pf.) Trivial from definition of normality.