What is the Maximum Value of $x^T Ax$? Let $A$ be the matrix $\begin{bmatrix}3 &1 \\  1&2\end{bmatrix}$. What is the maximum value of $x^T Ax$ where the maximum is taken over all $x$ that are the unit eigenvectors of $A$?


*

*$5$

*$\frac{(5 + \sqrt{5})}{2}$

*$3$

*$\frac{(5 - \sqrt{5})}{2}$



Eigenvalues of $A$ are $\frac{(5 \pm \sqrt{5})}{2}$

My question is: what is $x^T Ax$ ? Can you explain little bit please?

 A: 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 $M$.
A: This is an immediate application of this result from Raleigh.. https://en.m.wikipedia.org/wiki/Rayleigh_quotient
A: 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 $\frac{5 + \sqrt{5}}{2}$.
A: $$x^TAx=\begin{bmatrix} x_1 & x_2 \end{bmatrix}\begin{bmatrix} 3 & 1 \\ 1 & 2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= 3x_1^2 + 2x_1x_2 + 2x_2^2$$
