It is clear that if $ T: X \rightarrow X $ is a bijective compact operator, where $ X $ is a Banach space, then $ \dim(\text{Range}(T)) = \dim(X) $, which implies that $ \dim(X) $ must be $ < \infty $.
How do I prove the converse: If $ \dim(X) < \infty $, then there exists a bijective compact operator $ T: X \rightarrow X $?
Thank you!
On
Theorem, it is not true that $ \dim(\text{Range}(T)) = \dim(X) $ implies that $ \dim(X) < \infty $. Another way of reasoning is as follows. Let $ T: X \rightarrow X $ be a bijective compact operator. Then by the Bounded Inverse Theorem, $ T^{-1} $ exists and is continuous. Hence, $ T $ is also a homeomorphism. Let $ B_{X} $ be the closed unit ball of $ X $. Then $ T[B_{X}] $ is closed and has compact closure, which implies that $ T[B_{X}] $ is compact (a closed subset of a compact space is also compact). However, as $ T $ is a homeomorphism, $ B_{X} $ must then be compact. Hence, as a consequence of Riesz's Lemma, $ \dim(X) < \infty $.
Of course, Romeo's answer is very slick in the sense that reasoning with the ideal $ \mathcal{K}(X) $ saves us a lot of work.
I suppose the theorem you want to prove is this one:
One way (if) is clear: indeed, if $\dim X<+\infty$ then every operator $T \colon X \to X$ is compact (since its range is finite dimensional: this is a well-known sufficient condition for compactness). For example, take identity of $X$: it is bijective (obviously!) and compact.
Now, the other way (only if): suppose $T\colon X \to X$ is bijective and compact. There exists $T^{-1}$ and, moreover, it is continuous: so $TT^{-1}=\text{id}_X$ is compact, since $\mathcal K(X)$ is a closed ideal in $\mathcal L(X)$. In particular, the closure of the unit ball of $X$ is compact, hence the space $X$ is finite dimensional.
Hope this helps.