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.
Best Answer
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.