Linear Algebra| Basis & Dimension of a Homogeneous System — Step-by-Step Worked Example 4

Linear Algebra · Worked Example4

Solution Space, Basis & Dimension of a Homogeneous System

A complete row-reduction walkthrough for a 3×5 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 5 Variables

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The Problem

Find the dimension and a basis for the solution space \(W\) of the homogeneous system:

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

1 Set Up the System

In matrix form \(AX = 0\), with the augmented column of zeros dropped from here on:

\[ \begin{bmatrix} 1 & 2 & 2 & -1 & 3\\ 1 & 2 & 3 & 1 & 1\\ 3 & 6 & 8 & 1 & 5 \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix} \]

2 Row-Reduce to Echelon Form

Clear column 1 using row 1 as the pivot row:

R2 → R2 − R1 R3 → R3 − 3R1

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

↓ eliminate above/below the pivot in column 3 ↓

R1 → R1 − 2R2 R3 → R3 − 2R2

\[ \begin{bmatrix} 1 & 2 & 0 & -5 & 7\\ 0 & 0 & 1 & 2 & -2\\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]

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

3 Identify Pivot and Free Variables

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

Pivot Variables

• \(x_1\)
• \(x_3\)

Free Variables

• \(x_2 = s\)
• \(x_4 = t\)
• \(x_5 = u\)

4 Solve for the Pivot Variables

Reading each row of the echelon form as an equation:

\[ \begin{aligned} x_1 + 2x_2 - 5x_4 + 7x_5 &= 0\\ x_3 + 2x_4 - 2x_5 &= 0 \end{aligned} \]

Substituting \(x_2=s,\; x_4=t,\; x_5=u\) and solving each pivot variable in terms of the free ones:

\[ x_1 = -2s + 5t - 7u,\qquad x_3 = -2t + 2u \]

5 General Solution

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

\[ X=\begin{bmatrix} x_1\\x_2\\x_3\\x_4\\x_5\end{bmatrix} = s\begin{bmatrix}-2\\1\\0\\0\\0\end{bmatrix} + t\begin{bmatrix}5\\0\\-2\\1\\0\end{bmatrix} + u\begin{bmatrix}-7\\0\\2\\0\\1\end{bmatrix} \]

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

6 Basis and Dimension

The three vectors are linearly independent (each has a leading 1 in a position where the others have a 0) and they span \(W\), so together they form a basis.

\(\begin{bmatrix}-2\\1\\0\\0\\0\end{bmatrix}\)

\(\begin{bmatrix}5\\0\\-2\\1\\0\end{bmatrix}\)

\(\begin{bmatrix}-7\\0\\2\\0\\1\end{bmatrix}\)

Final Answer

The solution space \(W\) is the span of the three basis vectors above.

dim(W) = 3

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Worked example based on a 3×5 homogeneous system · rank = 2, so nullity = 5 − 2 = 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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