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Since you seem really interested in the inequality $\rho(A)\le\|A\|$ ($\rho$ the spectral radius), here is a simple and elegant proof. In the 2nd edition of my book Matrices (Springer Verlag GTM216), …

Here is a non trivial constraint : ${\rm Hom}(V,W)$ contains (is spanned by) rank one morphisms $$v\mapsto\ell(v)w,\qquad\ell\in V',w\in W.$$ If a given norm over ${\rm Hom}(V,W)$ is induced, then $$\ …

If $A$ is $n\times n$, then $$\frac1{\sqrt n}\|A\|_F\le\|A\|_1\le\sqrt n\,\|A\|_F,\qquad \|A\|_2\le\|A\|_F\le\sqrt n\,\|A\|_2.$$ More generally, if $A$ is $n\times m$, then $$\frac1{\sqrt m}\|A\|_F\le …

A good answer is given by R. Bhatia: Pinching, trimming, truncating, and averaging of matrices. Amer. Math. Monthly 107 (2000), no. 7, 602–608. If you consider the operator $T_r$ that retains the dia …

This is not the right way to think about operator norms. Instead, you can say that if $T:L^2\rightarrow L^2$ and $T:L^\infty\rightarrow L^1$ (as you consider) are bounded, then $T:L^p\rightarrow L^{p' …