# Linearly independent

(Redirected from Linearly dependent)

In linear algebra, a set of vectors $v_1, v_2, \ldots, v_n$ in a vector space $V$ over a field $K$ are linearly independent if there do not exist scalars $c_1, c_2, \ldots, c_n \in K$ not all equal to zero such that $$c_1v_1 + c_2v_2 + \cdots + c_nv_n = O.$$ Otherwise, the vectors are said to be linearly dependent.

In $\mathbb{R}^m$, vectors $\bold{v}_1, \ldots, \bold{v}_n$ are linearly independent iff their determinant $D(\bold{v}_1, \ldots, \bold{v}_n) = 0$.

## Examples

A basis of a vector space $V$ is a maximal set of linearly independent vectors, that is, if $\{v_1, \ldots, v_n\}$ are a basis, then $\{v_1, \ldots, v_n, w\}$ for any vector $w \in V$ are linearly dependent.

Any eigenvectors corresponding to different eigenvalues (with respect to a linear map $L$) are linearly independent. This can be proved by induction. Suppose $v_1, v_2, \ldots, v_n$ are eigenvectors corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$, and that there exists a statement of linear dependency $$\sum c_iv_i = O.$$ Multiplying both sides by $\lambda_1$ and applying $L$ to both sides, respectively, yields $$\sum \lambda_1c_iv_i = O,\qquad L\left(\sum c_iv_i\right) = \sum \lambda_ic_iv_i = O.$$ Subtracting the two equations yields $$\sum (\lambda_1-\lambda_i)c_iv_i,$$ which is a statement of linear dependency among $v_2, \ldots, v_n$.

## Problems

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