Editing Linear subspace (section)

===Independence, basis, and dimension===
{{main|Linear independence|Basis (linear algebra)|Dimension (vector space)}}
[[File:Basis for a plane.svg|thumb|280px|right|The vectors '''u''' and '''v''' are a basis for this two-dimensional subspace of '''R'''<sup>3</sup>.]]
In general, a subspace of ''K''<sup>''n''</sup> determined by ''k'' parameters (or spanned by ''k'' vectors) has dimension ''k''. However, there are exceptions to this rule. For example, the subspace of ''K''<sup>3</sup> spanned by the three vectors (1,&nbsp;0,&nbsp;0), (0,&nbsp;0,&nbsp;1), and (2,&nbsp;0,&nbsp;3) is just the ''xz''-plane, with each point on the plane described by infinitely many different values of {{nowrap| ''t''<sub>1</sub>, ''t''<sub>2</sub>, ''t''<sub>3</sub>}}.

In general, vectors '''v'''<sub>1</sub>,&nbsp;...&nbsp;,&nbsp;'''v'''<sub>''k''</sub> are called '''linearly independent''' if

:<math>t_1 \mathbf{v}_1 + \cdots + t_k \mathbf{v}_k \;\ne\; u_1 \mathbf{v}_1 + \cdots + u_k \mathbf{v}_k</math>

for
(''t''<sub>1</sub>,&nbsp;''t''<sub>2</sub>,&nbsp;...&nbsp;,&nbsp;''t<sub>k</sub>'')&nbsp;≠ (''u''<sub>1</sub>,&nbsp;''u''<sub>2</sub>,&nbsp;...&nbsp;,&nbsp;''u<sub>k</sub>'').<ref group="note">This definition is often stated differently: vectors '''v'''<sub>1</sub>, ..., '''v'''<sub>''k''</sub> are linearly independent if
{{nowrap| ''t''<sub>1</sub>'''v'''<sub>1</sub> + ··· + ''t<sub>k</sub>'''''v'''<sub>''k''</sub> ≠ '''0'''}} for {{nowrap| (''t''<sub>1</sub>, ''t''<sub>2</sub>, ..., ''t<sub>k</sub>'') ≠ (0, 0, ..., 0)}}. The two definitions are equivalent.</ref>
If {{nowrap| '''v'''<sub>1</sub>, ..., '''v'''<sub>''k''</sub> }} are linearly independent, then the '''coordinates''' {{nowrap| ''t''<sub>1</sub>, ..., ''t<sub>k</sub>''}} for a vector in the span are uniquely determined.

A '''basis''' for a subspace ''S'' is a set of linearly independent vectors whose span is ''S''. The number of elements in a basis is always equal to the geometric dimension of the subspace. Any spanning set for a subspace can be changed into a basis by removing redundant vectors (see [[#Algorithms|§ Algorithms]] below for more).

; Example
: Let ''S'' be the subspace of '''R'''<sup>4</sup> defined by the equations
::<math>x_1 = 2 x_2\;\;\;\;\text{and}\;\;\;\;x_3 = 5x_4.</math>
:Then the vectors (2,&nbsp;1,&nbsp;0,&nbsp;0) and (0,&nbsp;0,&nbsp;5,&nbsp;1) are a basis for ''S''. In particular, every vector that satisfies the above equations can be written uniquely as a linear combination of the two basis vectors:

::<math>(2t_1, t_1, 5t_2, t_2) = t_1(2, 1, 0, 0) + t_2(0, 0, 5, 1).</math>

:The subspace ''S'' is two-dimensional. Geometrically, it is the plane in '''R'''<sup>4</sup> passing through the points (0,&nbsp;0,&nbsp;0,&nbsp;0), (2,&nbsp;1,&nbsp;0,&nbsp;0), and (0,&nbsp;0,&nbsp;5,&nbsp;1).