TITLE: real symmetric matrix has real eigenvalues - elementary proof
QUESTION [43 upvotes]: Every real symmetric matrix has at least one real eigenvalue. Does anyone know how to prove this elementary, that is without the notion of complex numbers?

REPLY [3 votes]: Essentially the same as Marcos Cossarini's proof, but without Cauchy-Schwarz.
Let $A\in R^{n\times n}$ symmetric of rank $\ge 1$.
By compactness let $x\in R^n$ with $\|x\|=1$ be such that
$\|Ax \| = \max_{u\in R^n: \|u\|=1} \|Au\|$
and set $y = Ax/\lambda$ for $\lambda=\|Ax\|>0$. Then
\begin{align}
  \|A(x+ y) - \lambda(x+y)\|^2
&=\| Ay - \lambda x\|^2
\\&= \|A y\|^2 + \lambda^2 - 2 \lambda y^TA x
\\&= \|A y\|^2 + \lambda^2  - 2\lambda^2
\\&\le 0.
\end{align}
Either $x+y\ne 0$ is eigenvector for $\lambda$ or $x+Ax/\lambda=0$ and $x$ is is eigenvector for $-\lambda$.