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can anyone explain why it's obvious that a normal operator restricted to a invariant subspace is still normal?

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  • $\begingroup$ Define the restricted operator $M = TP$ where $P$ is the projection into the invariant subspace. Can you show that $M^*M = MM^*$? $\endgroup$ – Cameron Williams Sep 18 at 10:38
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    $\begingroup$ @cameron wlilliams sorry i cant. if the subspace is also T* invariant, i think the conclusion is obvious. but that's a corollary of what Artin prepare to prove. $\endgroup$ – ysTuan Sep 18 at 10:50
  • $\begingroup$ Just use prop. 8.6.3 (b). If T is normal the equality holds for all vectors in your space, does it also hold for all vectors in your invariant subspace? $\endgroup$ – Dasi Sep 18 at 10:51
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    $\begingroup$ @Dasi also has a good way to prove it. It's also the slightly easier way to do it. $\endgroup$ – Cameron Williams Sep 18 at 10:53
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    $\begingroup$ @Dasi i cant even see why the subspace is invariant under the restriction of conjugate, so how it act on the subspace directly $\endgroup$ – ysTuan Sep 18 at 11:09
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Let $T_0:=T_{|W}$. Now prove:

$$(T^*)_{|W}=T_0^*.$$

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