A compact operator $T$ such that $T^{2}=T$ is a finite rank operator

Question

Let $X$ be a normed linear space. Let $T: X \to X$ be a compact linear operator and $T^{2}=T$. Show that $T$ is a finite rank operator, that is, $Range(T)$ is finite dimensional in $X$.

I have no idea how to show that this operator is of finite rank. Any hint or solution will really be appreciated. Thanks in advance!

The restriction of a compact operator to a subspace is compact. Which subspace can you restrict $T$ to to exploit that fact? — Daniel Fischer, May 09 '17 at 08:26
@DanielFischer Although I have got an answer, I want to reflect on your hint!! Are you suggesting to restrict the operator to range T. But then it's clear how is it going to help to show that the operator is of finite rank — Riju, May 09 '17 at 08:52
Right. $T$ being idempotent means $T\lvert_{\operatorname{im} T} = \operatorname{id}_{\operatorname{im} T}$. And when is the identity of a space a compact operator? — Daniel Fischer, May 09 '17 at 09:06

score 0 · Accepted Answer · answered May 09 '17 at 08:27

0

Let $(y_n)$ be a bounded sequence in the range of $T$. From $T^2=T$ we get:

$Ty_n=y_n$ for all $n$. Since $T$ is compact, $(Ty_n)$ and hence $(y_n)$, contains a convergent subsequence.

Thus the range of $T$ has the Bolzano- Weierstrass -Property. Therefore the range has finite dimension.

answered May 09 '17 at 08:27

Fred

77,394

Thanks Fred.. Yeah this is a great solution. It also helps in a way that I can now try to show some operator is finite rank by showing that the range has Bolzano-Weierstrass property..I was not getting that point before.. – Riju May 09 '17 at 08:45

A compact operator $T$ such that $T^{2}=T$ is a finite rank operator

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