An idempotent operator is compact if and only if it is of finite rank

Question

Would you help me to solve this problem. Show that an idempotent operator on hilbert space is compact if and only if it has finite rank.

Answer 1

Okay. If you require an answer that uses the sequential definition of a compact operator, here it goes. :)

A finite-rank operator is clearly compact, so one direction is established.

Now, an operator $ T: \mathcal{H} \rightarrow \mathcal{H} $ is compact if and only if it maps every bounded sequence in $ \mathcal{H} $ to a sequence in $ \mathcal{H} $ that contains a convergent subsequence. This is the basic sequential characterization of compact operators.

Let $ T: \mathcal{H} \rightarrow \mathcal{H} $ be an idempotent and compact operator. We want to show that it is a finite-rank operator. By way of contradiction, assume that it is not finite-rank. Let $ R(T) $ denote the infinite-dimensional range space of $ T $. We claim that $ R(T) $ is a closed subspace of $ \mathcal{H} $. To prove this claim, let $ (T(\mathbf{x}_{n}))_{n \in \mathbb{N}} $ be a sequence in $ R(T) $ that converges to some $ \mathbf{y} \in \mathcal{H} $. Then $$ T(\mathbf{y}) = T \left( \lim_{n \rightarrow \infty} T(\mathbf{x}_{n}) \right) = \lim_{n \rightarrow \infty} T(T(\mathbf{x}_{n})) = \lim_{n \rightarrow \infty} T(\mathbf{x}_{n}) = \mathbf{y}. $$ Hence, $ \mathbf{y} \in R(T) $, and as $ (\mathbf{x}_{n})_{n \in \mathbb{N}} $ is arbitrary, we are done.

As $ R(T) $ is now seen to be a closed (hence complete) subspace of $ \mathcal{H} $, it follows that $ R(T) $ is a Hilbert subspace of $ \mathcal{H} $. By the infinite-dimensionality of $ R(T) $, pick an orthonormal sequence of vectors $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $ in $ R(T) $ (we can do so by extracting an orthonormal sequence from some infinite orthonormal basis for $ R(T) $). By construction, $ \mathbf{e}_{n} \in R(T) $ for each $ n \in \mathbb{N} $, so we can associate a $ \mathbf{v}_{n} \in \mathcal{H} $ such that $ T(\mathbf{v}_{n}) = \mathbf{e}_{n} $. Then $$ \forall n \in \mathbb{N}: \quad T(\mathbf{e}_{n}) = T(T(\mathbf{v}_{n})) = T(\mathbf{v}_{n}) = \mathbf{e}_{n}. $$ Hence, $ T $ maps the sequence $ (\mathbf{e}_{n})_{n \in \mathbb{N}} $ identically to itself. This sequence is clearly bounded. However, $ (T(\mathbf{e}_{n}))_{n \in \mathbb{N}} = (\mathbf{e}_{n})_{n \in \mathbb{N}} $ does not contain a convergent subsequence! (Simply observe that $ \| \mathbf{e}_{m} - \mathbf{e}_{n} \|_{\mathcal{H}} = \sqrt{2} $ for distinct $ m,n \in \mathbb{N} $.) It follows from the sequential definition of compactness that $ T $ is not compact, which contradicts the fact that we started our argument with $ T $ being compact.

Conclusion $ T $ must be a finite-rank operator.

I noticed that people have been asking questions about the exercises from the chapter on compact operators in Conway's A Course in Functional Analysis. For the benefit of the OP and also of other people who are interested in those exercises, the following link may be useful: A Question on Compact Operators. It contains the full solution to another exercise from the same chapter.

Answer 2

If $P$ is finite-rank, then it is compact.

If $P$ is compact, then its spectrum consists only of eigenvalues, with $0$ as the only possible accumulation point. The equality $P^2=P$ implies that the only eigenvalues are $0$ and $1$. Now take any $x$ in the range of $P$; then $$ Px=x, $$ so $x$ is an eigenvector with eigenvalue $1$. The multiplicity of $1$ has to be finite, as otherwise $P$ would not be compact. Thus the range of $P$ is finite-dimensional.

Answer 3

Let $ T \in B(\mathcal{H}) $ be idempotent.

If $ T $ is of finite rank, then it is clearly compact, so one direction is proven.

Conversely, suppose that $ T $ is compact. By way of contradiction, assume that $ T $ has infinite rank. Then $ T $ acts as the identity operator on its infinite-dimensional range space $ R(T) $. Letting $ \mathbb{B}_{R(T)} $ denote the open unit ball of $ R(T) $, we see that $ T $ maps $ \mathbb{B}_{R(T)} $ identically to itself. However, as $ \mathbb{B}_{R(T)} $ is not relatively compact in $ \mathcal{H} $ (i.e., $ {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) $ is not compact; see below), this contradicts the compactness of $ T $. Therefore, $ T $ must have finite rank.

Notes

As mentioned, $ {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) $ is not compact. If it were compact, then as $ R(T) $ is a closed subspace of $ \mathcal{H} $ (this follows from the idempotence of $ T $) and $ {\text{cl}_{R(T)}}(\mathbb{B}_{R(T)}) = {\text{cl}_{\mathcal{H}}}(\mathbb{B}_{R(T)}) \cap R(T) $, it would follow that $ {\text{cl}_{R(T)}}(\mathbb{B}_{R(T)}) $ is compact. However, this contradicts Riesz's Lemma, which implies that the closed unit ball in any infinite-dimensional normed space is never compact.

Answer 4

Let $P$ be idempotent on a Hilbert Space $H$. Put $R(x) = x - P(x)$; then $$R(R(x)) = R(x - P(x)) = R(x) - R(P(x)) = x - P(x) - (P(x) - P(P(x)) = x - P(x) - P(x) + P(P(x)) = x - P(x).$$

From this we see that $H = \ker(P) + \ker(I - P)).$ We have represented $H$ as the direct sum of two closed subspaces. If $P$ is compact, its range is a locally-compact Banach Space, so the range is finite dimensional.

Answer 5

If you are working on with the Conway's A Course in Functional Analysis, the proof gets even more easier.

Your question is in fact in this book, Exercise II.4.4. The right-to-left is immediate consequence of the Exercise I.4.3. For the left-to-right proof, by II.3.2 Proposition, $ranE=ker(I-E)=ker(E-I)$. Hence by II.4.13 Proposition, the proof is complete.