Unit vector that maximizes or minimizes

Question

I know by the Taylor expansion $f(x,y)$ that in order for the origin to be a minimum point, $f_{xx}$ and $f_{yy}$ have to be both positive. Which I know how to prove. I also know other methods like implicit differentiation and Lagrange multipliers.

Now what I having trouble proving is,

Showing that a unit vector $(x,y)$ that maximizes or minimizes $(x \ y )M(x \ y )^T$ is an eigenvector of $M$, which is a real, symmetric matrix.

How do I prove that?

Answer 1

In general, if $M$ is a symmetric $n\times n$ matrix, and $$ f(x)=\frac{x^TMx}{x^Tx}, $$ where $x=(x_1,\ldots,x_n)=x_1e_1+\cdots+x_ne_n$ (with $e_1,\ldots,e_n$ the standard basis of $\mathbb R^n$), then in a local extremum of $f$ we have $$ 0=\frac{\partial f(x)}{\partial x_j}=\frac{2e_j^TMx}{x^Tx}-\frac{2x_j(x^TMx)}{(x^Tx)^2}, $$ and hence $$ x_j(x^TMx)=e_j^TMx(x^Tx) $$ or $$ Mx=\frac{x^TMx}{x^Tx}x, $$ and hence $x$ is an eigenvector, with eigenvalue $\dfrac{x^TMx}{x^Tx}$.