09 Homogenous Systems

A homogenous system is just an equation where a matrix multiplied by some vector = 0.
Can also be interpreted as a set of solutions that make the system = 0.

$$ A\vec{x} = \vec{0}. $$

$A$ is some matrix, $\vec{0}$ is the Zero Vector.

This is always consistent because

$$ A\vec{0} = \vec{0} $$ always works. This is called the trivial solution, because it is trivial (duh).

The important question here is: Are there non-trivial solutions to a given system?

We simply set up an augmented matrix and solve for $x$'s that allow a solution of $\vec{0}$.

If there are any free variables, then we have infinitely many non-trivial solutions.

Example problem

For what values of $\vec{x}$ does the following equation have a solution?

$$ \begin{bmatrix} 1 & 3 & -1 & 3 \\ 2 & 6 & 2 & 2 \\ 1 & 3 & 1 & 1 \end{bmatrix}\vec{x}=0 $$

We will begin by looking at this equation as a solving system of equations:

$$ \begin{align*} \begin{bmatrix} 1 & 3 & -1 & 3 \\ 2 & 6 & 2 & 2 \\ 1 & 3 & 1 & 1 \end{bmatrix} \vec{x}=\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \\ \left[\begin{array}{cccc|c} 1 & 3 & -1 & 3 & 0 \\ 2 & 6 & 2 & 2 & 0 \\ 1 & 3 & 1 & 1 & 0 \end{array}\right] \\ \overset{ \text{rref} }{ \to } \left[\begin{array}{cccc|c} 1 & 3 & 0 & 2 & 0 \\ 0 & 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{align*} $$

Notice upon looking at the RREF that columns 2 and 4 do not have a pivot. This means that $x_{2}$ and $x_{4}$ are free variables.

This means that we can represent this equation as the following:

$$ \begin{align*} \begin{cases} x_{1}+3x_{2}+2x_{4}=0 \\ x_{2} \text{ is free.} \\ x_{3}-x_{4}=0 \\ x_{4} \text{ is free.} \\ \end{cases} \\ \text{let }x_{2}=s,x_{4}=t &\implies \begin{cases} x_{1}=-3s-2t \\ x_{3}=t \end{cases} \\ &\implies \vec{x}=\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{bmatrix} =\begin{bmatrix} -3s-2t \\ 1s+0t \\ 0s+1t \\ 0s+1t \end{bmatrix} \\ &\therefore \vec{x}= s\begin{bmatrix} -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+ t\begin{bmatrix} -2 \\ 0 \\ 1 \\ 1 \end{bmatrix}, s, t, \in \mathbb{R} \end{align*} $$