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 WilliamsSep 18, 2019 at 10:38
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1$\begingroup$ @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. $\endgroup$– ysTuanSep 18, 2019 at 10:50
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$\begingroup$ 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? $\endgroup$– DasiSep 18, 2019 at 10:51
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2$\begingroup$ @Dasi also has a good way to prove it. It's also the slightly easier way to do it. $\endgroup$– Cameron WilliamsSep 18, 2019 at 10:53
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1$\begingroup$ @Dasi i cant even see why the subspace is invariant under the restriction of conjugate, so how it act on the subspace directly $\endgroup$– ysTuanSep 18, 2019 at 11:09
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2 Answers
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Let $T_0:=T_{|W}$. Now prove:
$$(T^*)_{|W}=T_0^*.$$
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Other way to see it as follows: Suppose $T$ is normal and $$TT^*(v)=T^*T(v)$$, this is true for all $v \in V$, Now, Claim: $T_{|W}:W \rightarrow W$ is normal. So, $$\langle T^*Tw,w' \rangle=\langle TT^*w,w' \rangle $$ for all $w,w' \in W$. So, $$T^*Tw=TT^*w$$ for all $w \in W$. Hence $T_{|W}$ is normal.
