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10: Linear Independence

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  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    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 \(\textit{linearly independent.}\)

    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 \(\PageIndex{1}\):

    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
    Are these vectors linearly independent?

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


    This page titled 10: Linear Independence is shared under a not declared license and was authored, remixed, and/or curated by David Cherney, Tom Denton, & Andrew Waldron.

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