McAfee Virusscan Enterprise 8.7i (full) (Multilanguage) ((TOP))

McAfee Virusscan Enterprise 8.7i (full) (Multilanguage) ((TOP))



 
 
 
 
 
 
 

McAfee Virusscan Enterprise 8.7i (full) (Multilanguage)

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Can the operator norm of a matrix be defined using the supremum of all possible singular values?

I was studying the matrix formulation of the Frobenius norm
$$||A||_F = \sqrt{trace(A^TA)} = \sqrt{trace(AA^T)} = \sqrt{det(A^TA)} = \sup_{\lambda > 0} \lambda. $$
I try to use the inequality $\sqrt{a+b} \leq \sqrt{a} + \sqrt{b}$ to get
$$\|A\|_F \leq \|A\|_{2, \infty} \leq \|A\|_F + \|A\|_1, $$
since $\|A\|_{2, \infty}$ is the operator norm of $A$ and $\|A\|_1$ is the $\ell_1$ norm of the vectorization of $A$, but I wasn’t able to show the second inequality.
Thanks for your helps

A:
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