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!

Answer 1

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.