Solution Spaces of Homogenous Linear Systems

# 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 $\mathbb{R}^3$.