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Let $V$ be a finite dimensional complex inner product space, and $T,N:V \to V$, two linear transformations. I need to prove that if $N$ is normal and $TN=NT$, so does $T^*N=NT^*$ and $TN^*=N^*T$.

We know that if $N$ is normal so it can be diagonalized, but I don't know what else to do in order to prove this.

Thank you very much.

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1 Answer 1

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This looks like Fuglede's theorem. You can find the proof on Wikipedia. (Also note that $T^*N=NT^*$ if and only if $TN^*=N^*T$, just by taking the adjoint on both sides.)

A quick sketch of the proof for the impatient: Observe that $N$ can be diagonalized, so there is an orthgonal decomposition of your vector space $V$ into subspaces $E_1, \ldots, E_k$ such that for any vector $v\in E_j$, $Nv = \lambda_j v$. Consider $w = Tv$. That $NT = TN$ implies that $Nw = \lambda_j w$. So $T$ maps the eigenspaces $E_j$ to themselves. From this you see that $T^*$ also maps eigenspaces $E_j$ to themselves. Since $N$ restricted to $E_j$ is a multiple of the identity, we can say that $T^*$ and $N$ commute on $E)j$, and hence they commute on the whole vector space.

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I don't know this theorem and I can't use it.. – user6163 Aug 16 '11 at 9:22
@frog: cool! you gave me the same link that Per does! Thanks mate! – user6163 Aug 16 '11 at 12:34

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