# 10.2: Showing Linear Independence

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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 109**

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)