Section 4.2 Independence and Dimension
Definition: An indexed set of vectors [latex]\begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] in [latex]\mathbb{R}^n[/latex] is said to be linearly independent if the vector equation [latex]x_{1}\vec{v_{1}}+ \cdots+ x_{p}\vec{v_{p}} =\vec{0}[/latex] in [latex]\mathbb{R}^n[/latex] has only trivial solution. [latex]\begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] in [latex]\mathbb{R}^n[/latex] is said to be linearly dependent if there are [latex]c_{1}, \cdots, c_{p}[/latex] not all zero such that [latex]c_{1}\vec{v_{1}}, \cdots, c_{p}\vec{v_{p}} = \vec{0}[/latex]. [latex]c_{1}\vec{v_{1}}+ \cdots+c_{p}\vec{v_{p}} = \vec{0}[/latex] is called the linear dependence relation among [latex]\vec{v_{1}}, \cdots, \vec{v_{p}}[/latex].
Example 1: Determine if the set [latex]\begin{Bmatrix}\vec{v_{1}},\vec{v_{2}},\vec{v_{3}}\end{Bmatrix}[/latex] is linearly independent. If possible, find a linear dependence relation among [latex]\vec{v_{1}},\vec{v_{2}},\vec{v_{3}}[/latex]. [latex]\vec{v_{1}} = \begin{bmatrix}-1\\2\\3\end{bmatrix}[/latex], [latex]\vec{v_{2}} = \begin{bmatrix}4\\-1\\9\end{bmatrix}[/latex], and [latex]\vec{v_{3}} = \begin{bmatrix}2\\-4\\-6\end{bmatrix}[/latex].
Exercise 1: Determine if the set [latex]\begin{Bmatrix}\vec{v_{1}},\vec{v_{2}},\vec{v_{3}}\end{Bmatrix}[/latex] is linearly independent. If possible, find a linear dependence relation among [latex]\vec{v_{1}},\vec{v_{2}},\vec{v_{3}}[/latex]. [latex]\vec{v_{1}} = \begin{bmatrix}-2\\2\\-3\end{bmatrix}[/latex], [latex]\vec{v_{2}} = \begin{bmatrix}6\\-1\\4\end{bmatrix}[/latex], and [latex]\vec{v_{3}} = \begin{bmatrix}4\\-4\\6\end{bmatrix}[/latex].
Note: 1. Given a matrix [latex]A = \begin{bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{bmatrix}[/latex] with [latex]p[/latex] columns, the matrix equation [latex]A\vec{x} = \vec{0}[/latex] can be written as [latex]x_{1}\vec{v_{1}} + \cdots + x_{p}\vec{v_{p}} = \vec{0}[/latex]. Then each linear dependence relation among the columns of [latex]A[/latex] corresponds to a nontrivial solution of [latex]A\vec{x} = \vec{0}[/latex]. Hence the columns of matrix [latex]A[/latex] are linearly independent if and only if the equation [latex]A\vec{x} = \vec{0}[/latex] has only the trivial solution.
2. A set with only one vector is linearly independent if and only if it is not a zero vector. The zero vector is linearly dependent.
3. A set with two vectors is linearly independent if and only if they are not multiple of each other.
Example 2: Show that the column set of [latex]A[/latex] is a linearly independent set. [latex]A = \begin{bmatrix}2 & 0 & 1\\1 & -1 & 0\\-1 & 2 & 1\end{bmatrix}[/latex].
Exercise 2: Show that the column set of [latex]A[/latex] is a linearly independent set. [latex]A = \begin{bmatrix}1 & 2 & 1\\0 & -1 & 2\\-1 & -2 & 0\end{bmatrix}[/latex].
Theorem: If [latex]S = \begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] is an linear independent vectors in