compact operators and finite dimentional spaces

Question

Let $Q_n$ a finite dimentional space. Since any finite rank operator is compact, it's true that any linear operator $K:Q_n\to Q_n$ is compact 'cause $\dim(R(K))<\infty$?

Answer 1

A linear function from a finite dimensional Hausdorff topological vector space to itself is continuous, bounded, and compact. This is because a finite dimensional Hausdorff topological vector is isomorphic to $\mathbb{R}^n$ and a linear function from $\mathbb{R}^n$ to $\mathbb{R}^n$ is continuous. Now closed bounded subsets of $\mathbb{R}^n$ are compact and visa versa. So it takes bounded sets to sets whose closure is compact.