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Apr 5 at 9:18 comment added Geoff Robinson Notice that a nilpotent operator $A$ has $r(A) = 0,$ whereas $\|A\|$ can be arbitrarily large, so I wonder what sort of answer you would regard as useful in your context?
Apr 4 at 14:48 comment added Gerald Edgar I think you need to consider related operators $B = S^{-1}AS$, where $S$ is invertible. The spectral radius for $B$ is the same as for $A$, but the norm may be different. I believe (I read somewhere long ago?) the infimum of all these norms is the spectral radius.
Apr 4 at 14:10 history asked stoic-santiago CC BY-SA 4.0