# Finite rank operators

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## Main Question or Discussion Point

This is probably easy. It's really annoying that I don't see how to do this...

A finite rank operator (on a Hilbert space) is a bounded (linear) operator such that its range is a finite-dimensional subspace. I want to show that if T has finite rank, than so does T*.

I'm thinking that the theorem that says $$\operatorname{ker} T=(\operatorname{ran} T^*)^\perp$$ should be relevant, but I don't see how to use it. Feel free to use $\|T\|=\|T^*\|$ too if that helps.

Uh, maybe I should have posted this in homework. It's not homework, but a textbook-style question. Moderators, feel free to move it if you want to.

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Sigh... I couldn't even get to the bottom of the page before I ran into another thing that looks like it might be easy, but that I still can't figure out. Let E={ek} be an orthonormal basis. (In this problem we're dealing with a separable Hilbert space). Suppose that the set

$$\{\alpha_k\in\mathbb C|k\in\mathbb Z^+\}$$

is such that $\{|\alpha_k|\}$ is bounded from above. Define M to be the least upper bound. Now define

$$Ae_k=\alpha_k e_k$$

for k=1,2,... Show that that A extends by linearity to a bounded linear operator with norm M. This is how I interpret the problem: We need to show a) that there's a unique linear operator A' such that A' restricted to E is A, and b) that this A' is bounded. (Maybe this isn't what he meant?)

I will drop that prime right away, because I find it annoying. If we can find a bounded linear operator A that when restricted to E is the A we started with, then uniqueness follows from the formula

$$x=\sum_{k=1}^\infty \langle e_k,x\rangle e_k$$

because it means that when we express a vector as an infinite "linear combination" of basis vectors, the coefficients are unique. (Oh, yeah, my inner products are linear in the second variable).

$$Ax=A\left(\sum_{k=1}^\infty\langle e_k,x\rangle e_k\right)=\sum_{k=1}^\infty\langle e_k,x\rangle \alpha_k e_k$$

We can also take this as the definition of A, because when we restrict it to E, it has the same effect as the A we started with. The A defined this way is bounded: By Pythagoras, continuity of the norm, and $|\alpha_k|\leq M$,

$$\|Ax\|^2=\sum_{k=1}^\infty \|\langle e_k,x\rangle\alpha_k e_k\|^2=\sum_{k=1}^\infty |\langle e_k,x\rangle|^2|\alpha_k|^2 \leq M^2\sum_{k=1}^\infty |\langle e_k,x\rangle|^2 = M^2\|x\|^2$$

These calculations prove that there exists a unique bounded linear operator that when restricted to E is the A we started with. Hm, this is clearer to me now than when I started typing this post. I guess what remains is just to show that there's no unbounded operator that restricted to E is A. So how do we do that?

I can prove that if T is an operator that when restricted to E is A, the set $\|Tx\|/\|x\|$ with x a (finite) linear combination of basis vectors is bounded from above. So I just need to do the same for infinite "linear combinations". (I think most authors define a linear combination to have a finite number of terms, so I don't really know what to call the ones with infinitely many). This is the proof for finite linear combinations:

$$\|Tx\|\leq\sum_{k=1}^n\|\langle e_k,x\rangle Te_k\| =\sum_{k=1}^n|\langle e_k,x\rangle||\alpha_k| \leq \left(\sum_{k=1}^n|\langle e_k,x\rangle|^2\right)^{\frac{1}{2}}\left(\sum_{k=1}^n|\alpha_k|^2\right)^{\frac{1}{2}}\leq\sqrt{n}M\|x\|$$

The inequality in the middle is the Cauchy-Schwartz inequality for the dot product on ℝn.

Edit: The inequality $\|Ax\|^2 \leq M^2\|x\|^2$ implies $\|A\|\leq M$. I don't see how to prove that $\|A\|=M$.

This book is really frustrating. I feel like I have to spend hours, sometimes days, on every little comment he makes without supporting arguments, and I can still easily prove half the things he does prove.

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For your first question: If T: A -> B is a finite rank map of topological vector spaces, then it is a composition of A -> C -> B, where C = T(A) is finite-dimensional. Taking duals, you get that T* is a composition B* -> C* -> A*, where C* is finite-dimensional, so T* must have finite-dimensional image.

For your last part, $\|A\| \ge M$, because $\|A\| \ge \sup \|A e_k\| = \sup |\alpha_k| = M$.

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By the way, there are many unbounded linear operators T that agree with A on the ek, since you can extend ek to an algebraic (i.e. Hamel) basis and define T to be whatever you want on the new basis elements. It's just that there's a unique continuous one, since the (algebraic) span of the ek is dense.

(Whoops, meant this to be an edit of the last post, but I accidentally made a new reply apparently.)

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Thank you very much. This has been really frustrating, and your comments (plus half an hour of thinking) resolved every one of the issues I asked about in posts #1 and #2.

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I had another look at this problem, and the issues discussed above are all clear to me now, but there's another one that I overlooked before. At one point, I defined an operator A by

$$Ax=\sum_{k=1}^\infty\langle e_k,x\rangle \alpha_k e_k$$

but this only makes sense if the series is convergent. How do I show that it is?

The way the problem is stated strongly suggests that I don't have to use any assumptions about the $\alpha_k$, except that M is the least upper bound of the set of $|\alpha_k|$. The obvious thing to try is to show that the series converges absolutely, but

$$\|\langle e_k,x\rangle \alpha_k e_k\| =|\langle e_k,x\rangle| |\alpha_k|\leq M |\langle e_k,x\rangle|$$

and I doubt that

$$\sum_{k=1}^\infty|\langle e_k,x\rangle|$$

is convergent.

The series sure doesn't necessarily converge absolutely. However, you can show that it converges by showing that its sequence of partial sums is Cauchy. Specifically,
$$\left\lVert \sum_{k=m}^{n} \langle e_k, x \rangle \alpha_k e_k \right\rVert^2 \le M \sum_{k=m}^{n} \lvert \langle e_k, x \rangle \rvert^2 \to 0$$
as $m, n \to \infty$, because
$$\lVert x \rVert^2 = \sum_{k=1}^\infty \lvert \langle e_k, x \rangle \rvert^2$$
converges.

By the way, your problem has a more general version which is true in any normed vector space: Let V be a normed vector space with basis {ek}; by that I mean that they are linearly independent and finite linear combinations of the ek are dense in V. Given a bounded set {αk} of numbers, there is a unique bounded (= continuous) linear operator A on V such that Aek = αkek.

This can be proved as follows: Let U be the linear subspace of V consisting of finite linear combinations of the ek; thus, {ek} is an algebraic basis for U (that is, any element of U can be uniquely written as a finite linear combination of the ek). It is clear that there is a unique linear map A0: U -> V such that A0ek = αkek. Note that it is bounded (because of the triangle inequality) and therefore continuous. Since U is dense in V, by general topology A0 extends uniquely to a continuous function A: V -> V, which is also linear (since the vector space operations are continuous), and clearly satisfies the requirements.

If sup |αk| = M, then ||A|| = M also: By the triangle inequality, ||A0|| = M; then ||A|| = M since the norm ||.||: V -> C is a continuous function.

(Small issue: As far as I can tell, not every normed vector space (or even Banach space) has a basis as I have described. Hilbert spaces always have a basis, of course.)

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Wow, that was fast. Thank you again. I don't know how I could fail to see that trick. It's been a part of many other proofs involving series that I've done lately.

The more general theorem is interesting, but I don't see how to fill in the details you left out. For example, assuming that the ek are normalized, we have

$$\|A_0 x\|=\|A_0\Big(\sum_{k=1}^n x_k e_k\Big)\| \leq\sum_{k=1}^n |x_k||a_k|$$

I don't see how to get this calculation to end with ≤ something $\|x\|$.

The small issue at the end looks like it might be pretty big. This is related to an issue that I was finally able to figure out just a couple of days ago: If B={v1,v2,...} is a linearly independent set in an infinite-dimensional Banach space, then

$$\sum_{k=1}^\infty 2^{-k} \frac{v_k}{\|v_k\|}$$

converges (because it converges absolutely). If we call the limit v, then B⋃{v} is linearly independent. The conclusion is that no countable linearly independent set in an infinite-dimensional Banach space can be maximal. So any Hamel basis of an infinite-dimensional Banach space must be uncountable.

I guess we can define your V to be the closed linear span of a specific countable linearly independent set. Not sure if that solves all the problems.

Whoa, you're absolutely right, and I'm completely wrong. The proof fails at the triangle inequality part, and it can't be fixed.

Indeed, here's a counterexample, in a Hilbert space, even: In $\ell^2$, let the basis {ek} be defined by e1 = (1, 0, 0, ...), and ek = (1, 0, ..., 1/k, ...) (that is, a 1 in the first component, 1/k in the kth component, and 0 elsewhere). Then let α1 = 1, and αk = 0 for k > 1. If Aek = αkek, then for any integer n > 1 consider x = (0, ..., 1, ...), where 1 is in the nth component and 0 everywhere else. Then x = nen - ne1 and Ax = -ne1, so that while ||x|| = 1, we have ||Ax|| = n. Thus, A is unbounded, since n can be arbitrarily large.

I wonder if this will teach me not to post things at 3am without thinking through them properly. :)

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Indeed, here's a counterexample, in a Hilbert space, even: In $\ell^2$, let the basis {ek} be defined by e1 = (1, 0, 0, ...), and ek = (1, 0, ..., 1/k, ...) (that is, a 1 in the first component, 1/k in the kth component, and 0 elsewhere). Then let α1 = 1, and αk = 0 for k > 1. If Aek = αkek, then for any integer n > 1 consider x = (0, ..., 1, ...), where 1 is in the nth component and 0 everywhere else. Then x = nen - ne1 and Ax = -ne1, so that while ||x|| = 1, we have ||Ax|| = n. Thus, A is unbounded, since n can be arbitrarily large.
This is convincing except for one detail. The "basis" is neither an orthonormal basis, nor a Hamel basis. It's not an orthonormal set, because <ei,ej>≠0 for all i,j, and it's not a Hamel basis because all Hamel bases are uncountable.

I wonder if this will teach me not to post things at 3am without thinking through them properly. :)
Been there, done that. Landau
I haven't really read the discussion, so I apologize in advance if I am repeating someone or stating something obvious.
I'm thinking that the theorem that says $$\operatorname{ker} T=(T^*(H))^\perp$$ should be relevant, but I don't see how to use it.
Yes, this is useful! Let's call our Hilbert space H. Then $$\ker T^*=T(H)^\perp$$, so we have the orthogonal decomposition

$$H=T(H)\oplus \ker T^*.$$

Hence T*(H)=T*T(H). As T(H) has finite dimension, T*T(H) does too.

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Yes, this is useful! Let's call our Hilbert space H. Then $$\ker T^*=T(H)^\perp$$, so we have the orthogonal decomposition

$$H=T(H)\oplus \ker T^*.$$

Hence T*(H)=T*T(H). As T(H) has finite dimension, T*T(H) does too.
Thank you. I hadn't thought of that. Adriank showed me a different way to solve the problem, but I like this solution too.

Landau
You're welcome :) By the way, shouldn't this topic in Functional Analysis be in 'Calculus and Analysis'?

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You're welcome :) By the way, shouldn't this topic in Functional Analysis be in 'Calculus and Analysis'?
I've been trying to figure out where to put posts about functional analysis for years. I even wrote a couple of posts about it (this is the most recent one), but they got ignored, so I decided to put all of my FA posts in topology and geometry (since most proofs involve methods and results from topology, usually in the most difficult part of the proof).

The calculus & analysis forum seems to be a place where people discuss derivatives and integrals, and maybe series of real numbers, but not infinite-dimensional Hilbert spaces and operator algebras.

I noticed you've been away for a while, but wrote a bunch of posts today. Done with the thesis already, or just taking a break from it? Landau
I noticed you've been away for a while, but wrote a bunch of posts today. Done with the thesis already, or just taking a break from it? 