Let $$A = \begin{bmatrix}3 &1 \\ 1&2\end{bmatrix}$$ What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?
The eigenvalues of $A$ are $\frac{5 \pm \sqrt{5}}{2}$. What is $x^T Ax$? Can you explain a little bit, please?
On
$$x^TAx=\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$
On
This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient
On
OK, Let us say that the eigenvalue of $A$ is $\lambda$ and then call the corresponding eigenvector $x_{\lambda}$ and hence
$$Ax_{\lambda}=\lambda x_{\lambda}$$
and observe that
$$x_\lambda ^TA{x_\lambda } = \lambda x_\lambda ^T{x_\lambda } = \lambda (1) = \lambda $$
so you should find the maximum eigenvalue of $A$. So the maximum value of the quadratic form will be $\frac{5 + \sqrt{5}}{2}$.
If $x\neq 0$ is an eigenvector of $A$ with unit length and associated eigenvalue $\lambda$, then $x^TAx=x^T\lambda x=\lambda x^Tx = \lambda$. So the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$ is simply the value of the largest eigenvalue of $A$.
As pointed out by @Ant, the quantity $x^TAx$ arise in the Rayleigh quotient associated to $A$ (note that $A$ is symmetric). In fact, one can show that the eigenvectors (resp. eigenvalues) of a symmetric matrix $M$ are the critical points (resp. values) of the function $$R(x)=\frac{x^TMx}{x^Tx}$$ which is called the Rayleigh quotient associated to $M$.