# why the restriction of a normal operator is normal?

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

• Define the restricted operator $M = TP$ where $P$ is the projection into the invariant subspace. Can you show that $M^*M = MM^*$? Sep 18, 2019 at 10:38
• @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. Sep 18, 2019 at 10:50
• 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?
– Dasi
Sep 18, 2019 at 10:51
• @Dasi also has a good way to prove it. It's also the slightly easier way to do it. Sep 18, 2019 at 10:53
• @Dasi i cant even see why the subspace is invariant under the restriction of conjugate, so how it act on the subspace directly Sep 18, 2019 at 11:09

Let $$T_0:=T_{|W}$$. Now prove:
$$(T^*)_{|W}=T_0^*.$$
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.