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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 2.9.2.

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 2.9.3.

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*}

Activity 2.9.5.

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 2.9.6.

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 2.9.7.

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*}

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 \(\emptyset\text{.}\)

Subsection 2.9.1 Videos

Figure 2.9.9. Video: Polynomial and matrix calculations

Exercises 2.9.2 Exercises

Exercises available at checkit.clontz.org 1 .

https://checkit.clontz.org/#/banks/tbil-la/V9/