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Mathematics-Online course: Linear Algebra - Basic Structures - Bases

Basis

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A subset $ B$ of vector space $ V$ is called basis of $ V$, if $ B$ is linearly independent as well as a generating system for $ V$, that is, any vector $ v\in V$ has a unique representation as finite linear combination

$\displaystyle v = \sum_{i=1}^n \lambda_i b_i $

where $ b_i\in B$.

In a finite-dimensional vector space any vector can be described by its coordinates relative to a given basis:

$\displaystyle v_B = (\lambda_1,\ldots,\lambda_n)^{\operatorname t}\,. $

In an infinite-dimensional vectorspace a generating system $ S$ is a basis if and only if any finite subset of $ S$ consists of linearly independent vectors.

(Authors: App/Burkhardt/Höllig/Kimmerle)
By the so called canonical isomorphism

$\displaystyle v = \sum_{i=1}^n \lambda_i b_i \leftrightarrow v_B = (\lambda_1,\ldots,\lambda_n)^{\operatorname t}\, , $

a finite dimensional $ K$-vector space $ V$ can be identified with the vector space $ K^n$ of $ n$-tuples. Thus, in particular, studying real and complex vector spaces with finite basis, we can confine our consideration to $ \mathbb{R}^n$ and $ \mathbb{C}^n$.

As the example of the vector space of polynomials shows, vector spaces do not necessarily have finite bases. In the context of linear algebra, however, only finite linear combinations are considered. This implies that the sequences

$\displaystyle e_1 = (1,0,0,\ldots)^{\operatorname t},\, e_2 = (0,1,0,\ldots)^{\operatorname t},\, \ldots $

do not form a basis for the vector space of sequences. 'Infinite linear combinations', as for example Fourier series, require a notion of convergence the mere vector space structure does not provide.
(Authors: App/Burkhardt/Höllig)
  automatically generated 4/21/2005