10.2: Showing Linear Independence

We have seen two different ways to show a set of vectors is linearly dependent: we can either find a linear combination of the vectors which is equal to zero, or we can express one of the vectors as a linear combination of the other vectors. On the other hand, to check that a set of vectors is linearly $\textit{independent}$, we must check that every linear combination of our vectors with non-vanishing coefficients gives something other than the zero vector. Equivalently, to show that the set $v_{1}, v_{2}, \ldots, v_{n}$ is linearly independent, we must show that the equation $c_{1} v_{1}+c_{2} v_{2} + \cdots + c_{n} v_{n}=0$ has no solutions other than $c_{1}=c_{2}=\cdots=c_{n}=0.$

Example $\PageIndex{1}$:

Consider the following vectors in $\Re^{3}$:
$$v_{1}=\begin{pmatrix}0\\0\\2\end{pmatrix}, \qquad v_{2}=\begin{pmatrix}2\\2\\1\end{pmatrix}, \qquad v_{3}=\begin{pmatrix}1\\4\\3\end{pmatrix}.$$

Are they linearly independent?

We need to see whether the system

$$c^{1}v_{1} + c^{2}v_{2}+ c^{3}v_{3}=0$$

has any solutions for $c^{1}, c^{2}, c^{3}$. We can rewrite this as a homogeneous system:

$$\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}\begin{pmatrix}c^{1}\\c^{2}\\c^{3}\end{pmatrix}=0.$$

This system has solutions if and only if the matrix $M=\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}$ is singular, so we should find the determinant of $M$:

$$\det M = \det \begin{pmatrix} 0 & 2 & 1 \\ 0 & 2 & 4 \\ 2 & 1 & 3 \\ \end{pmatrix} = 2 \det \begin{pmatrix} 2 & 1 \\ 2 & 4 \\ \end{pmatrix} =12.$$

Since the matrix $M$ has non-zero determinant, the only solution to the system of equations

$$\begin{pmatrix}v_{1}&v_{2}&v_{3}\end{pmatrix}\begin{pmatrix}c^{1}\\c^{2}\\c^{3}\end{pmatrix}=0$$

is $c_{1}=c_{2}=c_{3}=0$. So the vectors $v_{1}, v_{2}, v_{3}$ are linearly independent.

Contributor

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