Learning Outcomes
-
Find a basis for the solution set of a homogeneous system of equations.
Section 2.7 Homogeneous Linear Systems (EV7)
Subsection 2.7.1 Warmup
Remark 2.7.1.
Recall from Section 2.3 that a homogeneous system of linear equations is one of the form:
\begin{alignat*}{5}
a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& 0 \\
a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& 0 \\
\vdots& &\vdots& && &\vdots&&\vdots\\
a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& 0
\end{alignat*}
This system is equivalent to the vector equation:
\begin{equation*}
x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}
\end{equation*}
and the augmented matrix:
\begin{equation*}
\left[\begin{array}{cccc|c}
a_{11} & a_{12} & \cdots & a_{1n} & 0\\
a_{21} & a_{22} & \cdots & a_{2n} & 0\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn} & 0
\end{array}\right].
\end{equation*}
Activity 2.7.2.
(a)
In Section 2.3, we observed that if
\begin{equation*}
x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}
\end{equation*}
is a homogeneous vector equation, then:
-
The zero vector \(\vec{0}\) is a solution;
-
The sum of any two solutions is again a solution;
-
Multiplying a solution by a scalar produces another solution.
(b)
Based on this recollection, which of the following best describes the solution set to the homogeneous equation?
-
A basis for \(\IR^n\text{.}\)
-
A subspace of \(\IR^n\text{.}\)
-
All of \(\IR^n\text{.}\)
-
The empty set.
Subsection 2.7.2 Class Activities
Activity 2.7.3.
Consider the homogeneous system of equations
\begin{alignat*}{5}
x_1&\,+\,&2x_2&\,\,& &\,+\,& x_4 &=& 0\\
2x_1&\,+\,&4x_2&\,-\,&x_3 &\,-\,&2 x_4 &=& 0\\
3x_1&\,+\,&6x_2&\,-\,&x_3 &\,-\,& x_4 &=& 0
\end{alignat*}
(a)
Find its solution set (a subspace of \(\IR^4\)).
(b)
Rewrite this solution space in the following forms
\begin{equation*}
\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right] }{a,b \in \IR}
\end{equation*}
\begin{equation*}
=
\vspan \left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}\text{.}
\end{equation*}
(c)
So, how can we use this to find a basis for the solution space?
-
Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot row.
-
Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot column.
-
Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot row or column.
-
This set cannot be a basis for the solution space because it will always have in a zero row in the RREF.
Answer.
B.
This is exactly the technique used in Section 2.6 to find a basis from a set of spanning vectors.
Observation 2.7.4.
To find a basis for the subspace
\begin{equation*}
\vspan \left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\}
\end{equation*}
we may compute
\begin{equation*}
\left[\begin{array}{cc} -2 & -1 \\ 1 & 0 \\ 0 & -4 \\ 0 & 1\end{array}\right] \sim
\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0\end{array}\right]\text{.}
\end{equation*}
Because all columns are pivot columns, the set
\begin{equation*}
\left\{\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right], \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]\right\}
\end{equation*}
is already linearly independent and therefore already a basis for the subspace.
Activity 2.7.5.
Consider the homogeneous system of equations
\begin{alignat*}{8}
2x_1&\,+\,&4x_2&\,+\,&2x_3 &\,-\,&3 x_4 &\,+\,&31x_5&\,+\,&2x_6&\,-\,&16x_7&=& 0\\
-1x_1&\,-\,&2x_2&\,+\,&4x_3 &\,-\,&x_4 &\,+\,&2x_5&\,+\,&9x_6&\,+\,&3x_7&=& 0\\
x_1&\,+\,&2x_2&\,+\,&x_3 &\,+\,& x_4 &\,+\,&3x_5&\,+\,&6x_6&\,+\,&7x_7&=& 0
\end{alignat*}
(a)
Find its solution set (a subspace of \(\IR^7\)).
(b)
Rewrite this solution space in the following forms:
\begin{equation*}
\setBuilder{ a \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] + b \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+c \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]+d \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right] }{a,b,c,d \in \IR}.
\end{equation*}
\begin{equation*}
=\vspan\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right],\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\\ \unknown\\ \unknown \\ \unknown\end{array}\right]\right\}\text{.}
\end{equation*}
(c)
Find a basis for this solution space.
Fact 2.7.6.
The coefficients of the free variables in the solution space of a linear system always yield linearly independent vectors that span the solution space.
Thus if
\begin{equation*}
\setBuilder{
a \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\\0\\0\\0\end{array}\right] +
b \left[\begin{array}{c} -7 \\ 0 \\ -1 \\ 5\\1\\0\\0 \end{array}\right]+
c \left[\begin{array}{c} -1 \\ 0 \\ -3 \\ -2\\0\\1\\0 \end{array}\right]+
d \left[\begin{array}{c} 1 \\ 0 \\ -2 \\ -6\\0\\0\\1 \end{array}\right]
}{
a,b,c,d \in \IR
} = \vspan\left\{ \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\\0\\0\\0\end{array}\right],
\left[\begin{array}{c} -7 \\ 0 \\ -1 \\ 5\\1\\0\\0 \end{array}\right],
\left[\begin{array}{c} -1 \\ 0 \\ -3 \\ -2\\0\\1\\0 \end{array}\right],
\left[\begin{array}{c} 1 \\ 0 \\ -2 \\ -6\\0\\0\\1 \end{array}\right] \right\}
\end{equation*}
is the solution space for a homogeneous system, then
\begin{equation*}
\setList{
\left[\begin{array}{c} -2 \\ \textcolor{blue}{1} \\ 0 \\ 0\\\textcolor{blue}{0}\\\textcolor{blue}{0}\\\textcolor{blue}{0}\end{array}\right],
\left[\begin{array}{c} -7 \\ \textcolor{blue}{0} \\ -1 \\ 5\\\textcolor{blue}{1}\\\textcolor{blue}{0}\\\textcolor{blue}{0} \end{array}\right],
\left[\begin{array}{c} -1 \\ \textcolor{blue}{0} \\ -3 \\ -2\\\textcolor{blue}{0}\\\textcolor{blue}{1}\\\textcolor{blue}{0} \end{array}\right],
\left[\begin{array}{c} 1 \\ \textcolor{blue}{0} \\ -2 \\ -6\\\textcolor{blue}{0}\\\textcolor{blue}{0}\\\textcolor{blue}{1} \end{array}\right]
}
\end{equation*}
is a basis for the solution space.
Activity 2.7.7.
Consider the homogeneous system of equations
\begin{alignat*}{5}
x_1&\,-\,&3x_2&\,+\,& 2x_3 &=& 0\\
2x_1&\,+\,&6x_2&\,+\,&4x_3 &=& 0\\
x_1&\,+\,&6x_2&\,-\,&4x_3 &=& 0
\end{alignat*}
(a)
Find its solution space.
(b)
Which of these is the best choice of basis for this solution space?
-
\(\displaystyle \{\}\)
-
\(\displaystyle \{\vec 0\}\)
-
The basis does not exist
Answer.
Activity 2.7.8.
To create a computer-animated film, an animator first models a scene as a subset of \(\mathbb R^3\text{.}\) Then to transform this three-dimensional visual data for display on a two-dimensional movie screen or television set, the computer could apply a linear transformation that maps visual information at the vector \(\left[\begin{array}{c}x\\y\\z\end{array}\right]\in\mathbb R^3\) onto the pixel located at \(\left[\begin{array}{c}x+y\\y-z\end{array}\right]\in\mathbb R^2\text{.}\)
(a)
What homogeneous linear system describes the positions \(\left[\begin{array}{c}x\\y\\z\end{array}\right]\) within the original scene that would be aligned with the pixel \(\left[\begin{array}{c}0\\0\end{array}\right]\) on the screen?
(b)
Find a basis for the solution set of this system. What best describes the shape of the data that gets projected onto this single point of the screen?
-
A point
-
A line
-
A plane
Subsection 2.7.3 Individual Practice
Activity 2.7.9.
Let \(S=\setList{
\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right],
\left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right],
\left[\begin{array}{c} 1 \\ 0 \\ -2 \\ 3 \end{array}\right]
}\) and \(A=\left[\begin{array}{ccc}
-2 & -1 &1\\
1 & 0 &0\\
0 & -4 &-2\\
0 & 1 &3
\end{array}\right];\) note that
\begin{equation*}
\RREF(A)=\left[\begin{array}{ccc}
1 & 0 &0\\
0 & 1 &0\\
0 & 0 &1\\
0 & 0 &0
\end{array}\right].
\end{equation*}
The following statements are all invalid for at least one reason. Determine what makes them invalid and, suggest alternative valid statements that the author may have meant instead.
(a)
(b)
(c)
The set of vectors \(S\) spans.
(d)
The set of vectors \(S\) is a basis.
Subsection 2.7.4 Videos
Subsection 2.7.5 Exercises
Exercises available at
https://tbil.org/preview/linear-algebra/exercises/#/bank/EV7/
Subsection 2.7.6 Mathematical Writing Explorations
Exploration 2.7.10.
An \(n \times n\) matrix \(M\) is non-singular if the associated homogeneous system with coefficient matrix \(M\) is consistent with one solution. Assume the matrices in the writing explorations in this section are all non-singular.
-
Prove that the reduced row echelon form of \(M\) is the identity matrix.
-
Prove that, for any column vector \(\vec{b} = \left[\begin{array}{c}b_1\\b_2\\ \vdots \\b_n \end{array}\right]\text{,}\) the system of equations given by \(\left[\begin{array}{c|c}M & \vec{b}\end{array} \right]\) has a unique solution.
-
Prove that the columns of \(M\) form a basis for \(\mathbb{R}^n\text{.}\)
-
Prove that the reduced row echelon form of \(M\) has \(n\) pivots.
Subsection 2.7.7 Sample Problem and Solution
Sample problem Exercise B.7.
