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Linear Algebra for Team-Based Inquiry Learning
2022 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
University of South Alabama |
|
|
August 2, 2022
Section 2.9: Homogeneous Linear Systems (VS9)
Definition 2.9.1
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 2.9.2 (~5 min)
Note that if \(\left[\begin{array}{c} a_1 \\ \vdots \\ a_n \end{array}\right] \) and \(\left[\begin{array}{c} b_1 \\ \vdots \\ b_n \end{array}\right] \) are solutions to \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\) so is \(\left[\begin{array}{c} a_1 +b_1\\ \vdots \\ a_n+b_n \end{array}\right] \text{,}\) since
\begin{equation*}
a_1 \vec{v}_1+\cdots+a_n \vec{v}_n = \vec{0}
\text{ and }
b_1 \vec{v}_1+\cdots+b_n \vec{v}_n = \vec{0}
\end{equation*}
implies
\begin{equation*}
(a_1 + b_1) \vec{v}_1+\cdots+(a_n+b_n) \vec{v}_n = \vec{0} .
\end{equation*}
Similarly, if \(c \in \IR\text{,}\) \(\left[\begin{array}{c} ca_1 \\ \vdots \\ ca_n \end{array}\right] \) is a solution. Thus the solution set of a homogeneous system is...
A basis for \(\IR^n\text{.}\)
A subspace of \(\IR^n\text{.}\)
The empty set.
Activity 2.9.3 (~10 min)
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*}
Part 1.
Find its solution set (a subspace of \(\IR^4\)).
Activity 2.9.3 (~10 min)
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*}
Part 2.
Rewrite this solution space in the form
\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*}
Activity 2.9.3 (~10 min)
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*}
Part 3.
Rewrite this solution space in the form
\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\}.
\end{equation*}
Fact 2.9.4
The coefficients of the free variables in the solution set of a linear system always yield linearly independent vectors.
Thus if
\begin{equation*}
\setBuilder{
a \left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right] +
b \left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]
}{
a,b \in \IR
} = \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*}
is the solution space for a homogeneous system, then
\begin{equation*}
\setList{
\left[\begin{array}{c} -2 \\ 1 \\ 0 \\ 0\end{array}\right],
\left[\begin{array}{c} -1 \\ 0 \\ -4 \\ 1 \end{array}\right]
}
\end{equation*}
is a basis for the solution space.
Activity 2.9.5 (~10 min)
Consider the homogeneous system of equations
\begin{alignat*}{5}
2x_1&\,+\,&4x_2&\,+\,& 2x_3&\,-\,&4x_4 &=& 0 \\
-2x_1&\,-\,&4x_2&\,+\,&x_3 &\,+\,& x_4 &=& 0\\
3x_1&\,+\,&6x_2&\,-\,&x_3 &\,-\,&4 x_4 &=& 0
\end{alignat*}
Find a basis for its solution space.
Activity 2.9.6 (~10 min)
Consider the homogeneous vector equation
\begin{equation*}
x_1 \left[\begin{array}{c} 2 \\ -2 \\ 3 \end{array}\right]+
x_2 \left[\begin{array}{c} 4 \\ -4 \\ 6 \end{array}\right]+
x_3 \left[\begin{array}{c} 2 \\ 1 \\ -1 \end{array}\right]+
x_4 \left[\begin{array}{c} -4 \\ 1 \\ -4 \end{array}\right]=
\left[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}\right]
\end{equation*}
Find a basis for its solution space.
Activity 2.9.7 (~5 min)
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*}
Find a basis for its solution space.
Observation 2.9.8
The basis of the trivial vector space is the empty set. You can denote this as either \(\emptyset\) or \(\{\}\text{.}\)
Thus, if \(\vec{0}\) is the only solution of a homogeneous system, the basis of the solution space is \(\emptyset\text{.}\)