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Linear Algebra for Team-Based Inquiry Learning
2023 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
|
|
|
August 24, 2023
Section 2.7: Homogeneous Linear Systems (EV7)
Definition 2.7.1
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 2.7.1 (~5 min)
Consider the homogeneous vector equation \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\text{.}\)
Part 1.
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 both solutions, we know that
\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} \text{.}
\end{equation*}
Therefore by adding these equations,
\begin{equation*}
(a_1 + b_1) \vec{v}_1+\cdots+(a_n+b_n) \vec{v}_n = \vec{0}
\end{equation*}
shows that \(\left[\begin{array}{c} a_1+ b_1 \\ \vdots \\ a_n+b_n \end{array}\right] \) is also a solution. Thus the solution set of a homogeneous system is...
Closed under addition.
Not closed under addition.
Linearly dependent.
Linearly independent.
Activity 2.7.1 (~5 min)
Consider the homogeneous vector equation \(x_1 \vec{v}_1 + \cdots+x_n \vec{v}_n = \vec{0}\text{.}\)
Part 2.
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 also closed under scalar multiplication, and therefore...
A basis for \(\IR^n\text{.}\)
A subspace of \(\IR^n\text{.}\)
All of \(\IR^n\text{.}\)
The empty set.
Activity 2.7.2 (~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 2.7.2 (~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 2.7.2 (~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*}
Activity 2.7.2 (~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 4.
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 2.7.2
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 2.7.3 (~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 2.7.4 (~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 2.7.5 (~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*}
Part 1.
Find its solution space.
Activity 2.7.5 (~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*}
Part 2.
Which of these is the best choice of basis for this solution space?
- \(\displaystyle \{\}\)
- \(\displaystyle \{\vec 0\}\)
- The basis does not exist
Activity 2.7.6 (~5 min)
Suppose that in a certain 3D video game, the “camera” aligns the position \((x,y,z)\) within the level onto the pixel located at \((x+y,y-z)\) on the television screen.
Part 1.
What homoegeneous linear system describes the positions within the level that would be aligned with the pixel \((0,0)\) on the screen?
Activity 2.7.6 (~5 min)
Suppose that in a certain 3D video game, the “camera” aligns the position \((x,y,z)\) within the level onto the pixel located at \((x+y,y-z)\) on the television screen.
Part 2.
Solve this system to describe these locations.