My problem is originated from the boxed inequality below. In the lemma, after Schur decomposition $D$ contains eigenvalues of $A$ along its diagonal.
I knew that $\|A \|_{2} =\|D\|_{2}$ as shown Norm of a symmetric matrix equals spectral radius.
And, $\|A \|_{2} = \sigma_{max}$ which is the maximum singular value of $A$.
So, $\|D \|_{2} =\|A\|_{2}=\sigma_{max} < \sqrt{\lambda_{max}} < \lambda_{max}$
I am not sure my argument is correct or not because if $0<\lambda_{max}<1$, e.g. $\lambda_{max} = 0.8$, then $\sigma_{max} = \sqrt{\lambda_{max}}=\sqrt(0.8) \approx 0.8944 > \lambda_{max}$.
So, my question is:
I do not understand why $\|D\|_2$ can be replaced by largest eigenvalue of $A$ (or absolute of the largest element of $D$ itself because of the result of Schur decomposition.)
Thanks in advance.
Best Answer
$D$ is a diagonal matrix whose entries are the eigenvalues of of $A$. The norm of any diagonal matrix (or for that matter, any normal matrix) is the maximum of the absolute values of its eigenvalues. So, $\|D\|_2 = |\lambda|$.
Note, however, that it is not necessarily true that $\|A\| = \|D\|$. In general, we have $\| A\| \geq \|D\|$.