# Solution Spaces of Homogenous Linear Systems

Before we look into what a solution space is, it is important to recall that a linear system in the form $Ax = b$ is said to be a **homogenous linear system** if $b = 0$, that is $Ax = 0$. We are now ready to proceed with an important definition to remember.

Definition: If $Ax = b$ is a linear system, then every vector $x$ which satisfies the system is said to be a Solution Vector of the linear system. The set of solution vectors of the system is called the Solution Space of the linear system. |

We will now look at an important theorem which relates a homogenous system $Ax = 0$ (where $A$ is an $m \times n$ matrix) solution space to the vector space $\mathbb{R}^n$.

Theorem 1: Let $Ax = 0$ be a homogenous linear system where $A$ is an $m \times n$ matrix, that is, the system contains $m$ linear equations of $n$ unknowns, then the solution space $W$ of the system is a subspace of $\mathbb{R}^n$. |

**Proof:**Suppose that $W$ is the solution space to the homogenous linear system $Ax = 0$. We know that $W$ is a nonempty set since $x = 0$ is in the solution space to the system. We need to verify axioms 5 (closure under addition) and axioms 10 (closure under multiplication) to verify that $W$ is a subspace of $\mathbb{R}^n$.

- Suppose that $x$ and $x’$ are solutions to the homogenous linear system, that is $Ax = 0$ and $Ax’ = 0$. If we add both of these equations together we obtain that $Ax + Ax’ = A(x + x’) = 0$. Therefore $x + x’$ is a solution to the system so $W$ is closed under addition.

- Now suppose that $k$ is a scalar and that $x$ is once again, a solution to our system. We obtain that $A(kx) = kAx = k0 = 0$. Therefore, $W$ is closed under multiplication.

- Since $W \subseteq \mathbb{R}^n$, the vector space $\mathbb{R}^n$ satisfies axioms 1-10, and $W$ satisfies axioms 5 and 10, we know that $W$ is a subspace of $\mathbb{R}^n$. $\blacksquare$

## Example 1

**Verify that the linear system $\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 3 & 3 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ has a solution space $W$ that is a subspace of $\mathbb{R}^3$.**

When we solve the system, we obtain that $x = -s -t$, $y = s$, $z = t$. We therefore get the equation of the plane when we substitute $y$ for $s$ and $z$ for $t$, that is $x = -y – z$ or rather $x + y + z = 0$. Recall that a plane in $\mathbb{R}^3$ that passes through the origin is a subspace of the vector space $\mathbb{R}^3$.

## Example 2

**Verify that the linear system $\begin{bmatrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$ has a solution space $W$ that is a subspace of $\mathbb{R}^3$.**

When we solve the system, we obtain that $x = s$, $y = -2s$ and $z = s$. We note that this solution space represents parametric equations for a line that passes through the origin and a line that is parallel to the vector $(1, -2, 1)$.

## Example 3

**Verify that an invertible $3 \times 3$ matrix $A$ in the linear system $Ax = 0$ is a subspace of $\mathbb{R}^3$.**

We note that if $A$ is invertible, then the homogenous linear system $Ax = 0$ has only the trivial solution, that is $(x, y, z) = (0, 0, 0)$ since $A^{-1}Ax = A^{-1}0 = 0$. Therefore, the only vector in the solution space of this linear system is the zero vector $\mathbf{0} = (0, 0, 0)$. But the zero vector itself is a subspace of $\mathbb{R}^3$.