Linear independence

In the theory of vector spaces, a set of vectors is said to be linearly independent if there exists no nontrivial linear combination of the vectors that equals the zero vector. If such a linear combination exists, then the vectors are said to be linearly dependent. These concepts are central to the definition of dimension.^[1]

For linear dependence of random variables, see Covariance.

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.

${\vec {u}}$ and ${\vec {v}}$ are independent and define the P.

plane

${\vec {u}}$ , ${\vec {v}}$ and ${\vec {w}}$ are dependent because all three are contained in the same plane.

${\vec {u}}$ and ${\vec {j}}$ are dependent because they are parallel to each other.

${\vec {u}}$ , ${\vec {v}}$ and ${\vec {k}}$ are independent because ${\vec {u}}$ and ${\vec {v}}$ are independent of each other and ${\vec {k}}$ is not a linear combination of them or, equivalently, because they do not belong to a common plane. The three vectors define a three-dimensional space.

The vectors ${\vec {o}}$ (null vector, whose components are equal to zero) and ${\vec {k}}$ are dependent since ${\vec {o}}=0{\vec {k}}$

Evaluating linear independence[edit]

The zero vector[edit]

If one or more vectors from a given sequence of vectors $\mathbf {v} _{1},\dots ,\mathbf {v} _{k}$ is the zero vector $\mathbf {0}$ then the vector $\mathbf {v} _{1},\dots ,\mathbf {v} _{k}$ are necessarily linearly dependent (and consequently, they are not linearly independent). To see why, suppose that $i$ is an index (i.e. an element of $\{1,\ldots ,k\}$ ) such that $\mathbf {v} _{i}=\mathbf {0} .$ Then let $a_{i}:=1$ (alternatively, letting $a_{i}$ be equal any other non-zero scalar will also work) and then let all other scalars be $0$ (explicitly, this means that for any index $j$ other than $i$ (i.e. for $j\neq i$ ), let $a_{j}:=0$ so that consequently $a_{j}\mathbf {v} _{j}=0\mathbf {v} _{j}=\mathbf {0}$ ). Simplifying $a_{1}\mathbf {v} _{1}+\cdots +a_{k}\mathbf {v} _{k}$ gives: