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 \( ( \vec{v_1}, \vec{v_2} ) \).