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Linear independence
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==Generalizations== ===Affine independence=== {{See also|Affine space}} A set of vectors is said to be '''affinely dependent''' if at least one of the vectors in the set can be defined as an [[affine combination]] of the others. Otherwise, the set is called '''affinely independent'''. Any affine combination is a linear combination; therefore every affinely dependent set is linearly dependent. Contrapositively, every linearly independent set is affinely independent. Note that an affinely independent set is not necessarily linearly independent. Consider a set of <math>m</math> vectors <math>\mathbf{v}_1, \ldots, \mathbf{v}_m</math> of size <math>n</math> each, and consider the set of <math>m</math> augmented vectors <math display="inline">\left(\left[\begin{smallmatrix} 1 \\ \mathbf{v}_1\end{smallmatrix}\right], \ldots, \left[\begin{smallmatrix}1 \\ \mathbf{v}_m\end{smallmatrix}\right]\right)</math> of size <math>n + 1</math> each. The original vectors are affinely independent if and only if the augmented vectors are linearly independent.<ref name="lp">{{Cite Lovasz Plummer}}</ref>{{Rp|256}} ===Linearly independent vector subspaces=== Two vector subspaces <math>M</math> and <math>N</math> of a vector space <math>X</math> are said to be {{em|linearly independent}} if <math>M \cap N = \{0\}.</math><ref name="BNFA">{{Bachman Narici Functional Analysis 2nd Edition}} pp. 3β7</ref> More generally, a collection <math>M_1, \ldots, M_d</math> of subspaces of <math>X</math> are said to be {{em|linearly independent}} if <math display=inline>M_i \cap \sum_{k \neq i} M_k = \{0\}</math> for every index <math>i,</math> where <math display=inline>\sum_{k \neq i} M_k = \Big\{m_1 + \cdots + m_{i-1} + m_{i+1} + \cdots + m_d : m_k \in M_k \text{ for all } k\Big\} = \operatorname{span} \bigcup_{k \in \{1,\ldots,i-1,i+1,\ldots,d\}} M_k.</math><ref name="BNFA" /> The vector space <math>X</math> is said to be a {{em|[[direct sum]]}} of <math>M_1, \ldots, M_d</math> if these subspaces are linearly independent and <math>M_1 + \cdots + M_d = X.</math>
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