Linear Algebra| Example 3| How to Find the Solution Space, Basis, and Dimension of a Homogeneous System (Step-by-Step)

Worked Example 3

Solution Space, Basis & Dimension of a Homogeneous System

A complete row-reduction walkthrough for a 4×6 coefficient matrix — finding the null space, extracting a basis, and reading off the dimension.

📐 Systems of Equations 🧮 Row Reduction 📚 Null Space

► Tutorial Video: Basis & Dimension of Homogeneous System

Homogeneous System Basis & Dimension Null Space Gaussian Elimination RREF Free Variables Rank–Nullity Theorem 6 Variables

The Problem 3

Find the solution space, a basis for that space, and its dimension for the homogeneous system \(AX = 0\), where

\[ A=\begin{bmatrix} 1 & -3 & 4 & -2 & 5 & 4 \\ 2 & -6 & 9 & -1 & 8 & 2 \\ 2 & -6 & 9 & -1 & 9 & 7 \\ -1 & 3 & -4 & 2 &-5 & -4 \end{bmatrix} \]

1 Set Up the System

Writing out \(AX = 0\) explicitly gives four equations in six unknowns:

\[ \begin{aligned} x_1 - 3x_2 + 4x_3 - 2x_4 + 5x_5 + 4x_6 &= 0 \\ 2x_1 - 6x_2 + 9x_3 - x_4 + 8x_5 + 2x_6 &= 0 \\ 2x_1 - 6x_2 + 9x_3 - x_4 + 9x_5 + 7x_6 &= 0 \\ -x_1 + 3x_2 - 4x_3 + 2x_4 - 5x_5 - 4x_6 &= 0 \end{aligned} \]

In augmented matrix form (the last column of zeros is dropped from here on, since it never changes):

\[ \left[\begin{array}{cccccc|c} 1 & -3 & 4 & -2 & 5 & 4 & 0\\ 2 & -6 & 9 & -1 & 8 & 2 & 0\\ 2 & -6 & 9 & -1 & 9 & 7 & 0\\ -1 & 3 & -4 & 2 & -5 & -4 & 0 \end{array}\right] \]

2 Row-Reduce to Echelon Form

Clear column 1 using row 1 as the pivot row:

R2 → R2 − 2R1 R3 → R3 − 2R1 R4 → R4 + R1

\[ \begin{bmatrix} 1 & -3 & 4 & -2 & 5 & 4 \\ 0 & 0 & 1 & 3 & -2 & -6 \\ 0 & 0 & 1 & 3 & -1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

↓ eliminate above/below the pivot in column 3 ↓

R1 → R1 − 4R2 R3 → R3 − R2

\[ \begin{bmatrix} 1 & -3 & 0 & -14 & 13 & 28 \\ 0 & 0 & 1 & 3 & -2 & -6 \\ 0 & 0 & 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

↓ eliminate above the pivot in column 5 ↓

R1 → R1 − 13R3 R2 → R2 + 2R3

\[ \begin{bmatrix} 1 & -3 & 0 & -14 & 0 & -37 \\ 0 & 0 & 1 & 3 & 0 & 4 \\ 0 & 0 & 0 & 0 & 1 & 5 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

This is the reduced row-echelon form of \(A\).

3 Identify Pivot and Free Variables

Pivots sit in columns 1, 3, and 5, so those variables are determined; the rest are free to choose.

Pivot Variables

• \(x_1\)
• \(x_3\)
• \(x_5\)

Free Variables

• \(x_2 = s\)
• \(x_4 = t\)
• \(x_6 = u\)

4 Solve for the Pivot Variables

Reading each row of the echelon form as an equation:

\[ \begin{aligned} x_1 - 3x_2 - 14x_4 - 37x_6 &= 0 \\ x_3 + 3x_4 + 4x_6 &= 0 \\ x_5 + 5x_6 &= 0 \end{aligned} \]

Substituting \(x_2=s,\; x_4=t,\; x_6=u\) and solving each pivot variable in terms of the free ones:

\[ x_1 = 3s + 14t + 37u,\qquad x_3 = -3t - 4u,\qquad x_5 = -5u \]

5 General Solution

Collecting every component into a single vector and splitting it by the parameters \(s, t, u\):

\[ X=\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\\x_6\end{bmatrix} = s\begin{bmatrix}3\\1\\0\\0\\0\\0\end{bmatrix} + t\begin{bmatrix}14\\0\\-3\\1\\0\\0\end{bmatrix} + u\begin{bmatrix}37\\0\\-4\\0\\-5\\1\end{bmatrix} \]

This expression is the solution space of \(AX=0\) — every solution vector is some linear combination of the three vectors above.

6 Basis and Dimension

Since the three vectors are linearly independent (each has a leading 1 in a position where the others have a 0) and they span the solution space, they form a basis.

\(\begin{bmatrix}3\\1\\0\\0\\0\\0\end{bmatrix}\)

\(\begin{bmatrix}14\\0\\-3\\1\\0\\0\end{bmatrix}\)

\(\begin{bmatrix}37\\0\\-4\\0\\-5\\1\end{bmatrix}\)

Final Answer

The solution space is the span of the three basis vectors above.

dim(Solution Space) = 3

Worked example based on a 4×6 homogeneous system · rank = 3, so nullity = 6 − 3 = 3, consistent with the Rank–Nullity Theorem.

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Related Topics

• Null Space & Column Space
• Rank of a Matrix
• Vector Subspaces
• Rank–Nullity Theorem
• Linear Independence
• Eigenvalues & Eigenvectors

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