Eigenvalues of symmetric matrices are real without (!) complex numbers

Solution 1:

Let $\rho=\max_{\|x\|=1}\|Ax\|$ and $u=\arg\max_{\|x\|=1}\|Ax\|$. Then $Au=\rho v$ for some unit vector $v$ and $\|Av\|\le\rho$. But we also have $\langle Av,u\rangle=\langle v,A^Tu\rangle=\langle v,Au\rangle=\rho$. Hence $Av=\rho u$. Therefore, either $(\rho,u)$ (when $u=v$) or $(-\rho,u-v)$ (when $u\ne v$) is an eigenpair of $A$. Since the orthogonal complement of this eigenspace is invariant under $A$ (because $A$ is symmetric), we have reduced the dimension of the problem by $1$. Proceed recursively, $A$ has $n$ real eigenvalues and an orthogonal eigenbasis.