eigenvector and spectral norm $x^TAx=\lambda$ with $A$ is a semi-positive definite matrix, $x$ is a unit vector, and $\lambda\ne 0$ is the largest eigenvalue of $A$. Then $x$ is an eigenvector of $A$? I.e., $Ax=\lambda x$?
 A: The conjecture holds. Here is a proof.
Positive semi-definiteness is defined for real symmetric matrices $\bf A$ which have real non-negative eigenvalues $\lambda_i $ and unit eigenvectors $\vec u_i$ forming an orthonormal basis. Let w.l.o.g. the largest eigenvalue be $\lambda_1$ with eigenvector $\vec u_1$.
Let $\vec x$ be an arbitrary  unit vector.  Decompose it into the basis {$\vec u_i$}, giving $\vec x = \sum_{i=1}^n a_i \vec u_i$ with real coefficients $a_i$. Then by unity,
$$
1 = \vec x^T \vec x =  \sum_{i=1}^n a_i^2
$$
and by the equation in question, 
$$
\vec x^T {\bf A}  \vec x=  \sum_{i=1}^n \lambda _i a_i^2 = \lambda_1
$$
These two equations give
$\sum_{i=2}^n (\lambda_1 - \lambda _i) a_i^2 = 0$. 
Obviously $a_i^2 \geq 0$.
By positive semi-definiteness, $\lambda_i \geq 0$, and since $\lambda_1$ is the largest eigenvalue, $(\lambda_1 - \lambda _i)\geq 0$. Hence all terms in the sum must be zero which requires that for any base direction $\vec u_i$ which constructs $\vec x$ (hence $a_i^2 > 0$) we must have that   $(\lambda_1 - \lambda _i) = 0$. Now  this entails two cases:
case 1: no eigenvalue equals $\lambda_1$. Then the condition $(\lambda_1 - \lambda _i) = 0$ can never be met, hence the sum $\sum_{i=2}^n (\lambda_1 - \lambda _i) a_i^2$ can only be zero if all terms are zero, i.e. all terms have $a_i = 0$, hence $\vec x = \vec u_1$ which confirms the conjecture.
case 2:  there are eigenvalues $\lambda_2 ... \lambda_k$ which equal $\lambda_1$. Then $\lambda_1$ is degenerate and any unit vector $\vec x = \sum_{i=1}^k a_i \vec u_i$ satisfies the equation in question. However, due to the degeneracy, such a vector $\vec x$ is an eigenvector, hence the conjecture is again confirmed. 
$\qquad \Box $ 
A: It can be solved by Lagrange multipler for the following optimization
$
\text{maximize}\{ x^TAx\}\quad s.t. \quad x^Tx=1 .
$
To solve the above optimization, using Lagrange multipler, it needs
$
\text{maximize}\{ x^TAx+\lambda(x^Tx-1) \}.
$
Taking the derivative of the objective function w.r.t. $x$ and then let it be $0$ yields $Ax=\lambda x$.
Thus the optimal $x$ must be the eigenvector of A.
