Is $r(T)=r(S^{-1}TS)\;$?

Question

Let $E$ a complex Hilbert space and $L(E)$ the algebra of all bounded operators on $E$. An operator $S\in L(E)$ is called positive if $\langle Sx, x\rangle\geq0$ for all $x\in E$, and we write $S\geq 0$.

Recall that the spectral radius of an operator $T\in L(E)$ is given by $$r(T)=\displaystyle\lim_{n\to\infty}\|T^n\|^{\frac{1}{n}}.$$

Let $S\geq 0$ be an invertible operator and $T\in L(E)$. Is $$r(T)=r(S^{-1}TS)\;??$$

Answer 1

This is true, irrespective of $S$ being positive. You have $$ S^{-1}TS-\lambda I=S^{-1}(T-\lambda I)S, $$ which shows that $\sigma(S^{-1}TS)=\sigma(T).$ In particular, $$r(S^{-1}TS)=r(T).$$