Theorem 1   A linear transformation is continuous if the domain is finite dimensional.

Proof. Suppose $L\colon X\to Y$ is the transformation, $\dim X = n$ , and $\Vert \cdot \Vert_X$ , $\Vert \cdot \Vert_Y$ are the norms on $X$ , $Y$ , respectively. By this result and this result, it suffices to prove that $L\colon X\to L(X)$ is continuous when $L(X)$ is equipped with the topology given by $\Vert \cdot \Vert_Y$ restricted onto $L(X)$ . Also, since continuity and boundedness are equivalent, it suffices to prove that $L$ is bounded. Let $e_1,\ldots, e_n$ be a basis for $X$ such that $L$ is invertible on $\operatorname{span} \{e_{1}, \ldots, e_k\}$ and $\operatorname{ker} L = \operatorname{span} \{e_{k+1}, \ldots, e_n\}$ for $k=1,\ldots, n$ . (The zero map is always continuous.) Let $f_i=L(e_i)$ for $i=1,\ldots, k$ , so that $\operatorname{span}\{f_1, \ldots, f_k\}=L(X)$ . Let us define new norms on $X$ and $L(X)$ , \begin{eqnarray*} \Vert x \Vert'_X &=& \sqrt{\sum_{i=1}^n \alpha_i^2},\\ \Vert y \Vert'_X &=& \sqrt{\sum_{i=1}^k \beta_i^2}, \end{eqnarray*}for $x=\sum_{i=1}^n \alpha_i e_i\in X$ and $y=\sum_{i=1}^k \beta_i f_i \in Y$ . Since norms on finite dimensional vector spaces are equivalent, it follows that \begin{eqnarray*} 1/C \Vert x \Vert'_X \le \Vert x \Vert_X \le C \Vert x \Vert'_X, \quad x\in X \\ 1/D \Vert y \Vert'_Y \le \Vert y \Vert_Y \le D \Vert y \Vert'_Y, \quad y\in L(X) \end{eqnarray*}for some constants $C,D>0$ . For $x=\sum_{i=1}^n \alpha_i e_i\in X$ , \begin{eqnarray*} \Vert L(x)\Vert_Y &\le & D \Vert \sum_{i=1}^k \alpha_i f_i \Vert'_Y \\ &=& D \sqrt{\sum_{i=1}^k \alpha_i^2} \\ &\le& D \sqrt{\sum_{i=1}^n \alpha_i^2} \\ &=& D \Vert x \Vert'_X \\ &=& CD \Vert x \Vert_X. \end{eqnarray*}Thus $L\colon X\to L(X)$ is bounded.