Is it possible for $A\in \mathbb{C}^{n\times n}$, that $$\frac{|Ax|}{|x|}>|\lambda_{max}|$$ where $\lambda_{max}$ is the biggest eigenvalue of A?
I know this can not happen, if there is a basis of eigenvectors, but in general?
EDIT: So from the comments so far, I understood that the way to go may be to rewrite the LHS to $$\sqrt{\frac{x^* A^*Ax}{x^*x}}$$ and then use, that $A^*A$ is diagonalizable. But how then do I see, that $\sqrt{\rho(A^*A)}\leq\lambda_{max}$ (with $\rho$ being the spectral radius) without assuming anything about $A$?
This can't happen. Note that $A^*A$ is diagonalizable- and hence there is a basis of eigenvectors, and the modulus of the maximum eigenvalue of $A$ is the (positive) square root of maximum eigenvalue of $A^*A$