Basis (linear Algebra) Article Index for
Basis 		Website Links For
Basis 		 

Information About

Basis (linear Algebra)
APPAREL
BABY
BEAUTY
BOOKS
CAR TOYS
CELL PHONES
DVD'S
ELECTRONICS
GOURMET FOOD
GROCERIES
HEALTH & PERSONAL
HOME & GARDEN
JEWELRY
MUSIC
MUSIC INSTRUMENTS
OFFICE PRODUCTS
SOFTWARE
SPORTING GOODS
TOOLS & HARDWARE
TOYS
VIDEO GAMES
SHOPPING HOME
MORE SHOPPING...




DEFINITION


A basis ''B'' of a Vector Space ''V'' is a Linearly Independent subset of ''V'' that Span s (or Generates ) ''V''.

In more detail, suppose that ''B'' = { ''v''1, …, ''v''''n'' } is a finite subset of a vector space ''V'' over a Field ''F''. Then, ''B'' is a basis, if it satisfies the following conditions:
  • the ''linear independence'' property,

    • :: for all ''a''1, …, ''a''''n'' ∈ ''F'', if ''a''1''v''1 + … + ''a''''n''''v''''n'' = 0, then necessarily ''a''1 = … = ''a''''n'' = 0; and

  • the ''spanning'' property,

    • :: for every ''x'' in ''V'' it is possible to choose ''a''1, …, ''a''''n'' ∈ ''F'' such that ''x'' = ''a''1''v''1 + … + ''a''''n''''v''''n''.


A vector space that admits a Finite basis is called Finite-dimensional . To deal with infinite dimensional spaces, we must generalize the above definition to include infinite basis sets. We therefore say that a set (finite or infinite) ''B'' ⊂ ''V'' is a basis, if
  • every finite subset ''B''0 ⊆ ''B'' obeys the independence property shown above; and

  • for every ''x'' in ''V'' it is possible to choose ''a''1, …, ''a''''n'' ∈ ''F'' and ''v''1, …, ''v''''n'' ∈ ''B'' such that ''x'' = ''a''1''v''1 + … + ''a''''n''''v''''n''.


The axioms of a below.

When we want to describe the Matrix Of A Linear Transformation and in some other situations,
it is convenient to list the basis vectors in a specific ''order''. We then speak of an ordered basis, which we define to be a Sequence (rather than a Set ) of linearly independent vectors that span ''V''. Here is another way to think about this:
: every ordered basis of a finite-dimensional vector space ''V'' corresponds to a linear isomorphism from ''f'':R''n'' → ''V'', and vice versa.
''Proof.'' Given such an isomorphism, the sequence ''f''(e1), ... , ''f''(en), where e1, ... en is the Standard Basis of R''n'', is an ordered basis of ''V''. Conversely, given an ordered basis, the mapping defined by
: f(a) = ''a''1''v''1 + … + ''a''''n''''v''''n'',
where a= ''a''1e1 + … + ''a''''n''e''n'' is an element of '''R'''''n'', is necessarily a linear isomorphism.


PROPERTIES


Again, ''B'' denotes a subset of a vector space ''V''. Then, ''B'' is a basis if and only if any of the following equivalent conditions are met:
  • ''B'' is a minimal generating set of ''V'', i.e., it is a generating set but no Proper Subset of ''B'' is.

  • ''B'' is a maximal set of linearly independent vectors, i.e., it is a linearly independent set but no other linearly independent set contains it as a proper subset.

  • Every vector in ''V'' can be expressed as a linear combination of vectors in ''B'' in a unique way.


The theorem that every vector space has a basis is implied by the . All bases of a vector space have the same Cardinality (number of elements), called the Dimension of the vector space. The latter result is known as the Dimension Theorem , and requires the Ultrafilter Lemma , a strictly weaker form of the axiom of choice.


EXAMPLES


  • Consider R2, the vector space of all co-ordinates (''a'', ''b'') where both ''a'' and ''b'' are '' further down).


  • More generally, the vectors e1, e2, ..., e''n'' are linearly independent and generate R''n''. Therefore, they form a basis for R''n'' and the dimension of R''n'' is ''n''. This basis is called the '' Standard Basis ''.


  • Let ''V'' be the Real vector space generated by the functions ''e''''t'' and ''e''2''t''. These two functions are linearly independent, so they form a basis for ''V''.




BASIS EXTENSION


Between any linearly independent set and any generating set there is a basis. More formally: if ''L'' is a linearly independent set in the vector space ''V'' and ''G'' is a generating set of ''V'' containing ''L'', then there exists a basis of ''V'' that contains ''L'' and is contained in ''G''. In particular (taking ''G'' = ''V''), any linearly independent set ''L'' can be "extended" to form a basis of ''V''. These extensions are not unique.


PROVING THAT A SET IS A BASIS


To prove that a set ''B'' is a basis for a (finite-dimensional) vector space ''V'', it is sufficient to show that the number of elements in ''B'' equals the dimension of ''V'', and one of the following:
  • ''B'' is linearly independent, or

  • span(''B'') = ''V''.



EXAMPLES OF PROOFS

As an easy example, let us show that the vectors (1,1) and (-1,2) form a basis for R2.
The following proof methods require increasing amounts of sophistication and decreasing amounts of effort.


By brute force

We have to prove that these two vectors are linearly independent and that they generate R2.

Part I: To prove that they are linearly independent, suppose that there are numbers a,b such that:
: a(1,1)+b(-1,2)=(0,0). \,
Then:

:

(a-b,a+2b)=(0,0) \,
  and  

a-b=0 \;
  and  

a+2b=0. \,

Subtracting the first equation from the second, we obtain:

:

3b=0 \;
  so  

b=0. \,

And from the first equation then:

: a=0. \,

Part II: To prove that these two vectors generate R2, we have to let (a,b) be an arbitrary element of R2, and show that there exist numbers x,y such that:

: x(1,1)+y(-1,2)=(a,b). \,
Then we have to solve the equations:

: x-y=a \,
: x+2y=b. \,
Subtracting the first equation from the second, we get:

:

3y=b-a, \,
          and then
:

y=(b-a)/3, \,
        and finally
: x=y+a=((b-a)/3)+a. \,


By the dimension theorem


Since (-1,2) is clearly not a multiple of (1,1) and since (1,1) is not the Zero Vector , these two vectors are linearly independent. Since the dimension of R2 is 2, the two vectors already form a basis of R2 without needing any extension.


By the invertible matrix theorem


Simply compute the Determinant
:\det\begin{bmatrix}1&-1\1&2\end{bmatrix}=3
eq0.
Since the above matrix has a nonzero determinant, its .


ORDERED BASES


A basis is just a ''set'' of vectors with no given ordering. For many purposes it is convenient to work with an ordered basis. For example, when working with a coordinate representation of a vector it is customary to speak of the "first" or "second" coordinate, which makes sense only if an ordering is specified for the basis. For finite-dimensional vector spaces one typically Indexes a basis {''v''''i''} by the first ''n'' integers.

Suppose ''V'' is an ''n''-dimensional vector space over a Field F. A choice of an ordered basis for ''V'' is equivalent to a choice of a Linear Isomorphism from the Coordinate Space F''n'', with its standard basis, to ''V''. To see this, let
A

be a linear isomorphism. Define an ordered basis {''v''''i''} for ''V'' by
: ''v''''i'' = ''A''(e''i'') for 1 ≤ ''i'' ≤ ''n''
where {e''i''} is the standard basis for '''F'''''n''. Conversely, given any ordered basis {''v''''i''} for ''V'' define a linear map ''A'' : '''F'''''n'' → ''V'' by
:A(x) = \sum_{i=1}^n x_i v_i
It is not hard to check that ''A'' is an isomorphism. Thus ordered bases for ''V'' are in 1-1 correspondence with linear isomorphisms F''n'' → ''V''.


RELATED NOTIONS


The phrase ''Hamel basis'' is sometimes used to refer to a basis as defined above, in which the fact that all linear combinations are ''finite'' is crucial. A set ''B'' is a Hamel basis of a vector space ''V'' if every member of ''V'' is a linear combination of just ''finitely'' many members of ''B''.

In Hilbert Space s and other Banach Space s, there is a need to work with linear combinations of infinitely many vectors. In an infinite-dimensional Hilbert space, a set of vectors orthogonal to each other can never span the whole space via their finite linear combinations. What is called an Orthonormal Basis is a set of mutually orthogonal unit vectors that "span" the space via sometimes-infinite linear combinations. Except in the finite-dimensional case, this concept is not purely algebraic, and is distinct from a Hamel basis; it is also more generally useful. ''An orthonormal basis of an infinite-dimensional Hilbert space is therefore not a Hamel basis.''

In Topological Vector Space s, quite generally, one may define ''infinite sums'' ( Infinite Series ) and express elements of the space as certain ''infinite linear combinations'' of other elements. To keep clear the distinction of bases using finite and infinite combination, the former ones are called ''Hamel bases'' and the latter ones '' Schauder Bases ,'' if the context requires it. The corresponding dimensions are also known as ''Hamel dimension'' and ''Schauder dimension.''


Example


In the study of -valued functions that are quadratically integrable on the interval 2π , i.e., functions ''f'' satisfying

  :<math>\lim {n Ightarrow\infty}\int 0^{2\pi}\left\left(a 0+\sum {k 1}^n a_k\cos(kx)+b_k\sin(kx) ight)-f(x) ight^2\,dx=0</math>