Suppose $\bf{A}$ is a symmetric positive-definite matrix and now we want to maximize function $f(\bf{x})=\bf{x}^\rm{T}\bf{A}\bf{x}$ with restriction $\bf{x}^\rm{T}\bf{x}=\rm{1}$. Using Lagrange multiplier we have $L(\bf{x})=\bf{x}^\rm{T}\bf{A}\bf{x}-\lambda(\bf{x}^\rm{T}\bf{x}-\rm{1})$ and by taking derivative of both sides we get $L'(\bf{x})=(\bf{A}-\lambda\bf{I})\bf{x}=\rm{0}$, whose solutions are eigenvectors of $\bf{A}$.
My question is how to prove the solutions (the eigenvectors) are indeed maxima of $f(\bf{x})$ rather than minima. I am not sure but I think this is related to Hessian matrix and I found here Hessian matrix of a quadratic form that the Hessian matrix of a quadratic form seems to be $\bf{A}+\bf{A}^\rm{T}$, but I don't know how to use it with a restriction. I post this question to ask for help with a complete proof. Thank you.
P.S. The background of this question is a the widely-used statistical model Principal Component Analysis. A related question is Why the principal components correspond to the eigenvalues? if you are interested.