# 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 norms
on $X$, $Y$, respectively.
By this result (http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended)
and this result (http://planetmath.org/SubspaceTopologyInAMetricSpace),
it suffices to prove that $L:X\to L(X)$ is continuous
when $L(X)$ is equipped with the topology given by $\parallel \cdot {\parallel}_{Y}$
restricted onto $L(X)$.
Also, since continuity and boundedness are equivalent^{}, it suffices to
prove 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
$\mathrm{ker}L=\mathrm{span}\{{e}_{k+1},\mathrm{\dots},{e}_{n}\}$ 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 follows
that

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

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

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. ∎

Title | linear transformation is continuous if its domain is finite dimensional |
---|---|

Canonical name | LinearTransformationIsContinuousIfItsDomainIsFiniteDimensional |

Date of creation | 2013-03-22 15:17:59 |

Last modified on | 2013-03-22 15:17:59 |

Owner | matte (1858) |

Last modified by | matte (1858) |

Numerical id | 7 |

Author | matte (1858) |

Entry type | Theorem |

Classification | msc 15A04 |