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My question is regarding this post : compact and normal operator is diagonalizable, there is a detail in the argument that I don't understand why is $u(K^{\perp}) \subseteq K^{\perp}$ ?

user715747
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  • I think I found the answer : $K$ is reduced by $u$ since $K$ is invariant by $u$ and $u^*$ – user715747 Apr 26 '23 at 07:15
  • @AnneBauval $K$ is reduced by $u$ means that both $K$ and $K^{\perp}$ are invariant by $u$, this would follow from the fact that $K$ is invariant by $u$ as well as $u^*$ – user715747 Apr 26 '23 at 12:02
  • @AnneBauval To see why take any $x \in K^{\perp}$ and any $y \in K$ then $ (u(x),y)=(x,u^(y))=0$ since $K$ is invariant by $u^$ – user715747 Apr 26 '23 at 12:07
  • Perfect. This is why I told you that the key point was "$K$ is invariant by $u^∗$" ($K$ invariant by $u$ is useless here). But now, are you able to prove this crucial point, or is it precisely for this that you need help? (This was my question.) – Anne Bauval Apr 26 '23 at 13:40
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    @AnneBauval thanks, I only needed help for this, the rest of the proof is straightforward. – user715747 Apr 26 '23 at 16:44
  • Since you only need help for this, the following post answers your question: Eigenvectors for normal operators and their adjoints. – Anne Bauval Apr 26 '23 at 16:51
  • @user715747 could you plz explain why K is invariant by $u^*$ ? I'm actually searching for a proof that any invariant subspace of a compact normal operator $T$ actually reduce $T$. This post is the closest one I found. – Westlifer Jan 01 '24 at 17:36
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    @Westlifer Here, $K$ is not any invariant subspace. It is the closure of a subspace spanned by eigenvectors for the compact normal operator $u$. In the link in the closure banner, it is proved that if $u(e)=\lambda e$ then $u^(e)=\bar\lambda e$. This proves that $K$ is invariant by $u^$. As for your problem, if $K$ is invariant by some normal operator $T$ then the restriction of $T$ to $K$ is still normal hence (by the spectral theorem for such operators) you can apply the previous argument. – Anne Bauval Jan 01 '24 at 19:16
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    @AnneBauval I've noticed that and figured it out. Many thanks! – Westlifer Jan 02 '24 at 16:30

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