Why is every linear operator between finite-dimensional normed spaces compact?

Conditions

• The operator is linear.
• The domain and codomain are finite-dimensional normed spaces.

Reasoning, step by step

1. A linear operator between finite-dimensional normed spaces is automatically continuous.
2. For linear operators, continuity is equivalent to boundedness.
3. Because the operator is bounded, the image of the unit ball under the operator is a bounded set.
4. Taking the closure of this bounded image produces a closed and bounded set.
5. In finite-dimensional normed spaces, every closed and bounded set is compact.
6. Therefore, the closure of the image of the unit ball is compact, satisfying the definition of a compact operator.

Example

The board builds the chain: T: $\mathbb{F}^n \longrightarrow \mathbb{F}^m$ linear $\Rightarrow T$ is continuous / bounded $\Rightarrow T[B_1(0)] \subseteq \mathbb{F}^m$ bounded $\Rightarrow \overline{T[B_1(0)]} \subseteq \mathbb{F}^m$ compact.

Common misconceptions

• Assuming that the boundedness of the image of the unit ball alone is sufficient for compactness; the closure must also be taken.
• Believing that the argument depends on choosing the standard Euclidean norm, whereas the lecturer notes the statements do not depend on the particular norm choice.

Watch the explanation

Explore next

• What exactly is the definition of a compact operator given in this clip?
• Why is the identity operator on $\ell^p(\mathbb{N})$ not compact?
• Why do we take the closure of the image of the unit ball when defining compact operators?