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 [latex]\mathbb{R}^n[/latex], then every vector in Span[latex]\begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] has a unique representation as a linear combination of [latex]\vec{v_{i}}'s[/latex].
Note: Geometrically, any two vectors in [latex]\mathbb{R}^n[/latex] with [latex]n > 1[/latex] that are not multiple of each other span a plane( they are not co-linear). Any three vectors form a linearly independent set if they are not co-planar.
Theorem: If a set contains more vectors than there are entries in each vector, then the set is linearly dependent. That is, any set [latex]S = \begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] in [latex]\mathbb{R}^n[/latex] is linearly dependent if [latex]p > n[/latex].
Theorem: If [latex]S = \begin{Bmatrix}\vec{v_{1}}, \cdots, \vec{v_{p}}\end{Bmatrix}[/latex] in [latex]\mathbb{R}^n[/latex] contains the zero vector then it is linearly dependent.
Example 3: Use inspection to decide if the vector set is linear independent. State the reasoning.
(a) [latex]\begin{bmatrix}2\\1\\-1\end{bmatrix}[/latex], [latex]\begin{bmatrix}-4\\-2\\2\end{bmatrix}[/latex].
(b) [latex]\begin{bmatrix}2\\1\\-1\end{bmatrix}[/latex], [latex]\begin{bmatrix}3\\4\\5\end{bmatrix}[/latex], [latex]\begin{bmatrix}0\\0\\0\end{bmatrix}[/latex].
(c) [latex]\begin{bmatrix}2\\1\\-1\end{bmatrix}[/latex], [latex]\begin{bmatrix}3\\4\\5\end{bmatrix}[/latex], [latex]\begin{bmatrix}-2\\6\\-7\end{bmatrix}[/latex], [latex]\begin{bmatrix}0\\2\\1\end{bmatrix}[/latex].
Exercise 3: Use inspection to decide if the vector set is linear independent. State the reasoning.
(a) [latex]\begin{bmatrix}2\\1\end{bmatrix}[/latex], [latex]\begin{bmatrix}-4\\-2\end{bmatrix}[/latex], [latex]\begin{bmatrix}4\\3\end{bmatrix}[/latex].
(b) [latex]\begin{bmatrix}2\\1\\0\\-1\end{bmatrix}[/latex], [latex]\begin{bmatrix}3\\2\\1\\5\end{bmatrix}[/latex], [latex]\begin{bmatrix}0\\0\\0\\0\end{bmatrix}[/latex], [latex]\begin{bmatrix}2\\1\\-1\\3\end{bmatrix}[/latex].
(c) [latex]\begin{bmatrix}2\\1\\-1\end{bmatrix}[/latex], [latex]\begin{bmatrix}3\\-6\\9\end{bmatrix}[/latex], [latex]\begin{bmatrix}-2\\4\\-6\end{bmatrix}[/latex].
Theorem: The following are equivalent for an [latex]n \times n[/latex] matrix [latex]A[/latex]:
1. [latex]A[/latex] is invertible.
2. The column set of [latex]A[/latex] is linearly independent.
3. The column space of [latex]A[/latex] is [latex]\mathbb{R}^n[/latex].
4. The row set of [latex]A[/latex] is linearly independent.
5. The row space of [latex]A[/latex] span the set of all [latex]1\times n[/latex] rows.
Example 4: Find the value of [latex]h[/latex] such that the columns of [latex]A = \begin{bmatrix}1 & -3 & 2\\1 & 2 & h\\-5 & -5 & 6\end{bmatrix}[/latex] is linearly dependent.
Exercise 4: Find the value of [latex]h[/latex] such that the columns of [latex]A = \begin{bmatrix}2 & -3 & h\\1 & -2 & 2\\-5 & 1 & 6\end{bmatrix}[/latex] is linearly dependent.
Definition: A basis for a subspace [latex]H[/latex] of [latex]\mathbb{R}^n[/latex] is a linearly independent set in [latex]H[/latex] that spans [latex]H[/latex].
Fact: The columns of an invertible [latex]n \times n[/latex] matrix form a basis of [latex]\mathbb{R}^n[/latex] because they are linearly independent and span [latex]\mathbb{R}^n[/latex].
Definition: The columns of [latex]n \times n[/latex] identity matrix [latex]\vec{e_{1}} = \begin{bmatrix}1\\0\\0\\\vdots\\0\end{bmatrix}, \vec{e_{2}} = \begin{bmatrix}0\\1\\0\\\vdots\\0\end{bmatrix}, \cdots, \vec{e_{n}} = \begin{bmatrix}0\\0\\0\\\vdots\\1\end{bmatrix}[/latex] form a basis of [latex]\mathbb{R}^n[/latex]. The set [latex]\begin{Bmatrix}\vec{e_{n}}, \cdots, \vec{e_{n}}\end{Bmatrix}[/latex] is called standard basis of [latex]\mathbb{R}^n[/latex].
Theorem: The pivot columns of a matrix [latex]A[/latex] form a basis for the column space of [latex]A[/latex].
Definition: The dimension of a nonzero subspace [latex]H[/latex], denoted by dim [latex]H[/latex], is the number of vectors in any basis for [latex]H[/latex]. The dimension of the zero subspace [latex]\begin{Bmatrix}\vec{0}\end{Bmatrix}[/latex] is defined to be zero.
The Basis Theorem: If [latex]H[/latex] is a [latex]p[/latex]-dimensional subspace of [latex]\mathbb{R}^n[/latex], then any linearly independent set of exactly [latex]p[/latex] elements in [latex]H[/latex] is automatically a basis for [latex]H[/latex]. Also, any set of [latex]p[/latex] elements of [latex]H[/latex] that spans [latex]H[/latex] is automatically a basis for [latex]H[/latex].
Example 5: Find a basis and calculate the dimension of the following
subspaces of [latex]\mathbb{R}^n[/latex]:
[latex]U=\left\{\begin{bmatrix}a\\a+b\\a-c\\b\end{bmatrix}|a,b,c\text{ in }\mathbb{R}\right\}[/latex],
[latex]V=\left\{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}|a + b - c + 2d = 0\text{ in }\mathbb{R}\right\}[/latex].
Exercise 5: Find a basis and calculate the dimension of the following
subspaces of [latex]\mathbb{R}^n[/latex]:
[latex]U=\left\{\begin{bmatrix}a\\a+b\\a-2c\\c\end{bmatrix}|a,b,c\text{ in }\mathbb{R}\right\}[/latex],
[latex]V=\left\{\begin{bmatrix}a\\b\\c\\d\end{bmatrix}|a - 2b + c + d = 0\text{ in }\mathbb{R}\right\}[/latex].
GroupWork 1: Mark each statement True or False. Justify each answer.
a. If [latex]B[/latex] is an echelon form of a matrix [latex]A[/latex], then the pivot columns of [latex]B[/latex] form a basis for Col[latex]A[/latex].
b. Row operations do not affect linear dependence relations among the
columns of a matrix.
c. The columns of a matrix [latex]A[/latex] are linearly independent if the equation [latex]A\vec{x} = \vec{0}[/latex] has trivial solution only.
d. The columns of any [latex]4 \times 5[/latex] matrix are linearly dependent.
e. If [latex]\vec{u}[/latex] and [latex]\vec{v}[/latex] are linearly independent and if [latex]\vec{w}[/latex] is in Span[latex]\left\{ \vec{u},\vec{v}\right \}[/latex] then [latex]\left\{ \vec{u},\vec{v},\vec{w}\right \}[/latex] is linearly dependent.
f. If three vectors in [latex]\mathbb{R}^3[/latex] lie on the same plane, then they are linearly dependent.
g. If a set contains fewer vectors than there are entries in the vectors, then
they are linearly independent.
h. If a set in [latex]\mathbb{R}^n[/latex] is linearly dependent, then it contains more than [latex]n[/latex] vectors.
GroupWork 2: Describe the possible echelon form of the matrix.
(a) [latex]A[/latex] is a [latex]2 \times 2[/latex] matrix with linearly independent columns.
(b) [latex]A[/latex] is a [latex]4 \times 2[/latex] matrix such that the first column is the multiple of the second column.
GroupWork 3: In each case show that the statement is true, or give an example showing that it is false.
a. If [latex]\left\{ \vec{u},\vec{v}\right \}[/latex] is independent, then [latex]\left\{ \vec{u},\vec{v}, \vec{u} + \vec{v}\right \}[/latex] is independent.
b. If [latex]\left\{ \vec{u},\vec{v},\vec{w}\right \}[/latex] is independent, then [latex]\left\{ \vec{u},\vec{v}\right \}[/latex] is independent.
c. If [latex]\left\{ \vec{u},\vec{v}\right \}[/latex] is dependent, then [latex]\left\{ \vec{u},\vec{v},\vec{w}\right \}[/latex] is dependent for any [latex]\vec{w}[/latex].
GroupWork 5: How many pivot columns must be a [latex]6 \times 4[/latex] matrix have if its columns are linearly independent.
GroupWork5: How many pivot columns must be a [latex]4 \times 6[/latex] matrix have if its columns span [latex]\mathbb{R}^4[/latex]? Why?