1
$\begingroup$

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

enter image description here

$\endgroup$
7
  • $\begingroup$ Define the restricted operator $M = TP$ where $P$ is the projection into the invariant subspace. Can you show that $M^*M = MM^*$? $\endgroup$ Sep 18, 2019 at 10:38
  • 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. $\endgroup$
    – ysTuan
    Sep 18, 2019 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, 2019 at 10:51
  • 2
    $\begingroup$ @Dasi also has a good way to prove it. It's also the slightly easier way to do it. $\endgroup$ Sep 18, 2019 at 10:53
  • 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 $\endgroup$
    – ysTuan
    Sep 18, 2019 at 11:09

2 Answers 2

1
$\begingroup$

Let $T_0:=T_{|W}$. Now prove:

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .