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Let $T$ and $S$ be two normal operators in a infinite dimensional inner-product complex vector space. If $ST=TS$, I want to show that $TS$ is normal.

For the finite-dimensional case, it went down to showing $T^*=f(T)$ for some polynomial $f$. But I have no clue on what to do on the infinite dimensional case. This is an exercise of Hoffman's Linear Algebra.

Jonas Gomes
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2 Answers2

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You can use Fuglede's theorem: http://en.wikipedia.org/wiki/Fuglede%27s_theorem which ensures, in your case that $ST^*=T^*S$ and $S^*T=ST^*$.

In view of the facts that $(TS)^*=S^*T^*$, $SS^*=S^*S$ and $TT^*=T^*T$, the conclusion follows by easy computations: $$ TS(TS)^*=T(SS^*)T^*=(TS^*)ST^*=S^*T(ST^*)=S^*(TT^*)S=(S^*T^*)TS=(TS)^*TS $$

The property is not longer true in the unbounded case. You can follow the paper "M. H. Mortad, On the closedness, the self-adjointness and the normality of the product of two unbounded operators, Demonstratio Math., 45 (2012), 161-167" for an example. It is also stated there that the normality of $TS$ is valid when, in addition, $S$ (or $T$) is unitary.

aly
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If bounded operators $T$ and $S$ on a Hilbert space commute and $S$ is normal, then $T$ and $S^*$ commute (Fuglede's theorem).

Robert Israel
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  • What if they are not bounded? – Jonas Gomes Feb 10 '15 at 19:33
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    Commutation of unbounded (densely-defined) operators can be tricky, because the domains of $TS$ and $ST$ might have non-dense (even trivial) intersection. If you assume the spectral projections of $S$ and $T$ commute, then Fuglede's theorem should go through. – Robert Israel Feb 10 '15 at 19:45