Homogeneous Equations

Linear Algebra

Homogeneous Equation

A system of equations with the variables $x_1,x_2,...,x_n$ is called [homogeneous] if all the constant terms are [zero].

That is, if each equation of the system has the form:

$a_1{x}_1+a_2{x}_2+...+a_n{x}_n=0$

A [trivial solution] is one where all all values are zero (in the augmented matrix?)

Show that the homogenous system has nontrivial solutions

$$\{\begin{array}{l}{x}_1-x_2+2x_3-x_4=0\\ \\ 2x_1+2x_2+x_4=0\\ \\ 3x_1+x_2+2x_3-x_4=0\end{array}$$ $$\left[\begin{array}{cccc}1& -1& 2& -1|0\\ & & & \\ 2& 2& 0& 1|0\\ & & & \\ 3& 1& 2& -1|0\\ & & & \end{array}\right]$$

r2:r2-2*r1 r3: r3-3*r1

$$\left[\begin{array}{cccc}1& -1& 2& -1|0\\ & & & \\ 0& 4& -4& 3|0\\ & & & \\ 0& 4& -4& 2|0\\ & & & \end{array}\right]$$

r2: 1/4 * r2

$$\left[\begin{array}{cccc}1& -1& 2& -1|0\\ & & & \\ 0& 1& -1& 3/4|0\\ & & & \\ 0& 4& -4& 2|0\\ & & & \end{array}\right]$$

r1: r1+r2 r3: r3-4*r2

$$\left[\begin{array}{cccc}1& 0& 1& -1/4|0\\ & & & \\ 0& 1& -1& 3/4|0\\ & & & \\ 0& 0& 0& -1|0\\ & & & \end{array}\right]$$

r3: -1 * r3

$$\left[\begin{array}{cccc}1& 0& 1& -1/4|0\\ & & & \\ 0& 1& -1& 3/4|0\\ & & & \\ 0& 0& 0& 1|0\\ & & & \end{array}\right]$$

r1=r1+1/4*r3 r2=r2-3/4*r3

$$\left[\begin{array}{cccc}1& 0& 1& 0|0\\ & & & \\ 0& 1& -1& 0|0\\ & & & \\ 0& 0& 0& 1|0\\ & & & \end{array}\right]$$ $$\{\begin{array}{l}{x}_1+x_3=0\\ \\ x_2-x_3=0\\ \\ x_4=0\end{array}$$

hence:

$$\{\begin{array}{l}{x}_1=-t\\ \\ x_2=t\\ \\ x_3=t\\ \\ x_4=0\end{array}$$

nontrivial solutions:

$$\{\begin{array}{l}{x}_1=-1\\ \\ x_2=1\\ \\ x_3=1\\ \\ x_4=0\end{array}$$

^^ not all zeros, so non-trivial.

Theorem 1.3.1

If a homogenous system of linear equations has more variables than equations, then it has [a nontrivial] solution (in fact, [infinitely many||amount])

Proof for theorem 1.3.1

n = number of variables m = number of equations $n>m\ge r$ n > r => infinitely many solutions

Linear combinations and basic solutions

if

$$x=\left[\begin{array}{c}{x}_1\\ \\ x_2\\ \\ ...\\ \\ x_n\end{array}\right]$$

and

$$y=\left[\begin{array}{c}{y}_1\\ \\ y_2\\ \\ ...\\ \\ y_n\end{array}\right]$$ $$x+y=\left[\begin{array}{c}{x}_1\\ \\ x_2\\ \\ ...\\ \\ x_n\end{array}\right]+\left[\begin{array}{c}{y}_1\\ \\ y_2\\ \\ ...\\ \\ y_n\end{array}\right]=\left[\begin{array}{c}{x}_1+y_1\\ \\ x_2+y_2\\ \\ ...\\ \\ x_n+y_n\end{array}\right]$$

Scalar multiple

$$\begin{array}{r}kx=\left[\begin{array}{c}k{x}_1\\ \\ k{x}_2\\ \\ ...\\ \\ k{x}_n\\ \end{array}\right]\end{array}$$

^ Called a scalar multiple.

Sum of columns

Definition

A sum of scalar multiples of several columns is called a [linear combination] of those columns.

For example, sx+ty is a linear combination of x and y. ($s,t\in \mathbb{R}$)

Example

If $x=\left[\begin{array}{c}-5\\ 2\end{array}\right]$ and $y=\left[\begin{array}{c}2\\ \\ -3\end{array}\right]$, then 3x+5y =

$$\begin{array}{r}=\left[\begin{array}{c}-15\\ \\ 6\end{array}\right]+\left[\begin{array}{c}10\\ \\ -15\end{array}\right]=\left[\begin{array}{c}-5\\ \\ -9\end{array}\right]\end{array}$$

—

$$\begin{array}{r}x=\left[\begin{array}{c}1\\ \\ 0\\ \\ 1\end{array}\right]y=\left[\begin{array}{c}2\\ \\ 1\\ \\ 0\end{array}\right]z=\left[\begin{array}{c}3\\ \\ 1\\ \\ 1\end{array}\right]\end{array}$$

if

$$\begin{array}{r}v=\left[\begin{array}{c}0\\ \\ -1\\ \\ 2\end{array}\right]\end{array}$$

determine whether v is a linear combination of x,y & z

Solution: We must determine whether r,s,t exist such that v=rx+sy+tz

$$\begin{array}{r}\left[\begin{array}{c}0\\ \\ -1\\ \\ 2\end{array}\right]=r\left[\begin{array}{c}1\\ \\ 0\\ \\ 1\end{array}\right]+s\left[\begin{array}{c}2\\ \\ 1\\ \\ 0\end{array}\right]+t\left[\begin{array}{c}3\\ \\ 1\\ \\ 1\end{array}\right].\end{array}$$

^^ Which is the same as:

$$\{\begin{array}{l}r+2s+3t=0\\ \\ s+t=-1\\ \\ r+t=2\end{array}$$

Augmented matrix:

$$\left[\begin{array}{ccccc}1& 2& 3& |& 0\\ & & & & \\ 0& 1& 1& |& -1\\ & & & & \\ 1& 0& 1& |& 2\end{array}\right]$$

r3 = r3-r1

$$\left[\begin{array}{ccccc}1& 2& 3& |& 0\\ & & & & \\ 0& 1& 1& |& -1\\ & & & & \\ 0& -2& -2& |& 2\end{array}\right]$$

r3 = r3+2*r2

$$\left[\begin{array}{ccccc}1& 2& 3& |& 0\\ & & & & \\ 0& 1& 1& |& -1\\ & & & & \\ 0& 0& 0& |& 0\end{array}\right]$$

RREF r1=r1-2*r2

$$\left[\begin{array}{ccccc}1& 0& 1& |& 2\\ & & & & \\ 0& 1& 1& |& -1\\ & & & & \\ 0& 0& 0& |& 0\end{array}\right]$$ $$\{\begin{array}{l}r+t=2\\ \\ s+t=-1\end{array}$$

t is a non-leading variable, so set it as a parameter. $t=k$.

$$\{\begin{array}{l}r=2-k\\ \\ s=-1-k\\ \\ t=k\end{array}$$

So let’s try $k=0$

$$\{\begin{array}{l}r=2\\ \\ s=-1\\ \\ t=0\end{array}$$

so V really is a linear combination of x,y & z.

More on basic solutions

When solving a system with n variables ($x_1...x_n$), write the variables as a column matrix: $\left[\begin{array}{c}{x}_1\\ \\ ...\\ \\ x_n\end{array}\right]$.

The trivial solution is noted by: $0=\left[\begin{array}{c}0\\ \\ 0\\ \\ ...\\ \\ 0\end{array}\right]$.

Remark: Let x 7 y be two solutions to a homogeneus system with n variables. Then any linear combination sx+ty of these solutions is a solution to the system. More generally, any linear combination of solutions to a homogenous system is a solution.

Definition

linear combinations of multiple solutions to a [homogenous systems] are also solutions.

Example

Solve the homogenous system w/ coefficient matrix:

$$\begin{array}{r}A=\left[\begin{array}{cccc}1& -2& 3& -2\\ & & & \\ -3& 6& 1& 0\\ & & & \\ -2& 4& 4& -2\\ & & & \end{array}\right]\end{array}$$

.

$$\left[\begin{array}{cccccc}1& -2& 3& -2& |& 0\\ & & & & & \\ -3& 6& 1& 0& |& 0\\ & & & & & \\ -2& 4& 4& -2& |& 0\\ & & & & & \end{array}\right]$$

r2=r2+3*r1; r3=r3+2*r1

$$\left[\begin{array}{cccccc}1& -2& 3& -2& |& 0\\ & & & & & \\ 0& 0& 10& -6& |& 0\\ & & & & & \\ 0& 0& 10& -6& |& 0\\ & & & & & \end{array}\right]$$

r3 = r3-r2

$$\left[\begin{array}{cccccc}1& -2& 3& -2& |& 0\\ & & & & & \\ 0& 0& 10& -6& |& 0\\ & & & & & \\ 0& 0& 0& 0& |& 0\\ & & & & & \end{array}\right]$$

RREF r2 /= 10

$$\left[\begin{array}{cccccc}1& -2& 3& -2& |& 0\\ & & & & & \\ 0& 0& 1& -3/5& |& 0\\ & & & & & \\ 0& 0& 0& 0& |& 0\\ & & & & & \end{array}\right]$$

r1= r1 - r2*3

$$\left[\begin{array}{cccccc}1& -2& 0& -1/5& |& 0\\ & & & & & \\ 0& 0& 1& -3/5& |& 0\\ & & & & & \\ 0& 0& 0& 0& |& 0\\ & & & & & \end{array}\right]$$

The reduced system is:

$$\{\begin{array}{l}{x}_1-2x_2-1/5x_4=0\\ \\ x_3-3/5x_4=0\\ \end{array}$$

parameterize $x_2$ and $x_4$.

$$\begin{array}{rl}{x}_1-2s-1/5t& =0\\ & \\ x_1& =2s+1/5t\\ & \\ & \\ & \\ x_3-3/5t& =0\\ & \\ x_3& =3/5t\\ & \end{array}$$ $$\begin{array}{r}x=\left[\begin{array}{c}{x}_1\\ \\ x_2\\ \\ x_3\\ \\ x_4\end{array}\right]=\left[\begin{array}{c}2s+1/5t\\ \\ s\\ \\ 3/5t\\ \\ t\end{array}\right]=s\left[\begin{array}{c}2\\ \\ 1\\ \\ 0\\ \\ 0\end{array}\right]+t\left[\begin{array}{c}1/5\\ \\ 0\\ \\ 3/5\\ \\ 1\end{array}\right]\\ \\ =s{x}_1+t{x}_2\text{ where }{x}_1=\left[\begin{array}{c}2\\ \\ 1\\ \\ 0\\ \\ 0\\ \end{array}\right]\text{ and }{x}_2=\left[\begin{array}{c}1/5\\ \\ 0\\ \\ 3/5\\ \\ 1\end{array}\right]\end{array}$$

Theorem

Let A be an $m\cdot n$ matrix of rank r, and consider the homogeneous system in n variables with A as a coefficient matrix. Then:

1. the system has exactly [$n-r$ basic solutions||count of basic solutions], one for each parameter.
2. every solution is a [linear combination] of these basic solutions