Linear maps on finite-dimensional spaces are continuous

Statement

Let $V$ be a finite-dimensional normed space and let $W$ be a normed space, both over $\mathbb{R}$ or over $\mathbb{C}$. Then each linear map $T \colon V \to W$ is continuous.

Remarks

Proof: choose a basis $e_1, \ldots, e_n$ of $V$. For $x = \sum_i a_i e_i$, the triangle inequality gives $\|Tx\|_W \le \left( \sum_i \|T e_i\|_W \right) \max_i |a_i|$. By equivalence of norms in finite dimension, $\max_i |a_i| \le c \|x\|_V$ for some $c$. Thus $T$ is bounded, and so continuous. The finite dimension of $W$ is not needed, only that of $V$. For example, each matrix gives a continuous map $\mathbb{R}^n \to \mathbb{R}^m$.

Explanations

L1 human

In finite dimensions, linear maps are automatically continuous, for each choice of norms. Problems with continuity only appear in infinite dimensions.