Definition 1.5.1.
A system \(AX = B\) of linear equations is called homogeneous if \(B = 0\text{.}\) Otherwise it is called a non homogeneous system.
Section 1.5 Homogeneous System of Linear Equations
Observe that any homogeneous system \(AX = 0\) of linear equations has at least one solution, namely the trivial solution (\(x_i = 0\text{,}\) for all \(i\)). The augmented matrix is got from adding one zero column to \(A\text{.}\) Hence \(.r(A) =
r(A^*)\text{.}\)
Theorem 1.5.2.
A homogeneous system \(AX = 0\) of \(n\) equations in the same number of unknowns (\(n\) unknowns) has a nontrivial solution if and only if \(A\) is singular, that is , if and only if \(\det(A)=0\text{.}\)
Proof.
The given homogeneous system has a nontrivial solution if and only if it has infinitely many solutions, if and only if \(r(A)\lt n\text{,}\) the number of variables (Theorem 1.4.8). But \(r(A)\lt n \text{,}\) if and only if \(A\) is singular.
Theorem 1.5.3.
A homogeneous system \(AX = 0\) of \(m\) equations in \(n\) unknowns, where \(m \lt n\) has infinitely many nontrivial solutions.
Proof.
We would like to get the connection between solutions of a homogeneous system \(AX = 0\) and the solutions of \(AX = B\text{,}\) \(B \neq 0\text{.}\) These are stated in the following Theorem.
Theorem 1.5.4.
Suppose the non homogeneous system \(AX
= B\) of linear equations has a solution say \(X = X_0\text{.}\) All the solutions of this system are of the form
\begin{equation*}
X_0 + X_h,
\end{equation*}
where \(X_h\) runs through all the solutions of the corresponding homogeneous system \(AX =0\text{.}\)
Proof.
Let \(X_h\) be a solution of \(AX=0\text{.}\) Then
\begin{equation*}
A(X_0 + X_h) = AX_0 + A
X_h = B + 0 = B\text{.}
\end{equation*}
Hence \(X_0 +X_h\) is a solution of \(AX =
B\text{.}\)
Suppose \(Y\) is a solution of \(AX = B\text{,}\) i.e. \(AY = B\text{.}\) Then \(A (Y - X_0) = 0\) and hence \(Y-X_0\) is a solution of \(AX = 0\) and \(Y = X_0+(Y-X_0)=X_0+X_h\text{,}\) where \(X_h=Y-X_0\text{.}\)
The above theorem can be written equivalently as
Theorem 1.5.5.
If \((\alpha_1 ,\ldots, \alpha_n)\) is a solution of a system of linear equations, then the complete solution is given by \(x_1 = \alpha_1 + y_1 ,\ldots , x_n = \alpha_n + y_n\) , where \((y_1 ,\ldots, y_n)\) is the general solution of the associated homogeneous system.
Example 1.5.6.
Let us consider one equation in two variables, \(2x_1+3x_2=5\text{.}\) The corresponding homogeneous system is \(2x_1+3x_2=0\text{.}\) It is easy see that \(\left\{\begin{pmatrix} \alpha\\-\frac{2}{3}\alpha\end{pmatrix}:\alpha\in \R\right\}\) is set of all solutions of \(2x_1+3x_2=0\text{.}\) Let us consider a solution \(X_0=\begin{pmatrix} 0\\\frac{5}{3}\end{pmatrix}\) of the non homogeneous system \(2x_1+3x_2=5\text{.}\) Let \(X=\begin{pmatrix} \alpha\\-\frac{2}{3}\alpha\end{pmatrix}\) be any solution of \(2x_1+3x_2=0\text{.}\) Then it is easy to check that \(X_0+X=\begin{pmatrix} \alpha\\\frac{5}{3}-\frac{2}{3}\alpha\end{pmatrix}\) is a solution of \(2x_1+3x_2=5\text{.}\) Alternatively, to solve \(2x_1+3x_2=5\text{,}\) we can free one of the variable say \(x_1\text{.}\) That is \(x_1\) can take any real value. Let \(x_1=\alpha\text{,}\) then \(x_2=\frac{5}{3}-\frac{2}{3}\alpha\text{.}\) Hence a generic solution of \(2x_1+3x_2=5\) is given by \(\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix} \alpha\\\frac{5}{3}-\frac{2}{3}\alpha\end{pmatrix}\text{,}\) which can be rewritten as \(\begin{pmatrix} \alpha\\\frac{5}{3}-\frac{2}{3}\alpha\end{pmatrix}=\begin{pmatrix} 0\\\frac{5}{3}\end{pmatrix}+\begin{pmatrix} \alpha\\-\frac{2}{3}\alpha\end{pmatrix}
\text{.}\) Thus all solutions of \(2x_1+3x_2=5\) are of the form
\begin{equation*}
\left\{\begin{pmatrix} 0\\\frac{5}{3}\end{pmatrix}+\begin{pmatrix} \alpha\\-\frac{2}{3}\alpha\end{pmatrix} :\alpha \in \R\right\}=\left\{\begin{pmatrix} 0\\\frac{5}{3}\end{pmatrix}+\alpha\begin{pmatrix} 1\\-\frac{2}{3}\end{pmatrix} :\alpha \in \R\right\}.
\end{equation*}
Notice that tho two lines, \(2x_1+3x_2=5\) and \(2x_1+3x_2=0\) are parallel to each other. Plot the figure in Sage using the following Sage sybtax.
Example 1.5.7.
Let consider 2 equations in three variables.
\begin{equation}
2x_1-x_2-x_3=5; x_1+x_2-2x_3=7 \tag{1.5.1}
\end{equation}
The corresponding homogeneous system is given by
\begin{equation}
2x_1-x_2-x_3=0; x_1+x_2-2x_3=0 \text{.}\tag{1.5.2}
\end{equation}
It is easy to see that the set of all solutions of the homogeneous system (1.5.2) is
\begin{equation*}
\begin{bmatrix} x_1\\x_2\\ x_3\end{bmatrix}=\left\{
\begin{bmatrix} \alpha\\ \alpha\\ \alpha\end{bmatrix}:
\alpha\in \R
\right\}
\end{equation*}
(1.5.1), we get \(3x_1-3x_3=12\) which implies \(x_1=4+x_3\text{.}\) Now substituting \(x_1=4+x_3\) in the second equation of (1.5.1), we get \(x_2=3+x_3\text{.}\) Here we can take \(x_3\) as a free variable, hence any solution of~(1.5.1) is of the form
\begin{equation*}
\begin{bmatrix}4+\alpha\\3+\alpha\\\alpha\end{bmatrix}
\end{equation*}
which can be written as
\begin{equation*}
\begin{bmatrix}4\\3\\0\end{bmatrix}+\begin{bmatrix}\alpha\\\alpha\\\alpha\end{bmatrix}\text{.}
\end{equation*}
Note that \(\begin{bmatrix}4\\3\\0\end{bmatrix}\) is a solution (1.5.1). Thus set of all solution of (1.5.1) is of the form
\begin{equation*}
\begin{bmatrix} x_1\\x_2\\ x_3\end{bmatrix}=\left\{
\begin{bmatrix}4\\3\\0\end{bmatrix}+\begin{bmatrix} \alpha\\ \alpha\\ \alpha\end{bmatrix}:
\alpha\in \R
\right\}
\end{equation*}
Geometrically, each equation in the non homogeneous system represents a plane and the solution is the line of intersection of the the two planes. Similarly, each equation of the corresponding homogeneous system represents a plane passing through the origin and parallel to the corresponding plane of the non homogeneous system. Thus the solution of the corresponsing homogeneosu system in the line of intersection of the the two planes passing through the origin. Thus the solutions of the system and the corresponsing homogeneous system are parallel lines. Visualize this using the following Sage syntax.
