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Mathematics LibreTexts

10: Linear Independence

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 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
Are these vectors linearly independent?

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