# Finite Rank Operator is Compact

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## Theorem

Let $H, K$ be Hilbert spaces.

Let $T \in B_{00} \left({H, K}\right)$ be a bounded finite rank operator.

Then $T \in B_0 \left({H, K}\right)$, i.e., $T$ is compact.

## Proof

Since $T$ is bounded, it maps bounded sets to bounded sets. Because its range is finite dimensional, the closure of an image of a bounded set is compact. Therefore, $T$ is a compact linear operator.

## Sources

- 1990: John B. Conway:
*A Course in Functional Analysis*... (previous) ... (next) $II.4 \text{ Exercise } 2$