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Linear Algebra for Team-Based Inquiry Learning PREVIEW Edition

Steven Clontz, Drew Lewis

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*}
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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*}
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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;
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  • The sum of any two solutions is again a solution;
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  • Multiplying a solution by a scalar produces another solution.
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(b)
Based on this recollection, which of the following best describes the solution set to the homogeneous equation?
  1. A basis for \(\IR^n\text{.}\)
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  2. A subspace of \(\IR^n\text{.}\)
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  3. All of \(\IR^n\text{.}\)
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  4. The empty set.
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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*}
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(a)
Find its solution set (a subspace of \(\IR^4\)).
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(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*}
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(c)
So, how can we use this to find a basis for the solution space?
  1. Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot row.
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  2. Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot column.
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  3. Take RREF of an appropriate matrix and remove the vectors that caused a non-pivot row or column.
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  4. This set cannot be a basis for the solution space because it will always have in a zero row in the RREF.
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Answer.
B.
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This is exactly the technique used in Section 2.6 to find a basis from a set of spanning vectors.
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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.
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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*}
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(a)
Find its solution set (a subspace of \(\IR^7\)).
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(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*}
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(c)
Find a basis for this solution space.
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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*}
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(b)
Which of these is the best choice of basis for this solution space?
  1. \(\displaystyle \{\}\)
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  2. \(\displaystyle \{\vec 0\}\)
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  3. The basis does not exist
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Answer.
A.
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In Section 2.4 we established that sets that contain the zero vector cannot be independent.
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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 \(\left[\begin{array}{c}x+y\\y-z\end{array}\right]\in\mathbb R^2\text{.}\)
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(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?
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Answer.
\begin{equation*} x+y=0 \end{equation*}
\begin{equation*} y-z=0 \end{equation*}
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(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?
  1. A point
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  2. A line
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  3. A plane
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Answer.
B.
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The basis \(\left\{\left[\begin{array}{c}-1\\1\\1\end{array}\right]\right\}\) describes a one-dimensional line.
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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.
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The matrix \(A\) is linearly independent because \(\RREF(A)\) has a pivot in each column.
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(b)
The matrix \(A\) does not span \(\IR^4\) because \(\RREF(A)\) has a row of zeroes.
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The set of vectors \(S\) spans.
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(d)
The set of vectors \(S\) is a basis.
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Subsection 2.7.4 Videos

Figure 23. Video: Polynomial and matrix calculations
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Subsection 2.7.5 Exercises

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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.
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  • 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.
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  • Prove that the columns of \(M\) form a basis for \(\mathbb{R}^n\text{.}\)
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  • Prove that the reduced row echelon form of \(M\) has \(n\) pivots.
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Subsection 2.7.7 Sample Problem and Solution

Sample problem Example B.1.11.
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