Homogeneous Systems

A system $A\mathbf{x} = \mathbf{0}$, with every constant equal to zero, is called homogeneous. It always has at least one solution, $\mathbf{x} = \mathbf{0}$, called the trivial solution. So existence is never in doubt; the real question is whether there are others. This uses free variables and the matrix equation.

Key idea

$A\mathbf{x} = \mathbf{0}$ has a nontrivial solution exactly when the equation has at least one free variable, that is, when some column of $A$ has no pivot. Its solution set is the span of one vector per free variable.

Worked example: Parametric vector form

Solve $A\mathbf{x} = \mathbf{0}$ for $A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 5 & 1 \end{bmatrix}$.

Row reduce $A$ (the column of zeros never changes, so leave it off): $R_2 \to R_2 - 2R_1$ gives $(0, 1, 3)$, then $R_1 \to R_1 - 2R_2$: $\begin{bmatrix} 1 & 0 & -7 \\ 0 & 1 & 3 \end{bmatrix}$.
$x_3$ is free; let $x_3 = t$. Row 1: $x_1 = 7t$. Row 2: $x_2 = -3t$.
$\mathbf{x} = t\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix}$. The solution set is $\operatorname{Span}\left\{\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix}\right\}$: a line through the origin in $\mathbb{R}^3$.
Check: $A\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$. ✓

Answer: $\mathbf{x} = t(7, -3, 1)$

This $A$ squashes the plane onto a line. Every vector on the line through $(2, -1)$ is sent to $\mathbf{0}$: those are the solutions of $A\mathbf{x} = \mathbf{0}$.

Homogeneous and nonhomogeneous solutions

Compare the general solution of $A\mathbf{x} = \mathbf{b}$ from the parametric topic, $\mathbf{x} = \mathbf{p} + t\mathbf{v}$, with the solution of $A\mathbf{x} = \mathbf{0}$: the same $t\mathbf{v}$ appears. That is no accident.

Why it works

Suppose $A\mathbf{p} = \mathbf{b}$. If $A\mathbf{v}_h = \mathbf{0}$, then $A(\mathbf{p} + \mathbf{v}_h)$ $= A\mathbf{p} + A\mathbf{v}_h$ $= \mathbf{b} + \mathbf{0}$ $= \mathbf{b}$, so $\mathbf{p} + \mathbf{v}_h$ is a solution. Conversely, if $A\mathbf{y} = \mathbf{b}$, then $A(\mathbf{y} - \mathbf{p}) = \mathbf{b} - \mathbf{b} = \mathbf{0}$, so $\mathbf{y}$ $= \mathbf{p} + (\text{a homogeneous solution})$. Every solution of $A\mathbf{x} = \mathbf{b}$ is one particular solution plus a solution of $A\mathbf{x} = \mathbf{0}$. Geometrically, the solution set of $A\mathbf{x} = \mathbf{b}$ is the solution set of $A\mathbf{x} = \mathbf{0}$ slid over by $\mathbf{p}$: a parallel line or plane.

Worked example: Using the structure

For the same $A$, one solution of $A\mathbf{x} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$ is $\mathbf{p} = \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix}$. Find all solutions.

Check $\mathbf{p}$: $A\mathbf{p} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$. ✓
All solutions are $\mathbf{p}$ plus the homogeneous solutions: $\mathbf{x} = \begin{bmatrix} 4 \\ -1 \\ 0 \end{bmatrix} + t\begin{bmatrix} 7 \\ -3 \\ 1 \end{bmatrix}$.

Answer: $\mathbf{x} = (4 + 7t,\; -1 - 3t,\; t)$

With one free variable the homogeneous solutions form a line through the origin; with two, a plane through the origin; in general, the span of one vector for each free variable. Those vectors come straight from the reduced echelon form: set one free variable to $1$ and the others to $0$, and read off the basic variables. We will meet this set again under the name null space.

If $A$ has more columns than rows, $A\mathbf{x} = \mathbf{0}$ always has a nontrivial solution: there are more unknowns than possible pivots, so something is free.

Homogeneous Systems

Homogeneous and nonhomogeneous solutions

Practice

Builds on

Leads to

Sources and further reading