Rayleigh quotient for real matrices, critical points

I have found an exercice in a calculus book, which I have problems to solve.

$$\text{Let}\, R:\mathbb{R}^n-{0}\to \mathbb{R},\quad R(x)=\frac{x^tAx}{x^tx} = \frac{\langle x,Ax\rangle}{\langle x,x\rangle},$$ where $\langle \cdot,\cdot \rangle$ is the euclidean inner product, $A\in\mathbb{R}^{n\times n}$.

$1)$ $R$ has a minimum in $\mathbb{R}-{0}$.

$2)$ Every critical point of $R$ is an Eigenvector of $\frac{1}{2}(A^t+A)$ corresponding to an Eigenvalue of $A$. In particular, every symmetric real matrix has real Eigenvalues.

I have solved $1)$ (by seeing that it is sufficient to study $R|S^n$ and $S^n$ is compact, so $R$ has a minimum and maximum which it attains.) For $2)$ I can't find the desired result. I know that $\frac{1}{2}(A^t+A)$ is symmetric, and that $\langle x,Ax \rangle=\langle x,A^tx \rangle$.

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Did you try to compute $$\left. \frac{d}{d\epsilon} \right|_{\epsilon =0} R(x+\epsilon v)$$ at a critical point $x$ of $R$? –  Siminore Sep 16 '12 at 11:53
Ah yes this is the idea. Now I don't really know how to conclude from this. –  Lucien Sep 16 '12 at 12:19
Of course the derivative that I wrote is zero, since $x$ is a critical point of $R$. Write down the derivative, and you will find a relation that must hold for every $v$. This will imply that $x$ is an eigenvalue. Anyway, this is an easy case of the Lagrange multiplier rule. –  Siminore Sep 16 '12 at 12:46
Thank you for your hint, I have solved it. I will post the answer later. –  Lucien Sep 16 '12 at 15:57

Let $x\in\mathbb{R}$ be an critical point. Then $\left. \frac{d}{d\epsilon} \right|_{\epsilon =0} R(x+\epsilon v)=0$ $(*)$.
Computing $R(x+\epsilon v)$ explicitly, deriving with respect to $\epsilon$, and putting $\epsilon=0$ we find, using $(*)$ for all $v\in\mathbb{R}$ we find:
$\langle v, \frac{A+A^t}{2}\; x\rangle\langle x,x\rangle=\langle x, Ax\rangle\langle x,v\rangle$.
For all $v\bot\, x$ we find that $\langle v, \frac{A+A^t}{2}\; x\rangle=0$ which shows that $\frac{A+A^t}{2}\;x=\lambda x$ for some $\lambda\in \mathbb{R}$. Moreover $\lambda$ is an Eigenvalue of $A$: with the choice $x=v$ we get: $\lambda \langle x,x\rangle=\langle x,Ax\rangle \iff \langle x,(A-\lambda Id)x\rangle=0$, and so $Ax=\lambda x$ $(x\neq 0)$.