I have some difficulties proving this question: Let $A \in\mathbb{R}^{n\times n}$ be symmetric, and consider critical points of $f(x) = Ax\cdot x$ subject to the constraint $\|x\| = 1.$ Prove that critical points are precisely the unit eigenvectors of $A$.
I have a hint saying that: You may find it easier to write the constraint as $\|x\|^2 = 1$.
I tried to use Lagrange multipliers to prove this:
$\nabla f(x)= \lambda \nabla g(x)$, where $g(x)=\|x\|^2$. And I get $\nabla g(x)=2x$. But then I don't know how to continue my proof. Can anyone help me? Thanks!