Homogeneous System of Equations
A system of equations is called homoegenous if each of the contstant terms is equal to 0. A homogeneous system therefore has the form: \begin{aligned} a_{11} x_1 + a_{12} x_2 + \ldots + a_{1n}x_n &= 0 \\ a_{12} x_1 + a_{22} x_2 + \ldots + a_{2n}x_n &= 0 \\ \vdots \\ a_{m1} x_1 + a_{m2} x_2 + \ldots + a_{mn}x_n &= 0 \\ \end{aligned} where $$a_{ij}$$ are coefficients and $$x_j$$ are variables
Encoding a system of Equations into Matrix Vector Multiplication
The solution set of the following system of equations is the same as the solution set for $$Ax = b$$
TODO
Solution space of a homogeneous system of equations
Given a homogenous system of equations embedded in $$Ax = 0$$, where $$A$$ is an $$m \times n$$ matrix. We Define the solution space to be $$S = \{ x \in \mathbb{R}^n : Ax = 0\}$$
The solution space of a homogeneous system of equations forms a subspace
of $$\mathbb{R}^n$$
TODO
Basis for a Solution Space
Find a basis for the solution space of the following matrix encoded system of equations $\begin{bmatrix} 2 & 4 & -2 & 0 \\ 3 & 6 & -3 & 0 \\ 5 & 8 & -8 & 1 \end{bmatrix} \vec{v} = \vec{0}$

The matrix can be row reduced to the following form:

$\begin{bmatrix} 1 & 0 & -4 & 1 \\ 0 & 1 & \frac{3}{2} & - \frac{1}{2} \\ 0 & 0 & 0 & 0 \end{bmatrix}$

Which represents the following system of equations

\begin{aligned} 1x + 0y + -4 z + 1w &= 0 \\ 0x + 1y + \frac{3}{2}z - \frac{1}{2}w &= 0 \\ 0x + 0y + 0z + 0w &= 0 \end{aligned}

Since $$z$$ and $$w$$ are free variables, we let $$s, t \in \mathbb{R}$$ and set $$z = s$$ and $$w = t$$ , now using back substitution, we can deduce the following

\begin{aligned} 0x + 1y + \frac{3}{2}z - \frac{1}{2}t = 0 &\implies y = \frac{1}{2}t - \frac{3}{2}s \\ 1x + 0y + -4 z + 1w = 0 &\implies x = 4s - t\\ \end{aligned}

Therefore the solution set is given by

$\left \{ \begin{bmatrix} 4s - t \\ - \frac{3}{2}s + \frac{1}{2}t \\ s \\ t \end{bmatrix} : s, t \in \mathbb{R} \right\} = \left \{ s \begin{bmatrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} -1 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix} : s, t \in \mathbb{R} \right\}$

It can be confirmed that the two vectors $$\vec{v_1} = \begin{bmatrix} -1 \\ \frac{1}{2} \\ 0 \\ 1 \end{bmatrix}$$ and $$\vec{v_2} = \begin{bmatrix} 4 \\ - \frac{3}{2} \\ 1 \\ 0 \end{bmatrix}$$ are linearly independent due to the non-matching zero in the third component, thus the solution space can be written as $$\operatorname{span} ( \{ \vec{v_1}, \vec{v_2} \} )$$, therefore a possible basis for the solution set is $$( \vec{v_1}, \vec{v_2} )$$.