In linear algebra, a set of vectors is said to be linearly independent if there exists no vector in the set that is equal to a linear combination of the other vectors in the set. If such a vector exists, then the vectors are said to be linearly dependent. Linear independence is part of the definition of linear basis. A vector space can be of finite dimension or infinite dimension depending on the maximum number of linearly independent vectors. The definition of linear dependence and the ability to determine whether a subset of vectors in a vector space is linearly dependent are central to determining the dimension of a vector space.
Definition A sequence of vectors v 1 , v 2 , … , v k {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{k}} from a vector space V is said to be linearly dependent, if there exist scalars a 1 , a 2 , … , a k , {\displaystyle a_{1},a_{2},\dots ,a_{k},} not all zero, such that
a 1 v 1 + a 2 v 2 + ⋯ + a k v k = 0 , {\displaystyle a_{1}\mathbf {v} _{1}+a_{2}\mathbf {v} _{2}+\cdots +a_{k}\mathbf {v} _{k}=\mathbf {0} ,}
where 0 {\displaystyle \mathbf {0} } denotes the zero vector. If k = 1 {\displaystyle k=1} , this implies that a single vector is linear dependent if and only if it is the zero vector. If k > 1 {\displaystyle k>1} , this implies that at least one of the scalars is nonzero, say a 1 ≠ 0 {\displaystyle a_{1}\neq 0} , and the above equation is able to be written as
v 1 = − a 2 a 1 v 2 + ⋯ + − a k a 1 v k . {\displaystyle \mathbf {v} _{1}={\frac {-a_{2}}{a_{1}}}\mathbf {v} _{2}+\cdots +{\frac {-a_{k}}{a_{1}}}\mathbf {v} _{k}.}
Thus, a set of vectors is linearly dependent if and only if one of them is zero or a linear combination of the others. A sequence of vectors v 1 , v 2 , … , v n {\displaystyle \mathbf {v} _{1},\mathbf {v} _{2},\dots ,\mathbf {v} _{n}} is said to be linearly independent if it is not linearly dependent, that is, if the equation
a 1 v 1 + a 2 v 2 + ⋯ + a n v n = 0 , {\displaystyle a_{1}\mathbf {v} _{1}+a_{2}\mathbf {v} _{2}+\cdots +a_{n}\mathbf {v} _{n}=\mathbf {0} ,}
… excerpt ends here. Continue reading the full article.




