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Let $H$ be a separable Hilbert space and $A$ is an invertible bounded operator on $H$. Can we approximate $A$ with an invertible operator $B$ such that $sp(B)$ is a countable set?


If the answer is yes, this would give's us an alternative proof of connected ness of $GL(H)$. This alternative proof is identical to a short and interesting proof of connectedness of $GL(n,\mathbb{C})$, in page 19 of "Introduction to the Baums Connes conjecture" by Alain Valette

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Should the inverse also be bounded? – Joonas Ilmavirta Aug 19 '14 at 9:16
@JoonasIlmavirta yes all things are bounded operators? – Ali Taghavi Aug 19 '14 at 9:38
Ok. The inverse of a bounded invertible operator need not be bounded, so I wanted to check. – Joonas Ilmavirta Aug 19 '14 at 9:49
Hmm... In a complex Hilbert space you can write $A=B+iC$ with $B$ and $C$ selfadjoint, and selfadjoint operators are easy to approximate since they diagonalize. I don't know what happens for invertibility and the spectrum of the sum though. – Joonas Ilmavirta Aug 19 '14 at 9:55
@JoonasIlmavirta I don't understand your comment about inverses, unless you are dealing with densely-defined operators. In the case of bounded operators on a Banach space we have the Banach isomorphism theorem, unless I have misread or misunderstood something – Yemon Choi Aug 24 '14 at 19:25
up vote 7 down vote accepted

I think the answer is "no"; here is a sketch of an argument though I will have to go back and check the details. First, there is nothing special about invertibility here; if every invertible operator is approximable by operators with countable spectrum, then every operator is, just by translation. I claim the unilateral shift $S$ should not be approximable, because it is a Fredholm operator of nonzero index. The set of Fredholm operators is open, and the index is a continuous function on this set, so any sufficiently good approximant would have to be Fredholm of index -1. But I think that if a Fredholm operator has countable spectrum, then the index must be 0 (reason: if 0 is not isolated in the spectrum, then the range won't be closed, and if 0 is isolated (or the operator is invertible) then the index is clearly 0).

EDIT: Here is a proof that a Fredholm operator $T$ with countable spectrum must have index 0. We may clearly suppose $T$ is not invertible, so 0 is in the spectrum. The first claim is that 0 must be isolated--if not, then since the spectrum is a countable, compact set, 0 must be the limit of a sequence of isolated points in the spectrum. But each isolated point must be an eigenvalue, which means that $T$ would then have a sequence of eigenvalues tending to 0, and would not have closed range and not be Fredholm. So, 0 is isolated. But then $T+\epsilon I$ is invertible for all small $\epsilon$, and by the continuity of the index we have that $ind(T)=0$.

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thank you very much for your interesting answer. I have a question on your statement "Isolated points are eigenvalue" In this statement, is not necessary that we assume T is normal? – Ali Taghavi Aug 19 '14 at 16:09
I thought this should just follow from the Fredholm assumption and the Riesz functional calculus, but perhaps it is more subtle than that. Let me think about it some more. – Mike Jury Aug 19 '14 at 16:26
OK, I no longer trust my claim about eigenvalues; however it is true that if 0 is in the boundary of the spectrum of a Fredholm operator (which is of course the case here), then it must be isolated and the operator must have index 0. This follows from a theorem of Putnam; see Conway's "A Course in Functional Analysis", Theorems XI.6.8 and XI.6.9. – Mike Jury Aug 19 '14 at 16:37
Thank you very much for your beautiful answer and your reference to Putnam theorem. However, regarding the eigenvalue problem, i think that $I+V$ is a counter example where $V$ is the Voltra integral operator. It has a single spectrum $\{1\}$ which is not eigenvalue. – Ali Taghavi Aug 19 '14 at 19:29

The question has already been answered above (by Mike Jury). Here is another way of arguing: Any operator with countable spectrum is in the closure of the invertible operators. So are their limits. But again, the unilateral shift provides an example of an operator not in the closure of the invertibles. For if $\|S-X\|<1$ then $\|I-S^*X\|<1$, and so $S^*X$ is invertible. $X$ cannot be invertible, since otherwise $S^*$ would also be invertible. So no translate $S+\lambda I$ can be approximated by operators with countable spectrum.

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Nice! Indeed, this shows that if $\|S - X\| = a < 1$ then the spectrum of $X$ contains ${\rm ball}(0;1-a)$. – Nik Weaver Aug 20 '14 at 4:30
Very nice! More elegant than my answer, clearly. – Mike Jury Aug 20 '14 at 7:48
@LeonelRobert Thanks for your answer. Assume that $S^{*}X$ is invertible. How you conclude that $S^{*}$ is not right invertible/(How you obtained a contradiction?0 – Ali Taghavi Aug 20 '14 at 12:50
@AliTaghavi. The argument shows that $X$ within a distance less than 1 from $S$ cannot be invertible. For if it were, that together with $S^*X$ being invertible would imply that $S^*$ is invertible (which is not). – Leonel Robert Aug 22 '14 at 15:31

It is well-known that you can approximate a selfadjoint operator $A$ with an operator $B=A+V$ such that $B$ has pure point spectrum. The selfadjoint operator $V$ can have arbitrarily small Hilbert-Schmidt norm.

This is a theorem of Weyl-von Neumann and can be found e.g. in Kato.

But I don't think this works for the nonselfadjoint case.

EDIT: Christian Remling mentioned and is right, of course, that the point spectrum of $B$ has to be dense in the continuous spectrum of $A$ since the essential spectrum is conserved. So the pure point spectrum of $B$ don't have to be countable.

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The spectrum of $B$ is the closure of the eigenvalues, so this won't give countable spectrum. Quite on the contrary, you know that the essential spectrum is preserved. – Christian Remling Aug 19 '14 at 15:10
On the other hand, it is easy to give an approximation directly if $A$ is self-adjoint: just approximate $\int_I t\, dE(t)$ by a suitable multiple of the identity on $E(I)$ for small intervals $I$, and put these pieces back together to obtain $B$. (This $B$ will have finite spectrum.) – Christian Remling Aug 19 '14 at 15:21

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