Eigenvectors of real symmetric matrices are orthogonal 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.
For any real matrix $A$ and any vectors $\mathbf{x}$ and $\mathbf{y}$, we have
$$\langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle.$$
Now assume that $A$ is symmetric, and $\mathbf{x}$ and $\mathbf{y}$ are eigenvectors of $A$ corresponding to distinct eigenvalues $\lambda$ and $\mu$. Then
$$\lambda\langle\mathbf{x},\mathbf{y}\rangle = \langle\lambda\mathbf{x},\mathbf{y}\rangle = \langle A\mathbf{x},\mathbf{y}\rangle = \langle\mathbf{x},A^T\mathbf{y}\rangle = \langle\mathbf{x},A\mathbf{y}\rangle = \langle\mathbf{x},\mu\mathbf{y}\rangle = \mu\langle\mathbf{x},\mathbf{y}\rangle.$$
Therefore, $(\lambda-\mu)\langle\mathbf{x},\mathbf{y}\rangle = 0$. Since $\lambda-\mu\neq 0$, then $\langle\mathbf{x},\mathbf{y}\rangle = 0$, i.e., $\mathbf{x}\perp\mathbf{y}$.

Now find an orthonormal basis for each eigenspace; since the eigenspaces are mutually orthogonal, these vectors together give an orthonormal subset of $\mathbb{R}^n$. Finally, since symmetric matrices are diagonalizable, this set will be a basis (just count dimensions). The result you want now follows.
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 $v_2, v_2$, $$(\lambda_1-\lambda_2)&lt;v_1, v_2&gt;=&lt;Tv_1,v_2&gt;-&lt;v_1,\bar{\lambda_2} v_2&gt;=&lt;Tv_1,v_2&gt;-&lt;v_1,T^* v_2&gt;=0$$ 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.