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.
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 homogenous 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.
Based on this recollection, which of the following best describes the solution set to the homogenous 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 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*}
(c)
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*}
(d)
Which of these choices best describes the set of two vectors \(\left\{\left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown\end{array}\right], \left[\begin{array}{c} \unknown \\ \unknown \\ \unknown \\ \unknown \end{array}\right]\right\}\) used in this span?
- The set is linearly dependent.
- The set is linearly independent.
- The set spans all of \(\IR^4\text{.}\)
- The set fails to span the solution space.
Fact 2.7.4.
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\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 2.7.5.
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 2.7.6.
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 2.7.7.
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
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 tranformation that maps visual information at the point \((x,y,z)\in\mathbb R^3\) onto the pixel located at \((x+y,y-z)\in\mathbb R^2\text{.}\)
(a)
What homoegeneous linear system describes the positions \((x,y,z)\) within the original scene that would be aligned with the pixel \((0,0)\) on the screen?
(b)
Solve this system to describe these locations.
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)
The matrix \(A\) is linearly independent because \(\RREF(A)\) has a pivot in each column.
(b)
The matrix \(A\) does not span \(\IR^4\) because \(\RREF(A)\) has a row of zeroes.
(c)
The set of vectors \(S\) spans.
(d)
The set of vectors \(S\) is a basis.
Subsection 2.7.4 Videos
Exercises 2.7.5 Exercises
Exercises available at
https://tbil.org/linear-algebra/preview/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 rank of \(M\) is \(n\text{.}\)
Subsection 2.7.7 Sample Problem and Solution
Sample problem Example C.1.11.
