Why $\max \left\{ {{x^T}Ax:x \in {R^n},{x^T}x = 1} \right\}$ is the largest real eigenvalue of A? Let $A \in {M_n}(R)$ and A is symmetric.Why $\max \left\{ {{x^T}Ax:x \in {R^n},{x^T}x = 1} \right\}$ is the largest real eigenvalue of A?

So, here's an intuitive and somewhat geometric explaination of what is happening. 
$1:$ For $\lambda$ to be an eigenvalue, we need to be able to find a vector $v$ such that $Av=\lambda v$ 
$2:$ We observe that $v$ is an eigenvector iff $v\over{||v||}$ is an eigenvector.  So, we reduce the search space to the unit sphere.ie., vectors with norm $=1$. So, now we have to find $\lambda$ and $v$ such that $Av=\lambda v$ and $||v||=1$
$3:$ Since we are talking about symmetric matrices existence of eigenvalues is guaranteed. Our claim now is that the largest eigenvalue is given by $max_{||x||=1} x^TAx$.  
$4:$Observe here that $y=Ax$ is a vector. It need not lie on the unit sphere. But we are projecting $y$ onto a vector $x$ in the unit sphere by taking $x^Ty$. $5: $ When is this maximized ?  By the definition of standard dot product, if $\theta$ is the angle between $x$ and $y$,  $x^Ty=||x||||y||cos(\theta)=||y||cos(\theta)$. Here, for a given $x$, this would have been maximized when $cos(\theta)=1$. Since the existence of eigenvalues is guaranteed, this definitely occurs at the eigenvectors.The maximum among these is the eigenvector corresponding to the largest eigenvalue. As for vectors which are not eigenvectors,we can directly see that $x^Ty<||y||$. Now, because $A$ is diagonalizable and the eigenvectors $v_1,...,v_n$ form a basis, we can write $x=\alpha_1v_1 + ... + \alpha_nv_n$.Here, we denote the largest eigenvalue by $\lambda_1$ and the corresponding eigenvector to be $v_1$ . So, $Ax=\alpha_1\lambda_1v_1 + ... + \alpha_n\lambda_nv_n$. $||y||=||Ax||\leq||\alpha_1\lambda_1v_1 + ... + \alpha_n\lambda_1v_n||=\lambda_1||x||=\lambda_1$. So, those which are not eigenvectors donot exceed this value.
