Skip to main content

Section 2.9 Homogeneous Linear Systems (V9)

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 2.9.1.

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 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...

  1. A basis for \(\IR^n\text{.}\)

  2. A subspace of \(\IR^n\text{.}\)

  3. The empty set.

Activity 2.9.2.

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*}
Activity 2.9.3.

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 2.9.4.

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 2.9.5.

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 2.9.3.

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 \(\emptyset\text{.}\)

Exercises 2.9.1 Exercises

1.

Consider the homogeneous system of equations

\begin{alignat*}{6} x_{1} &-& 2 \, x_{2} &-& x_{3} &+& 3 \, x_{4} &+& 4 \, x_{5} &=& 0\\ 2 \, x_{1} &-& 3 \, x_{2} &-& 3 \, x_{3} &+& 5 \, x_{4} &+& 6 \, x_{5} &=& 0\\ &-& 5 \, x_{2} &+& 5 \, x_{3} &+& 5 \, x_{4} &+& 10 \, x_{5} &=& 0 \end{alignat*}
  1. Find the solution space of this system.
  2. Find a basis of the solution space.
Answer
\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccccc|c} 1 & -2 & -1 & 3 & 4 & 0 \\ 2 & -3 & -3 & 5 & 6 & 0 \\ 0 & -5 & 5 & 5 & 10 & 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & -3 & 1 & 0 & 0 \\ 0 & 1 & -1 & -1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  1. The solution space is \(\left\{ \left[\begin{array}{c} 3 \, a - b \\ a + b + 2 \, c \\ a \\ b \\ c \end{array}\right] \middle|\,a\text{\texttt{,}}b\text{\texttt{,}}c\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \(\left\{ \left[\begin{array}{c} 3 \\ 1 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{array}\right] , \left[\begin{array}{c} 0 \\ 2 \\ 0 \\ 0 \\ 1 \end{array}\right] \right\} \text{.}\)
2.

Consider the homogeneous system of equations

\begin{alignat*}{5} & & & & x_{3} &-& 2 \, x_{4} &=& 0\\ x_{1} &-& 3 \, x_{2} &+& 5 \, x_{3} &-& 12 \, x_{4} &=& 0\\ & & &-& 4 \, x_{3} &+& 8 \, x_{4} &=& 0\\ x_{1} &-& 3 \, x_{2} &+& 3 \, x_{3} &-& 8 \, x_{4} &=& 0 \end{alignat*}
  1. Find the solution space of this system.
  2. Find a basis of the solution space.
Answer
\begin{equation*} \operatorname{RREF} \left[\begin{array}{cccc|c} 0 & 0 & 1 & -2 & 0 \\ 1 & -3 & 5 & -12 & 0 \\ 0 & 0 & -4 & 8 & 0 \\ 1 & -3 & 3 & -8 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & -3 & 0 & -2 & 0 \\ 0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  1. The solution space is \(\left\{ \left[\begin{array}{c} 3 \, a + 2 \, b \\ a \\ 2 \, b \\ b \end{array}\right] \middle|\,a\text{\texttt{,}}b\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \(\left\{ \left[\begin{array}{c} 3 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \\ 1 \end{array}\right] \right\} \text{.}\)
3.

Consider the homogeneous system of equations

\begin{alignat*}{4} 3 \, x_{1} &-& 10 \, x_{2} & & &=& 0\\ -2 \, x_{1} &+& 7 \, x_{2} & & &=& 0\\ x_{1} &+& 2 \, x_{2} & & &=& 0\\ 2 \, x_{1} &-& 7 \, x_{2} & & &=& 0\\ 3 \, x_{1} &-& 9 \, x_{2} & & &=& 0 \end{alignat*}
  1. Find the solution space of this system.
  2. Find a basis of the solution space.
Answer
\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc|c} 3 & -10 & 0 & 0 \\ -2 & 7 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 2 & -7 & 0 & 0 \\ 3 & -9 & 0 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  1. The solution space is \(\left\{ \left[\begin{array}{c} 0 \\ 0 \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \(\left\{ \left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right] \right\} \text{.}\)
4.

Consider the homogeneous system of equations

\begin{alignat*}{5} x_{1} &-& x_{2} &-& x_{3} &-& 8 \, x_{4} &=& 0\\ 2 \, x_{1} &-& x_{2} &-& 5 \, x_{3} &-& 12 \, x_{4} &=& 0\\ &-& 3 \, x_{2} &+& 9 \, x_{3} &-& 11 \, x_{4} &=& 0\\ -x_{1} &+& x_{2} &+& x_{3} &+& 7 \, x_{4} &=& 0 \end{alignat*}
  1. Find the solution space of this system.
  2. Find a basis of the solution space.
Answer
\begin{equation*} \operatorname{RREF} \left[\begin{array}{cccc|c} 1 & -1 & -1 & -8 & 0 \\ 2 & -1 & -5 & -12 & 0 \\ 0 & -3 & 9 & -11 & 0 \\ -1 & 1 & 1 & 7 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & -4 & 0 & 0 \\ 0 & 1 & -3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  1. The solution space is \(\left\{ \left[\begin{array}{c} 4 \, a \\ 3 \, a \\ a \\ 0 \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \(\left\{ \left[\begin{array}{c} 4 \\ 3 \\ 1 \\ 0 \end{array}\right] \right\} \text{.}\)
5.

Consider the homogeneous system of equations

\begin{alignat*}{4} 2 \, x_{1} &-& 2 \, x_{2} & & &=& 0\\ -5 \, x_{1} &+& x_{2} &+& 4 \, x_{3} &=& 0\\ -2 \, x_{1} &+& 3 \, x_{2} &-& x_{3} &=& 0\\ -5 \, x_{1} &-& 2 \, x_{2} &+& 7 \, x_{3} &=& 0\\ x_{1} &-& 2 \, x_{2} &+& x_{3} &=& 0 \end{alignat*}
  1. Find the solution space of this system.
  2. Find a basis of the solution space.
Answer
\begin{equation*} \operatorname{RREF} \left[\begin{array}{ccc|c} 2 & -2 & 0 & 0 \\ -5 & 1 & 4 & 0 \\ -2 & 3 & -1 & 0 \\ -5 & -2 & 7 & 0 \\ 1 & -2 & 1 & 0 \end{array}\right] = \left[\begin{array}{ccc|c} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}
  1. The solution space is \(\left\{ \left[\begin{array}{c} a \\ a \\ a \end{array}\right] \middle|\,a\in\mathbb{R}\right\} \)
  2. A basis of the solution space is \(\left\{ \left[\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right] \right\} \text{.}\)

Additional exercises available at checkit.clontz.org.