equality for the spectral radius and operator norm in $B(X)$

Question

small question!

It can be proved that $\rho(A)=\|A\|$ if and only if $\|A^n\|=\|A\|^n$, now literature says that for normal operators or for $B(X)$ (the set of lineal and bounded operators in $X$ a Banach space) the equality holds.

It is clear that for $B(X)$, $\|T^n\| \leq \|T\|^n$, the thing is that for the converse it's where i'm having difficulties and for the normal operator can't figure it out either.

Thanks!

Answer 1

If you mean $\|A^n\| = \|A\|^n$ in $B(X)$, that's not true. For example, in a two-dimensional Banach space let $A$ be the operator corresponding to the matrix $\pmatrix{0 & 1\cr 0 & 0\cr}$. Then $A^2 = 0$ so $\|A^2\| < \|A\|^2$.

EDIT: Note also that $\rho(A) = 0 < \|A\|$.

For normal operators, use the Spectral Theorem.