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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.

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  • @uniquesolution yes it is. Here $|\cdot|2$ is the Euclidean operator norm and $\sigma{max}$ is the greatest singular value. – Ben Grossmann Sep 14 '15 at 17:56
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    reread the question that you've linked: the norm of a symmetric matrix is equal to the magnitude of its largest eigenvalue. – Ben Grossmann Sep 14 '15 at 18:02

1 Answers1

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$D$ 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 $\| A\| \geq \|D\|$.

Ben Grossmann
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