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10: Linear Independence

Last updated

Jul 27, 2023
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- 9.3: Review Problems
- 10.1: Showing Linear Dependence

( \newcommand{\kernel}{\mathrm{null}\,}\)

Consider a plane $P$ that includes the origin in $ℜ^{3}$ and a collection ${u, v, w}$ of non-zero vectors in $P$ :

$151257541312468.png$

If no two of $u, v$ and $w$ are parallel, then $P = s p a n {u, v, w}$ . But any two vectors determines a plane, so we should be able to span the plane using only two of the vectors $u, v, w$ . Then we could choose two of the vectors in ${u, v, w}$ whose span is $P$ , and express the other as a linear combination of those two. Suppose $u$ and $v$ span $P$ . Then there exist constants $d^{1}, d^{2}$ (not both zero) such that $w = d^{1} u + d^{2} v$ . Since $w$ can be expressed in terms of $u$ and $v$ we say that it is not independent. More generally, the relationship

$\begin{matrix} (10.1) & c^{1} u + c^{2} v + c^{3} w = 0 c^{i} \in ℜ, some c^{i} \neq 0 \end{matrix}$

expresses the fact that $u, v, w$ are not all independent.

Definition (Independent)

We say that the vectors $v_{1}, v_{2}, \dots, v_{n}$ are $linearly dependent$ if there exist constants (usually our vector spaces are defined over $R$ , but in general we can have vector spaces defined over different base fields such as $C$ or $Z_{2}$ . The coefficients $c^{i}$ should come from whatever our base field is (usually $R$ ).} $c^{1}, c^{2}, \dots, c^{n}$ not all zero such that

$\begin{matrix} (10.2) & c^{1} v_{1} + c^{2} v_{2} + \dots + c^{n} v_{n} = 0. \end{matrix}$

Otherwise, the vectors $v_{1}, v_{2}, \dots, v_{n}$ are $linearly independent.$

Remark
The zero vector $0_{V}$ can $never$ be on a list of independent vectors because $α 0_{V} = 0_{V}$ for any scalar $α$ .

Example $10.1$ :

Consider the following vectors in $ℜ^{3}$ :
$\begin{matrix} (10.3) & v_{1} = (\begin{matrix} 4 \\ - 1 \\ 3 \end{matrix}), v_{2} = (\begin{matrix} - 3 \\ 7 \\ 4 \end{matrix}), v_{3} = (\begin{matrix} 5 \\ 12 \\ 17 \end{matrix}), v_{4} = (\begin{matrix} - 1 \\ 1 \\ 0 \end{matrix}) . \end{matrix}$
Are these vectors linearly independent?

No, since $3 v_{1} + 2 v_{2} - v_{3} + v_{4} = 0$ , the vectors are linearly $dependent$ .

Contributor

David Cherney, Tom Denton, and Andrew Waldron (UC Davis)

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