Does the spectral norm of a square matrix equal its largest eigenvalue in absolute value? I have one simple question. 

Given the spectral norm $\left \| \cdot \right \| _2$ of a matrix $A$,
  which is equal to the square root of the largest eigenvalue of $A^{^*}A$
$$\left \| A \right \| _2=\sqrt{\lambda_{\text{max}}(A^{^*}A)}=\sigma_{\text{max}}(A)$$
for a square matrix $A$, is $\left \| A \right \| _2$ equal to the largest eigenvalue of $A$ in absolute value?

I know it is true for a symmetric matrix but I don't know for a random square one.

For any square $A$, $\rho(A)\leq\|A\|_2$, where $\rho(A)$ is the spectral radius of $A$, with the equality (but not necessarily) if $A$ is normal. Besides the general inequality, $\rho(A)$ and $\|A\|_2$ can be completely unrelated. Consider, e.g., 
$$
A_\alpha:=\pmatrix{0&\alpha\\0&0}
$$
with $\rho(A_\alpha)=0$ but $\|A_\alpha\|_2=|\alpha|$. All the eigenvalues are zero but the 2-norm can be an arbitrary non-negative number (depending on $\alpha$).
