# 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 $L\colon X\to Y$ is the transformation, $\dim X=n$, and $\|\cdot\|_{X}$, $\|\cdot\|_{Y}$ are the norms on $X$, $Y$, respectively. By this result (http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended) and this result (http://planetmath.org/SubspaceTopologyInAMetricSpace), it suffices to prove that $L\colon X\to L(X)$ is continuous when $L(X)$ is equipped with the topology given by $\|\cdot\|_{Y}$ 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 $L(X)$,

 $\displaystyle\|x\|^{\prime}_{X}$ $\displaystyle=$ $\displaystyle\sqrt{\sum_{i=1}^{n}\alpha_{i}^{2}},$ $\displaystyle\|y\|^{\prime}_{X}$ $\displaystyle=$ $\displaystyle\sqrt{\sum_{i=1}^{k}\beta_{i}^{2}},$

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 follows that

 $\displaystyle 1/C\|x\|^{\prime}_{X}\leq\|x\|_{X}\leq C\|x\|^{\prime}_{X},\quad x\in X$ $\displaystyle 1/D\|y\|^{\prime}_{Y}\leq\|y\|_{Y}\leq D\|y\|^{\prime}_{Y},\quad y% \in L(X)$

for some constants $C,D>0$. For $x=\sum_{i=1}^{n}\alpha_{i}e_{i}\in X$,

 $\displaystyle\|L(x)\|_{Y}$ $\displaystyle\leq$ $\displaystyle D\|\sum_{i=1}^{k}\alpha_{i}f_{i}\|^{\prime}_{Y}$ $\displaystyle=$ $\displaystyle D\sqrt{\sum_{i=1}^{k}\alpha_{i}^{2}}$ $\displaystyle\leq$ $\displaystyle D\sqrt{\sum_{i=1}^{n}\alpha_{i}^{2}}$ $\displaystyle=$ $\displaystyle D\|x\|^{\prime}_{X}$ $\displaystyle=$ $\displaystyle CD\|x\|_{X}.$

Thus $L\colon X\to L(X)$ is bounded. ∎

Title linear transformation is continuous if its domain is finite dimensional LinearTransformationIsContinuousIfItsDomainIsFiniteDimensional 2013-03-22 15:17:59 2013-03-22 15:17:59 matte (1858) matte (1858) 7 matte (1858) Theorem msc 15A04