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Suppose that $H$ is a finite-dimensional Hilbert space over $\mathbb{C}$ and $A$ is a normal linear operator on $H$. Let $V$ be a closed subspace of $H$ that is invariant under $A$. Show that there exists an orthonormal basis for $V$ consisting of simultaneous eigenvectors for $A$ and $A^*$.

HINT: Show that since A is normal, then there exists a simultaneous eigenvector of A and $A^*$. Furthermore, show that if V is invariant under A, then $V^\perp$ is invariant under $A^*$.

I have proven both of the hints. From here, it seems like we should take the set of simultaneous eigenvectors, which we know is not empty by the hint, and show that it spans all of V. However, I don’t know how the second part of the hint would come into play here.

Any help would be greatly appreciated!

slowspider
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  • If you want us to abide by the hint, it would be good to state the hint, verbatim, as it was stated to you. It's also possible that the hint is not just a stepping stone along the way to the solution, but that it's encouraging you towards a particular method/technique, so you should also share your proof so far. – Theo Bendit Jun 16 '22 at 00:22
  • @TheoBendit thank you for the comment. I updated the question. – slowspider Jun 16 '22 at 00:45
  • The problem I've having is that the hint is subtly deficient. If I were writing a hint like this for the given problem, I'd suggest finding an eigenvector in $V$ common to $A$ and $A^*$, not just any old common eigenvector. With this extra information, an induction argument presents itself, but without it, I wouldn't know how to proceed purely using the information in the hint. Giving an alternate solution would not answer the question. Maybe your solution finds this eigenvector in $V$? I don't know, because I can't see your work. It's difficult to answer this question as is. – Theo Bendit Jun 16 '22 at 01:18
  • @TheoBendit In my solution to the hint, I showed that if $\lambda$ is an eigenvalue of A and $E_\lambda$ is it’s corresponding eigenspace, then A and $A^*$ have a shared eigenvector in $E_\lambda$. – slowspider Jun 16 '22 at 01:24
  • Unfortunately, that piece of tantalising information is also just shy of being sufficient to push my solution along. Basically, the approach I see is, if you can find a common e-vector $w \in V$, then $W = \operatorname{span}{w}$ is closed and invariant w.r.t. $A$ and $A^$, so $W^\perp$ is invariant to $A^$ (and $A$, by switching $A$ and $A^*$). This makes $W^\perp \cap V$ invariant as well, with one fewer dimension. Stripping off vectors in this way inductively builds an orthogonal basis of eigenvectors, which can be normalised. – Theo Bendit Jun 16 '22 at 01:32
  • Without having $w \in V$, I can't proceed like this. We could find a common orthonormal basis of eigenvectors for all of $H$, but we can't really tailor it for $V$. Your result is oh-so-close to what we need, if we mix in the fact that $A$ must have an eigenvector in any non-trivial invariant $V$. However, there's nothing to say that the eigenspace $E_\lambda$ is entirely contained in $V$, or that the common eigenvector must lie in $V$. So, it's close, but not quite enough. We'll have to see if someone else has a better idea. – Theo Bendit Jun 16 '22 at 01:35
  • @TheoBendit I think I was able to show that a shared eigenvector is in $V$. That being said, thank you!! – slowspider Jun 16 '22 at 02:20
  • I’m glad that my angst could help in the end. :) – Theo Bendit Jun 16 '22 at 03:07

1 Answers1

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The two hints put together can solve the problem for $V=H$. Indeed, just do induction on $\dim H$: For the inductive step, find a common eigenvector $v$ of $A, A^*$, and since $\mathbb Cv$ is both $A$ and $A^*$-invariant, thus by the second hint, we have $(\mathbb Cv)^{\perp}$ must also be $A$ and $A^*$-invariant, for which the induction hypothesis can be applied.

The question now can be settled, if we can show that any $A$-invariant subspace $V$ is also $A^*$-invariant. By the second hint, it suffices to show $V^{\perp}$ is also $A$-invariant. This is not super easy and deserves a hint by itself. A proof can be found on Wikipedia with trace and Frobenius norm or in a computational manner Invariant subspace of $T$ (normal) is also an invariant subspace of $T^\ast$.. (I often find it hard to reproduce the proof. The major difficulty is perhaps this is a special fact about only finite dimensional spaces.)

Just a user
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