# A complete eigenvector basis for the restricted operator

Let $X$ be a (not necessarily bounded) selfadjoint linear operator on a Hilbert space $H$ and let $M$ be a closed subspace such that $X(M) \subset M$.

Suppose that $X$ admits an orthonormal basis of eigenvectors. Does it follow that $X|_M$ admits an orthonormal basis of eigenvectors too?

I think that the answer is 'yes' if $X$ commutes with the orthoprojection of range $M$:

$$PX \subset XP,$$

but I don't know what happens without this hypothesis. What do you think?

Thank you.

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It seems to me that the hypothesis that $PX\subseteq XP$ should always hold, but I don't have proof. What do you think? – Jonas Meyer Dec 14 '11 at 0:41
@Jonas: Motivated by your comment I started a search. Looks like you're right, at least for bounded $X$. Indeed, as I see in Kubrulsky's book link, Proposition 5.74, if $X$ is bounded then $M$ reduces $X$ (which is the same as $PX=XP$, if I'm not mistaken) iff $X(M)\subset M$ and $X^\star(M)\subset M$. For bounded selfadjoint operators, then, reducing subspaces and invariant subspaces are the same thing. – Giuseppe Negro Dec 14 '11 at 13:35
Errata corrige: In previous comment I mistakenly wrote "Kubrulsky" instead of the correct "Kubrusly". – Giuseppe Negro Dec 14 '11 at 14:05
I'm a little confused about the unbounded case. Then do you require that $M\subseteq D(X)$ (so that $M$ being closed seems quite a strong condition) or merely that $X( M\cap D(X)) \subseteq M$?? – Matthew Daws Dec 14 '11 at 20:37
Giuseppe: Yes, I was motivated by knowing it is true in the bounded case. In any case, note that $PXP=XP$. If $X$ is bounded, taking adjoints of the last equation shows that $PX=PXP=XP$. It is just that with unbounded operators, I am not sure how much of that carries through. If $A$ and $B$ are densely defined operators, then (I think) $B^*A^*\subseteq (AB)^*$, and I was thinking that this could be used to show that $PX\subseteq XP$ still holds. I don't plan to try to check the details soon. – Jonas Meyer Dec 15 '11 at 5:03

Here are the details of Jonas's suggestion. Let $X$ be self-adjoint with dense domain $D(X)$, and let $M$ be a closed subspace with $M\subseteq D(X)$ and $X(M)\subseteq M$. Let $P$ be the orthogonal projection onto $M$. Then $D(PX) = D(X)$ while $D(XP)=H$ as $P$ has range $M\subseteq D(X)$. For $\xi\in H$, let $\xi_0=P(\xi), \xi_1=\xi-P(\xi)$, and let $\eta\in D(X)$. Then $$(PX\eta|\xi) = (PX\eta|\xi_0) + (PX\eta|\xi_1) = (PX\eta|\xi_0) = (X\eta|P\xi_0) = (X\eta|\xi_0)$$ as $PX\eta\in M$ and $\xi_1\in M^\perp$. As $\xi_0\in M\subseteq D(X)$ and $X$ is self-adjoint, $$(X\eta|\xi_0) = (\eta|X\xi_0) = (\eta|PX\xi_0) = (P\eta|X\xi_0) = (XP\eta|\xi_0),$$ as $X\xi_0 \in X(M) \subseteq M$. Similarly, $(XP\eta|\xi_1) = 0$ as $XP\eta \in X(M) \subseteq M$ and $\xi_1\in M^\perp$. As $\xi$ was arbitrary, this shows that $XP\eta=PX\eta$. So we've shown that $$D(PX) \subseteq D(XP), PX\eta=XP\eta \ (\eta\in D(X)) \implies PX \subseteq XP.$$
As the OP claimed, this is enough. If $e$ is an eigenvector of $X$ then $PXe = P\lambda e = \lambda Pe$, and by the above, this also is equal to $XPe$. So $Pe\in M$ is an eigenvector for $X$ (unless it's zero!) As the original collection of eigenvectors has desne linear span, so does the projection onto $M$.