$2$-norm of a normal matrix

Question

I just proved that if a real $n \times n$ matrix $A$ is normal then its $2$-norm is equal to the maximum eigenvalue of $A$, by using $A= U\Lambda U^*$. Is this still true (that the $2$-norm of an $n \times n$ matrix is equal to the maximum eigenvalue of $A$) without the assumption that $A$ is normal? If so, wow we can prove it or disprove it?

Answer 1

No: if $A$ is not normal, then its $2$-norm is not necessarily equal to the maximum absolute value of its eigenvalues. As an example, consider $$ A = \pmatrix{1 & 1\\0 & 0}. $$ Its eigenvalues are $0$ and $1$, so its spectral radius is $\rho(A) = 1$. On the other hand, we have $\|A\|_2 = \sqrt{2}$.

On the other hand, it is possible for a matrix $A$ to fail to be normal but also have $\|A\|_2 = \rho(A)$. For example, consider $$ A = \pmatrix{2&0&0\\0&0&1\\0&0&0}. $$ In this case, $AA^T \neq A^TA$, but we have $\|A\|_2 = \rho(A) = 2$.

Answer 2

"Second norm" is not standard terminology. I suspect you mean the norm induced by the Euclidean norm. Also, it's not the maximum eigenvalue, it's the maximum absolute value of eigenvalues. And for the answer, you might try

$$ \pmatrix{0 & 1\cr 0 & 0\cr} $$