Is standard eigenvalue optimization problem convex For any arbitrary  symmetric matrix A , is the standard eigenvalue problem convex 
$ \lambda_{max}(A)= \max_{\|x\| \leq1} x^{T}Ax$ 
 A: Thanks to Michael for catching my previous errors.
You need to be careful with terminology here.
In the following, I am assuming that $A$ is symmetric.
The function $\phi(A) = \max_{\|x\|\le 1} \langle x, Ax \rangle$ is always convex
because it is the $\sup$ of a collection of convex (in fact, linear) functions
$A \mapsto \langle x, Ax \rangle$.
However, the function $\phi$ is not necessarily equal to the maximum eigenvalue function. In general, $\phi(A) = \max(0, \lambda_\max(A))$.
If you restrict the $x$s, then we have
$\lambda_\max(A) = \max_{\|x\| = 1} \langle x, Ax \rangle$ (which is a 
convex function of $A$ for the same reasons as above).
However, the optimisation problem $\max_{\|x\|\le 1} \langle x, Ax \rangle =
-\min_{\|x\|\le 1} \langle x, (-A)x \rangle$ is not necessarily a convex program.
 If you restrict $A$ to the 
cone of negative semidefinite matrices, then the optimisation problem is convex,
but the solution is trivially $0$.
For example, if $A=+1$, then the problem is $\max \{ x^2 | |x| \le 1 \}$
which is not a convex program.
