If $T$ normal invertible operator, then $\sigma(ST)=\sigma(TS).$

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

It is well known that if $T,S\in\mathcal{B}(F)$, then $$\{0\}\cup\sigma(ST)=\{0\}\cup\sigma(TS).$$

If $T$ is normal and invertible, why $$\sigma(ST)=\sigma(TS).$$

I see in a lecture note the following proof:

Since $T$ is normal and invertible, then $$\sigma(TS) = \sigma(T^{-1} TS T) = \sigma(ST).$$ I don't understand, why $$\sigma(TS) = \sigma(T^{-1} TS T).$$ holds.

Answer 1

Actually, if $T$ is invertible, then we have $$ \sigma(ST)=\sigma(TS). $$ Since we already know that $ \sigma(ST)\setminus\{0\}=\sigma(TS)\setminus\{0\}, $ it suffices to show that $$ 0\in \sigma(ST) \ \ \ \Longleftrightarrow \ \ \ 0\in \sigma(TS). $$ To show this, let us prove that $ ST$ is invertible if and only if $TS$ is invertible.
($\Longrightarrow$) Suppose that $ST$ is invertible and let $R=(ST)^{-1}$ so that $STR=RST = I$. Since $T$ is assumed to be invertible, we also find that $RST=TRS = I$. From this, we know that $STR=TRS$, i.e. $S^{-1} = TR$. This implies $TS$ is invertible with $(TS)^{-1}=S^{-1}T^{-1}$.
($\Longleftarrow$) To prove converse, assume $TS$ is invertible with $R=(TS)^{-1}$. Then it holds $TSR=RTS=I$ and $TSR=SRT=I$. Hence $S^{-1} =RT$ and $(ST)^{-1} = T^{-1}S^{-1}$.

As a result, we find that $\sigma(ST) = \sigma(TS)$ as wanted.

Answer 2

In fact, for any invertible $A \in \mathcal B(H)$ and any $B \in \mathcal B(H)$, we have $\sigma(B) = \sigma(A^{-1}BA)$. This is because of the following : $$A^{-1}(B - \lambda I)A = A^{-1}BA - \lambda I$$

Using the above, it is easy to see that for any $\lambda \in \mathbb C$ ,we have $(A^{-1}BA - \lambda I)$ is invertible if and only if $(B - \lambda I)$ is. For example, if $C = (B-\lambda I)^{-1}$ then (one side)$C(B-\lambda I) = I$ so we have : $$ (A^{-1}CA)(A^{-1}BA - \lambda I) = (A^{-1})C(B - \lambda I)A = A^{-1}A = I $$

Similarly if $(B-\lambda I)C = I$, and similarly if $A^{-1}BA - \lambda I$ is invertible, use $(B-\lambda I) = A(A^{-1}BA - \lambda I)A^{-1}$ to get that $(B-\lambda I)$ is invertible.

Hence, the spectra of $B$ and $A^{-1}BA$ coincide.

So, in fact one does not need to normality of $A$ over here at all.