linear transformation is continuous if its domain is finite dimensional

Theorem 1.

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

Proof.

Suppose $L:X\to Y$ is the transformation, $dimX=n$,and $\parallel \cdot {\parallel }_X$, $\parallel \cdot {\parallel }_Y$ are the normson $X$, $Y$, respectively.By this result (http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended)and this result (http://planetmath.org/SubspaceTopologyInAMetricSpace),it suffices to prove that $L:X\to L(X)$ is continuouswhen $L(X)$ is equipped with the topology given by $\parallel \cdot {\parallel }_Y$restricted onto $L(X)$.Also, since continuity and boundedness are equivalent, it suffices toprove that $L$ is bounded.Let $e_1,\mathrm{\dots },e_n$ be a basis for $X$ such that$L$ is invertible on $\mathrm{span}\{e_1,\mathrm{\dots },e_k\}$ and$\ker L=\mathrm{span}\{e_{k+1},\mathrm{\dots },e_n\}$ for$k=1,\mathrm{\dots },n$. (The zero map is always continuous.)Let $f_i=L(e_i)$ for $i=1,\mathrm{\dots },k$, so that$\mathrm{span}\{f_1,\mathrm{\dots },f_k\}=L(X)$.Let us define new norms on $X$ and $L(X)$,

${\parallel x\parallel }_X^{\prime }$$=$$\sqrt{{\displaystyle \sum _{i=1}^n}{\alpha }_i^2},$

${\parallel y\parallel }_X^{\prime }$$=$$\sqrt{{\displaystyle \sum _{i=1}^k}{\beta }_i^2},$

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 followsthat

$1/C{\parallel x\parallel }_X^{\prime }\le {\parallel x\parallel }_X\le C{\parallel x\parallel }_X^{\prime },x\in X$

$1/D{\parallel y\parallel }_Y^{\prime }\le {\parallel y\parallel }_Y\le D{\parallel y\parallel }_Y^{\prime },y\in L(X)$

for some constants $C,D>0$.For $x={\sum }_{i=1}^n{\alpha }_i{e}_i\in X$,

${\parallel L(x)\parallel }_Y$$\le$$D{\parallel {\displaystyle \sum _{i=1}^k}{\alpha }_i{f}_i\parallel }_Y^{\prime }$

$=$$D\sqrt{{\displaystyle \sum _{i=1}^k}{\alpha }_i^2}$

$\le$$D\sqrt{{\displaystyle \sum _{i=1}^n}{\alpha }_i^2}$

$=$$D{\parallel x\parallel }_X^{\prime }$

$=$$CD{\parallel x\parallel }_X.$

Thus $L:X\to L(X)$ is bounded.∎