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Mathematics-Online course: Prepcourse Mathematics - Linear Algebra and Geometry - Vector spaces

Dimension

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If vector space $ V$ has a basis consisting of a finite number of vectors $ \{b_1, \ldots ,b_n\}$, then $ n$ is called the dimension of $ V$ (or $ V$ is said to have dimension $ n$)(notation: $ {\rm dim} V = n$).

If $ V = 0$, that is, the only element in $ V$ is the zero vector, then we set $ {\rm dim} V = 0 .$

If a vector space has no finite basis, then it is called infinite-dimensional (notation: $ {\rm dim} V = \infty$).

Observe that according to the general Basis Teorem every vector space has a basis.

All bases of a finite-dimensional vector space have the same length, that is, the same number of basis vectors.

There exist bijections between different bases of a given infinite-dimensional vector space.

(Authors: App/Burkhardt/Höllig/Kimmerle)
To prove that any two bases of a finitely generated vector space (that is, the vector space is spanned by a finite set of vectors) has the same number of elements, it is sufficient to show the following:

If a vector space has an $ n$-element basis

$\displaystyle b_1,\ldots,b_n\, , $

then any $ n+1$ vectors $ v_1,\ldots,v_{n+1}$ (and, consequently, as well as any more than $ n+1$ vectors) are linearly dependent.

If there existed two bases with different numbers of vectors, we would obtain a contradiction to the linear independence of the basis vectors.

The above assertion can be proved by induction on $ n$.

For the inductive step (the base case $ n=1$ is trivial) let us consider the representation of the vectors $ v_i$ as linear combinations of the basis vectors:

$\displaystyle v_i = \sum_{j=1}^n \gamma_{i,j} b_j,\quad i=1,\ldots,n+1\, .$ (1)

If

$\displaystyle \gamma_{i,1}=\cdots=\gamma_{i,n}=0 $

then $ v_i=0$ and, thus, linear dependence is proved. Hence, by appropriate numbering we can assume that $ \gamma_{n+1,n}\ne 0$. Now, starting from the above equation we define vectors which can be represented as linear combinations of the $ n-1$ vectors $ b_1,\ldots,b_{n-1}$:

$\displaystyle v'_i = v_i - \frac{\gamma_{i,n}}{\gamma_{n+1,n}} v_{n+1}, \quad i=1,\ldots,n\, . $

It can easily be seen that the coefficient of $ b_n$ vanishes. Since

$\displaystyle v'_1,\ldots,v'_n \in V' =$   span$\displaystyle \,\{b_1,\ldots,b_{n-1}\}\, , $

we can apply the inductive hypothesis and we get a non-trivial linear combination

$\displaystyle \lambda_1 v'_1 + \cdots + \lambda_n v'_n = 0\, . $

By transformation we obtain a linear combination of $ v_i$, what proves the asserted linear dependence.
(Authors: App/Burkhardt/Höllig)
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  automatically generated 10/23/2009