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# 10: Linear Independence

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Consider a plane $$P$$ that includes the origin in $$\Re^{3}$$ and a collection $$\{u,v,w\}$$ of non-zero vectors in $$P$$: If no two of $$u, v$$ and $$w$$ are parallel, then $$P= span \{u,v,w\}$$. 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

$c^{1}u+c^{2}v+c^{3}w=0 \qquad c^{i} \in \Re, \textit{ some $$c^{i}\neq 0$$}$

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

Definition (Independent)

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

$c^{1}v_{1} + c^{2}v_{2}+ \cdots +c^{n}v_{n}=0.$

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

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

Example 106

Consider the following vectors in $$\Re^{3}$$:
$v_{1}=\begin{pmatrix}4\\-1\\3\end{pmatrix}, \qquad v_{2}=\begin{pmatrix}-3\\7\\4\end{pmatrix}, \qquad v_{3}=\begin{pmatrix}5\\12\\17\end{pmatrix}, \qquad v_{4}=\begin{pmatrix}-1\\1\\0\end{pmatrix}.$
Are these vectors linearly independent?

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