Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

10.2: Showing Linear Independence

  • Page ID
    2037
  • [ "article:topic", "authortag:waldron", "authorname:waldron" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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