PlanetMath: every finite dimensional normed vector space is a Banach space

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (194)
Orphanage
Unclass'd
Unproven (498)
Corrections (26)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

every finite dimensional normed vector space is a Banach space

(Theorem)

Theorem 1 Every finite dimensional normed vector space is a Banach space.

Proof. Suppose $(V,\Vert\cdot\Vert)$ is the normed vector space, and $(e_i)_{i=1}^N$ is a basis for . For $x=\sum_{j=1}^N \lambda_j e_j$ , we can then define

$\displaystyle \Vert x \Vert' = \sqrt{\sum_{j=1}^N \vert\lambda_j\vert^2}$

whence $\Vert\cdot\Vert'\colon V\to \mathbb{R}$ is a norm for

. Since all norms on a finite dimensional vector space are equivalent, there is a constant

such that

$\displaystyle \frac{1}{C} \Vert x \Vert' \le \Vert x \Vert \le C \Vert x \Vert', \quad x\in V.$

To prove that

is a Banach space, let $x_1,x_2,\ldots$ be a Cauchy sequence in $(V,\Vert\cdot \Vert)$ . That is, for all $\varepsilon>0$ there is an $M\ge 1$ such that

$\displaystyle \Vert x_j-x_k \Vert <\varepsilon, \$ for all $\displaystyle j,k\ge M.$

Let us write each

in this sequence in the basis

as $x_k=\sum_{j=1}^N \lambda_{k,j} e_j$ for some constants $\lambda_{k,j}\in \mathbbmss{C}$ . For $k,l\ge 1$ we then have

$\displaystyle \Vert x_k-x_l\Vert$	$\displaystyle \ge$	$\displaystyle \frac{1}{C} \Vert x_k-x_l \Vert'$
	$\displaystyle \ge$	$\displaystyle \frac{1}{C} \sqrt{\sum_{j=1}^N \vert\lambda_{k,j}-\lambda_{l,j}\vert^2}$
	$\displaystyle \ge$	$\displaystyle \frac{1}{C} \vert\lambda_{k,j}-\lambda_{l,j}\vert$

for all $j=1,\ldots, N$ . It follows that $(\lambda_{k,1})_{k=1}^\infty, \ldots, (\lambda_{k,N})_{k=1}^\infty$ are Cauchy sequences in $\mathbbmss{C}$ . As $\mathbbmss{C}$ is complete, these converge to some complex numbers $\lambda_1, \ldots, \lambda_N$ . Let $x=\sum_{j=1}^N \lambda_j e_j$ .

For each $k=1,2,\ldots$ , we then have

$\displaystyle \Vert x-x_k\Vert$	$\displaystyle \le$	$\displaystyle C \Vert x-x_k\Vert'$
	$\displaystyle \le$	$\displaystyle C \sqrt{\sum_{j=1}^N \vert\lambda_{j}-\lambda_{k,j}\vert^2}.$