Homogeneous Equations

Linear Algebra

Homogeneous Equation :drill:

Scheduled: 2022-10-22 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:r2-2r1 r3: r3-3r1 $\left[\begin{array}{cccc}1& -1& 2& -1|0\\ 0& 4& -4& 3|0\\ 0& 4& -4& 2|0\end{array}\right]$

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+1/4r3 r2=r2-3/4r3 $\left[\begin{array}{cccc}1& 0& 1& 0|0\\ 0& 1& -1& 0|0\\ 0& 0& 0& 1|0\end{array}\right]$

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

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 :drill: Scheduled: 2022-10-22 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 :drill: 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 =

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

—

z= \begin{bmatrix} 3\\ 1\\ 1 \end{bmatrix} \end{aligned} $$ if $$ \begin{aligned} v= \begin{bmatrix} 0\\ -1\\ 2 \end{bmatrix} \end{aligned} $$ determine whether v is a linear combination of x,y & z Solution: We must determine whether r,s,t exist such that v=rx+sy+tz $$ \begin{aligned} \begin{bmatrix} 0\\ -1\\ 2 \end{bmatrix} = r \begin{bmatrix} 1\\ 0\\ 1 \end{bmatrix} + s \begin{bmatrix} 2\\ 1\\ 0 \end{bmatrix} + t \begin{bmatrix} 3\\ 1\\ 1 \end{bmatrix}. \end{aligned} $$ ^^ Which is the same as: $$ \begin{cases} r+2s+3t=0\\ s+t=-1\\ r+t=2 \end{cases} $$ Augmented matrix: $$ \begin{bmatrix} 1&2&3&|&0\\ 0&1&1&|&-1\\ 1&0&1&|&2 \end{bmatrix} $$ r3 = r3-r1 $$ \begin{bmatrix} 1&2&3&|&0\\ 0&1&1&|&-1\\ 0&-2&-2&|&2 \end{bmatrix} $$ r3 = r3+2*r2 $$ \begin{bmatrix} 1&2&3&|&0\\ 0&1&1&|&-1\\ 0&0&0&|&0 \end{bmatrix} $$ RREF r1=r1-2*r2 $$ \begin{bmatrix} 1&0&1&|&2\\ 0&1&1&|&-1\\ 0&0&0&|&0 \end{bmatrix} $$ $$ \begin{cases} r+t=2\\ s+t=-1 \end{cases} $$ t is a non-leading variable, so set it as a parameter. $t=k$. $$ \begin{cases} r=2-k\\ s=-1-k\\ t=k \end{cases} $$ So let's try $k=0$ $$ \begin{cases} r=2\\ s=-1\\ t=0 \end{cases} $$ so V really is a linear combination of x,y & z. ## More on basic solutions When solving a system with n variables ($x_1...x_n$), write the variables as a column matrix: $\begin{bmatrix} x_1\\ ...\\ x_n \end{bmatrix}$. The trivial solution is noted by: $0= \begin{bmatrix} 0\\ 0\\ ...\\ 0 \end{bmatrix}$. Remark: Let x 7 y be two solutions to a homogeneus system with n variables. Then any linear combination sx+ty of these solutions is a solution to the system. More generally, any linear combination of solutions to a homogenous system is a solution. #### Definition :drill: linear combinations of multiple solutions to a [homogenous systems] are also solutions. #### Example Solve the homogenous system w/ coefficient matrix: $$ \begin{aligned} A= \begin{bmatrix} 1&-2&3&-2\\ -3&6&1&0\\ -2&4&4&-2\\ \end{bmatrix} \end{aligned} $$ . $$ \begin{bmatrix} 1&-2&3&-2&|&0\\ -3&6&1&0&|&0\\ -2&4&4&-2&|&0\\ \end{bmatrix} $$ r2=r2+3*r1; r3=r3+2*r1 $$ \begin{bmatrix} 1&-2&3&-2&|&0\\ 0&0&10&-6&|&0\\ 0&0&10&-6&|&0\\ \end{bmatrix} $$ r3 = r3-r2 $$ \begin{bmatrix} 1&-2&3&-2&|&0\\ 0&0&10&-6&|&0\\ 0&0&0&0&|&0\\ \end{bmatrix} $$ RREF r2 /= 10 $$ \begin{bmatrix} 1&-2&3&-2&|&0\\ 0&0&1&-3/5&|&0\\ 0&0&0&0&|&0\\ \end{bmatrix} $$ r1= r1 - r2*3 $$ \begin{bmatrix} 1&-2&0&-1/5&|&0\\ 0&0&1&-3/5&|&0\\ 0&0&0&0&|&0\\ \end{bmatrix} $$ The reduced system is: $$ \begin{cases} x_1 - 2x_2 - 1/5 x_4 = 0\\ x_3 - 3/5 x_4 = 0\\ \end{cases} $$ parameterize $x_2$ and $x_4$. $$ \begin{aligned} x_1-2s-1/5t &= 0\\ x_1 &= 2s+1/5t\\ \\ x_3 - 3/5t &= 0\\ x_3 &= 3/5t\\ \end{aligned} $$ $$ \begin{aligned} x = \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} = \begin{bmatrix} 2s+1/5t\\ s\\ 3/5t\\ t \end{bmatrix} = s\begin{bmatrix} 2\\ 1\\ 0\\ 0 \end{bmatrix} + t \begin{bmatrix} 1/5\\ 0\\ 3/5\\ 1 \end{bmatrix} \\ =sx_1 + tx_2 \text{ where } x_1 = \begin{bmatrix} 2\\ 1\\ 0\\ 0\\ \end{bmatrix} \text{ and } x_2= \begin{bmatrix} 1/5\\ 0\\ 3/5\\ 1 \end{bmatrix} \end{aligned} $$ #### Theorem :drill: Scheduled: 2022-10-22 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