Worked Example 1
Basis & Dimension of a
Homogeneous Linear System
Tutorial Video: Basis & Dimension of Homogeneous System
📌 Problem Statement
Determine the basis and the dimension of the solution space of the homogeneous system:
\[
\begin{aligned}
2x_1 + 2x_2 - x_3 + x_5 &= 0 \\
-x_1 - x_2 + 2x_3 - 3x_4 + x_5 &= 0 \\
x_1 + x_2 - 2x_3 - x_5 &= 0 \\
x_3 + x_4 + x_5 &= 0
\end{aligned}
\]
1
Set Up the Augmented Matrix
We write the system as $AX = 0$ and form the augmented matrix $[A \mid 0]$. Since the right-hand side is always zero for a homogeneous system, we work with $A$ alone:
\[
A = \begin{bmatrix}
2 & 2 & -1 & 0 & 1 \\
-1 & -1 & 2 & -3 & 1 \\
1 & 1 & -2 & 0 & -1 \\
0 & 0 & 1 & 1 & 1
\end{bmatrix}
\]
2
Row Reduce to Reduced Row Echelon Form
We apply elementary row operations systematically to reach RREF.
Operation: $R_1 \leftrightarrow R_3$ (bring a leading 1 to the top)
\[
\begin{bmatrix}
1 & 1 & -2 & 0 & -1 \\
-1 & -1 & 2 & -3 & 1 \\
2 & 2 & -1 & 0 & 1 \\
0 & 0 & 1 & 1 & 1
\end{bmatrix}
\]
Operations: $R_2 \leftarrow R_2 + R_1$, $R_3 \leftarrow R_3 - 2R_1$
\[
\begin{bmatrix}
1 & 1 & -2 & 0 & -1 \\
0 & 0 & 0 & -3 & 0 \\
0 & 0 & 3 & 0 & 3 \\
0 & 0 & 1 & 1 & 1
\end{bmatrix}
\]
Operation: $R_2 \leftrightarrow R_4$ (reorder for a better pivot pattern)
\[
\begin{bmatrix}
1 & 1 & -2 & 0 & -1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & -3 & 0
\end{bmatrix}
\]
Operations: $R_1 \leftarrow R_1 + 2R_2$, $R_3 \leftarrow R_3 - R_2$
\[
\begin{bmatrix}
1 & 1 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & -1 & 0 \\
0 & 0 & 0 & -3 & 0
\end{bmatrix}
\]
Operation: $R_3 \leftarrow R_3 \div (-1)$ (make pivot positive)
\[
\begin{bmatrix}
1 & 1 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & -3 & 0
\end{bmatrix}
\]
Operations: $R_1 \leftarrow R_1 - 2R_3$, $R_2 \leftarrow R_2 - R_3$, $R_4 \leftarrow R_4 + 3R_3$
\[
A \approx \begin{bmatrix}
1 & 1 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\quad \textbf{(RREF)}
\]
3
Identify Pivot and Free Variables
From the RREF, the pivot columns correspond to leading 1s, while columns without a pivot give free variables:
$x_1$ — pivot
$x_2$ — free ($= t$)
$x_3$ — pivot
$x_4$ — pivot
$x_5$ — free ($= s$)
Rank–Nullity Theorem check: rank$(A) = 3$ (three pivot rows) + nullity $= 2$ (two free variables) $= 5$ (total variables) ✓
4
Express Pivot Variables in Terms of Free Variables
Reading directly from the RREF rows, with $x_2 = t$ and $x_5 = s$:
\[
\begin{aligned}
x_1 + x_2 + x_5 &= 0 &\implies x_1 &= -t - s \\
x_3 + x_5 &= 0 &\implies x_3 &= -s \\
x_4 &= 0 &\implies x_4 &= 0
\end{aligned}
\]
5
Write the General Solution
Assembling all components into vector form and separating by parameters $t$ and $s$:
\[
\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix}
=
\begin{bmatrix} -t - s \\ t \\ -s \\ 0 \\ s \end{bmatrix}
= \;
t \begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}
+ \;
s \begin{bmatrix} -1 \\ 0 \\ -1 \\ 0 \\ 1 \end{bmatrix},
\quad t, s \in \mathbb{R}
\]
✅ Final Answer
The basis of the null space (solution space) is:
$\mathbf{v}_1$
\[\begin{bmatrix}-1\\1\\0\\0\\0\end{bmatrix}\]
，
$\mathbf{v}_2$
\[\begin{bmatrix}-1\\0\\-1\\0\\1\end{bmatrix}\]
The dimension of the solution space is:
dim = 2
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