Homogeneous and non-homogeneous systems of lineair equations

Systems of linear equations: Systems of linear equations

Homogeneous and non-homogeneous systems of lineair equations

A system of $m$ linear equations with $n$ unknowns $x_1, \ldots, x_n$ can be reduced by means of elementary operations to the following basic form

$\left\{\;\begin{array}{llllllllllll} a_{11}x_1 \!\!\!\!&+&\!\!\!\! a_{12}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{1n}x_n\!\!\!\!&=&\!\!\!\!b_1\\ a_{21}x_1 \!\!&+&\!\! a_{22}x_2 \!\!&+&\!\! \cdots \!\!&+&\!\!\!\! a_{2n}x_n\!\!\!\!&=&\!\!\!\!b_2\\ \vdots &&\vdots &&&& \vdots&&\!\!\!\!\vdots\\ a_{m1}x_1 \!\!\!\!&+&\!\!\!\! a_{m2}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{mn}x_n\!\!\!\!&=&\!\!\!\!b_m\end{array}\right.$ Here, all $a_{ij}$ and $b_i$ with $1\le i\le m$ and $1\le j\le n$ are real or complex numbers. The numbers $a_{ij}$ are called the coefficients of the system. The numbers $b_i$ are called the right members of the system.

A system of linear equations in which the right-hand sides of the above basic shape are all equal to zero is called a homogeneous system; a general system is called non-homogeneous. The system of equations that is obtained from an inhomogeneous system by replacing the right-hand sides by zero is called the associated homogeneous system. Thus, the associated homogeneous system of the above system of equations is

$\left\{\;\begin{array}{llllllllll} a_{11}x_1 \!\!\!\!&+&\!\!\!\! a_{12}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{1n}x_n\!\!\!\!&=&\!\!\!\!0\\ a_{21}x_1 \!\!&+&\!\! a_{22}x_2 \!\!&+&\!\! \cdots \!\!&+&\!\!\!\! a_{2n}x_n\!\!\!\!&=&\!\!\!\!0\\ \vdots &&\vdots &&&& \vdots&&\!\!\!\!\vdots\\ a_{m1}x_1 \!\!\!\!&+&\!\!\!\! a_{m2}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{mn}x_n\!\!\!\!&=&\!\!\!\!0\end{array}\right.$

Consider the following system of linear equations in the unknowns $x$ , $y$ , and $z$ :

$\lineqs{x-y +z &=& 9\cr x+y+z &=& 1\cr}$

This is a system of $2$ linear equations with $3$ unknowns. Instead of $x_1$ , $x_2$ , $x_3$ we have named the unknowns $x$ , $y$ , $z$ . Here, the order is somewhat less obvious. In particular, it must be indicated.

The associated homogeneous system is

$\lineqs{x-y +z &=& 0\cr x+y+z &=& 0\cr}$

Each equation is written in a form which differs only from the previously discussed basic form in that the constants are now on the right-hand side of the equation, whereas before they were on the left.

Moreover, in each equation of the system, the unknowns appear in the same order.

General and particular solution

The solutions of an inhomogeneous system of linear equations are of the form $p+H$ , where $p$ is a fixed solution of the inhomogeneous system (the so-called particular solution) and $H$ runs over all solutions of the associated homogeneous system (that is, $H$ is the general solution of the associated homogeneous system).

Consider the following system of linear equations in the unknowns $x$ , $y$ , and $z$ :

$\lineqs{x-y +z \!\!\!\!&=&\!\!\!\! 9\cr x+y+z \!\!\!\!&=&\!\!\!\! 1\cr}$ The system is non-homogeneous because the right people are not (all) equal to $0$ .

A particular solution of this system is $\rv{x,y,z} = \rv{5,-4,0}$ .

The associated homogeneous system is

$\lineqs{x-y +z \!\!\!\!&=&\!\!\!\! 0\cr x+y+z \!\!\!\!&=&\!\!\!\! 0\cr}$ The solution of this homogeneous system is $\rv{x,y,z} = \rv{-z,0,z}$ .

According to the proposition, the (overall) solution of the inhomogeneous system

$\rv{x,y,z} = \rv{5,-4,0}+\rv{-z,0,z} = \rv{5-z,-4,z}$

For the math enthusiast we present the proof for the proposition that every solution of a non-homogeneous system, to write linear equations is as the sum of a particular solution of the given system and a solution of the associated homogeneous system.

Suppose that $O$ is the general solution of the non-homogeneous system

$\left\{\;\begin{array}{lllllllll} a_{11}x_1 \!\!\!\!&+&\!\!\!\! a_{12}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{1n}x_n\!\!\!\!&=&\!\!\!\!b_1\\ a_{21}x_1 \!\!&+&\!\! a_{22}x_2 \!\!&+&\!\! \cdots \!\!&+&\!\! a_{2n}x_n\!\!\!\!&=&\!\!\!\!b_2\\ \vdots &&\vdots &&&& \vdots&&\!\!\!\!\vdots\\ a_{m1}x_1 \!\!\!\!&+&\!\!\!\! a_{m2}x_2 \!\!\!\!&+&\!\!\!\! \cdots \!\!\!\!&+&\!\!\!\! a_{mn}x_n\!\!\!\!&=&\!\!\!\!b_m\end{array}\right.$ Suppose that $p\i O$ and $p=(p_1,p_2,\cdots,p_n)$ . Then for each $i=1,2,\ldots,m$ :

$a_{i1}p_1+a_{i2}p_2+\cdots+a_{in}p_n=b_i$ If $h=\rv{h_1,h_2,\ldots,h_n}$ belongs to $H$ , then, for each $i$ with $1\le i \le m$ ,

$a_{i1}h_1+a_{i2}h_2+\cdots+a_{in}h_n=0$ So, for each $i=1,2,\ldots,m$ :

$\begin{array}{ll} a_{i1}(p_1+h_1)+a_{i2}(p_2+h_2)+\cdots+a_{in}(p_n+h_n) & \\ \quad = a_{i1}p_1+a_{i1}h_1+a_{i2}p_2+a_{i2}h_2+\cdots+a_{in}p_n+a_{in}h_n & \\ \quad = (a_{i1}p_1+a_{i2}p_2+\cdots+a_{in}p_n) +(a_{i1}h_1+a_{i2}h_2+\cdots+a_{in}h_n) & \\ \quad = b_i+0 & \\ \quad = b_i & \end{array}$ In other words, $p+h$ is a solution of the non-homogeneous system, i.e., $p+h\in O$ . Thus:

$p+H\subseteq O$ In words, this proves that each sum of a particular solution and a homogeneous solution is a solution of the original system.

Conversely, suppose that $o=\rv{o_1, o_2, \ldots, o_n}$ is an arbitrary solution of the non-homogeneous system. Then for each $i$ with $1\le i \le m$ :

$a_{i1}o_1+a_{i2}o_2+\cdots+a_{in}o_n=b_i$ Note that $o=p+(o-p)$ , where $p$ is the particular solution. We now prove that $o-p$ to $H$ , that is to say, that $o-p$ is a solution of the associated homogeneous system. For each $i$ with $1\le i\le m$ we have

$\begin{array}{ll} a_{i1}(o_1-p_1)+a_{i2}(o_2-p_2)+\cdots+a_{in}(o_n-p_n) & \\ \quad = a_{i1}o_1-a_{i1}p_1+a_{i2}o_2-a_{i2}p_2+\cdots+a_{in}o_n-a_{in}p_n & \\ \quad = (a_{i1}o_1+a_{i2}o_2+\cdots+a_{in}o_n) -(a_{i1}p_1+a_{i2}p_2+\cdots+a_{in}p_n) & \\ \quad = b_i-b_i & \\ \quad = 0 & \end{array}$ It follows that $o\in p+H$ and thus:

$O\subseteq p+H$ The final conclusion is:

$O=p+H$ In words, each solution of the non-homogeneous system is the sum of the particular solution $p$ and a homogeneous solution.

Linear structure of the solution of a homogeneous system of linear equations The sum of two solutions of a homogeneous system of linear equations is also a solution of the same system.

If a solution of a homogeneous system of linear equations is multiplied by a factor, it is also a solution of the same system.

Consider again, the following homogeneous set of linear equations with unknowns $x$ , $y$ , and $z$ :

$\lineqs{x-y +z &=& 0\cr x+y+z &=& 0\cr}$ The solution of this system is $\rv{x,y,z} = \rv{-z,0,z}$ . In particular, each solution is of the form $z\cdot \rv{-1,0,1}$ .

Addition of solutions $z_1\cdot \rv{-1,0,1}$ and $z_2\cdot \rv{-1,0,1}$ gives the solution $(z_1+z_2)\cdot \rv{-1,0,1}$ .

Multiplying the solution $z\cdot \rv{-1,0,1}$ by the scalar $\lambda$ gives the solution $(\lambda \cdot z)\cdot \rv{-1,0,1}$ .

The first statement follows from the previous theorem. After all, if there are two solutions of the homogeneous system, then we can regard the first solution as a particular solution and the second as a solution of the associated homogeneous system. The theorem then indicates that the sum is a solution of the original, in this case homogeneous system.

The second statement also makes use of the fact that the constant term is equal to $0$ . We give the proof for the case of one equation with two unknowns $x$ and $y$ . The general proof is very similar. So, we start from a linear equation of the form $a\cdot x + b\cdot y=0$ , where $a$ and $b$ are constants. If $\rv{x,y}$ is a solution and $\lambda$ constant, then we must derive that $\lambda\cdot \rv{x,y}$ , which equals $\rv{\lambda\cdot x,\lambda\cdot y}$ , is a solution of the given equation. This is indeed the case:

$a\cdot(\lambda\cdot x) + b\cdot (\lambda\cdot y)=\lambda\cdot(a\cdot x + b\cdot y)=\lambda\cdot 0=0$