Homogeneous systems

Definition

A homogeneous system of linear equations is a system that can be written in the form:

Ax = 0

Where:

A is an m×n coefficient matrix
x is an n×1 vector of variables
0 is an m×1 zero vector

In component form, a homogeneous system looks like: a₁₁x₁ + a₁₂x₂ + … + a₁ₙxₙ = 0 a₂₁x₁ + a₂₂x₂ + … + a₂ₙxₙ = 0 … aₘ₁x₁ + aₘ₂x₂ + … + aₘₙxₙ = 0

Key Properties

Trivial Solution: Every homogeneous system always has at least one solution: the trivial solution where x = 0 (all variables equal zero).
Solution Structure: If x₁ and x₂ are solutions to a homogeneous system, then:
- Any linear combination c₁x₁ + c₂x₂ is also a solution
- The set of all solutions forms a subspace of ℝⁿ, called the null space or kernel of A
Solution Uniqueness: A homogeneous system Ax = 0 has:
- Only the trivial solution if and only if rank(A) = n (full column rank)
- Infinitely many solutions if and only if rank(A) < n
Solution Dimension: The dimension of the solution space equals n - rank(A)

Solution Methods

Method 1: Gaussian Elimination

Form the augmented matrix [A 0]
Perform row operations to get row echelon form
Back-substitute to find the general solution

Method 2: Null Space Calculation

Transform A to reduced row echelon form (RREF)
Identify the free variables (corresponding to non-pivot columns)
Express basic variables in terms of free variables
Write the general solution as a linear combination of basis vectors for the null space

General Solution Format

If the system has infinitely many solutions, the general solution can be written as:

x = c₁v₁ + c₂v₂ + … + cₖvₖ

Where:

v₁, v₂, …, vₖ form a basis for the null space of A
c₁, c₂, …, cₖ are arbitrary constants (free parameters)
k = n - rank(A) (the number of free variables)

Examples

Example 1: Simple Homogeneous System

Consider the system:

2x - y = 0
3x + y = 0
5x + 2y = 0

Step 1: Form the augmented matrix

[2  -1 | 0]
[3   1 | 0]
[5   2 | 0]

Step 2: Row reduction R₂ = R₂ - (3/2)R₁:

[2  -1 | 0]
[0  5/2 | 0]
[5   2 | 0]

R₃ = R₃ - (5/2)R₁:

[2  -1 | 0]
[0  5/2 | 0]
[0  9/2 | 0]

R₃ = R₃ - (9/5)R₂:

[2  -1 | 0]
[0  5/2 | 0]
[0   0 | 0]

Step 3: Back-substitution From the second row: 5/2 y = 0, so y = 0 From the first row: 2x - 0 = 0, so x = 0

The only solution is the trivial solution (0,0)

Example 2: System with Infinitely Many Solutions

Consider the system:

2x - 3y - 2z - t = 0
4x + y + 4z - 2t = 0
x + y + z + t = 0

Step 1: Form the augmented matrix

[2  -3  -2  -1 | 0]
[4   1   4  -2 | 0]
[1   1   1   1 | 0]

Step 2: Row reduction (to RREF) After full row reduction:

[1   0   1   0 | 0]
[0   1   0   1 | 0]
[0   0   0   0 | 0]

Step 3: Identify free variables The pivot columns are 1 and 2, so x and y are basic variables The non-pivot columns are 3 and 4, so z and t are free variables

Step 4: Express basic variables in terms of free variables From the RREF: x = -z y = -t

Step 5: Write the general solution For any values of z and t:

x = -z
y = -t
z = z (free)
t = t (free)

The general solution can be written as:

⎡x⎤   ⎡-z⎤   ⎡-1⎤     ⎡ 0⎤
⎢y⎥ = ⎢-t⎥ = ⎢ 0⎥·z + ⎢-1⎥·t
⎢z⎥   ⎢ z⎥   ⎢ 1⎥     ⎢ 0⎥
⎣t⎦   ⎣ t⎦   ⎣ 0⎦     ⎣ 1⎦

The solution space is a 2-dimensional subspace of ℝ⁴ with basis {(-1,0,1,0), (0,-1,0,1)}.

Special Cases and Techniques

1. Homogeneous Systems with Parameters

For systems with parameters:

Perform row reduction keeping parameters symbolic
Analyze how the rank of the coefficient matrix changes with different parameter values
Determine conditions on parameters for unique or infinitely many solutions

2. Large Systems with Patterns

Look for patterns that allow simplification:

Equations that are linear combinations of others (redundant equations)
Variables that appear in specific patterns
Symmetries in the coefficient matrix

3. Connection to Linear Transformations

A homogeneous system Ax = 0 can be interpreted as:

Finding vectors x in the domain of the linear transformation T(x) = Ax that map to the zero vector
Finding the kernel (null space) of the transformation T

Solving Techniques for Exercises

Strategy 1: Row-Echelon Form Approach

Convert to row-echelon form using elementary row operations
Count the number of non-zero rows (r) to find rank(A)
The dimension of the solution space is n - r
Back-substitute to find the general solution

Strategy 2: Determinant Approach (for Square Matrices)

For a square n×n homogeneous system:

If det(A) ≠ 0, there is only the trivial solution
If det(A) = 0, there are infinitely many solutions

Strategy 3: Matrix Properties

Use matrix properties to determine the nature of solutions:

If columns of A are linearly independent, only the trivial solution exists
If A has a zero row, the system always has infinitely many solutions
If there are more variables than equations (n > m), the system always has infinitely many solutions

Example of Solving Exercises

Example: Homogeneous System with Parameter

Consider the system:

x + (h - 1)y - 2z = 0
2x + 3hy - 4z = 0
6x + (h + 5)y + 2z = 0

Step 1: Form the augmented matrix

[1   h-1  -2 | 0]
[2   3h   -4 | 0]
[6  h+5    2 | 0]

Step 2: Row reduction (keeping h as a parameter) R₂ = R₂ - 2R₁:

[1   h-1  -2 | 0]
[0   h+2   0 | 0]
[6  h+5    2 | 0]

R₃ = R₃ - 6R₁:

[1   h-1  -2 | 0]
[0   h+2   0 | 0]
[0  -6h+11  14 | 0]

Step 3: Analyze how rank changes with h

Case 1: If h ≠ -2, then the second row is non-zero

R₃ = R₃ - [(-6h+11)/(h+2)]R₂ gives a third independent equation
Rank = 3, so there’s only the trivial solution

Case 2: If h = -2, then the second row becomes all zeros

For the third row, we get [-6(-2)+11, 14] = [23, 14]
Rank = 2, so there’s a 1-dimensional solution space

Step 4: Find the solution for h = -2 For h = -2:

[1  -3  -2 | 0]
[0   0   0 | 0]
[0  23  14 | 0]

Further row reduction:

[1  -3  -2 | 0]
[0   1   14/23 | 0]
[0   0   0 | 0]

Back-substitution: y = -14z/23 x = 3y + 2z = 3(-14z/23) + 2z = -42z/23 + 2z = (2 - 42/23)z = (46 - 42)/23 · z = 4z/23

The general solution for h = -2 is:

x = 4z/23
y = -14z/23
z = z (free)

or written as:

⎡x⎤   ⎡ 4/23⎤
⎢y⎥ = ⎢-14/23⎥·z
⎣z⎦   ⎣   1  ⎦

Common Mistakes to Avoid

Forgetting that a homogeneous system always has the trivial solution
Incorrectly identifying basic and free variables
Errors in row operations, especially with parameters
Forgetting to check special cases when parameters equal specific values
Not verifying solutions by substituting back into the original equations

Connection to Other Topics

Homogeneous systems are fundamental to many concepts in linear algebra:

Null Space: The solution set is exactly the null space of the coefficient matrix
Linear Dependence: A set of vectors {v₁, v₂, …, vₙ} is linearly dependent if and only if the homogeneous system c₁v₁ + c₂v₂ + … + cₙvₙ = 0 has non-trivial solutions
Eigenvalues and Eigenvectors: Finding eigenvectors involves solving the homogeneous system (A - λI)v = 0
Differential Equations: Homogeneous systems arise naturally when solving systems of linear differential equations
Basis for Subspaces: The null space basis vectors form a basis for the solution subspace

Exercises

Solve the following homogeneous linear systems:
- $\begin{cases} 2x - y = 0 \ 3x + y = 0 \ 5x + 2y = 0 \end{cases}$
- $\begin{cases} 3x + 4y - z = 0 \ 2x - y - 5z = 0 \ 7x + 13y + 2z = 0 \end{cases}$
- $\begin{cases} 2x - 3y - 2z - t = 0 \ 4x + y + 4z - 2t = 0 \ x + y + z + t = 0 \end{cases}$
- $\begin{cases} x + y + 5z + t = 0 \ 6x - 6y + z - 8t = 0 \ 3x - 9y - 14z - 11t = 0 \end{cases}$
- $\begin{cases} 6x - 3y - 2z + 7t = 0 \ x + 4z - t = 0 \ y - z + 3t = 0 \ x + y + t = 0 \end{cases}$
- $\begin{cases} 3x + 4y - 5z + 2t = 0 \ 2x - 5y + z - t = 0 \ -2x + y + 6z - 2t = 0 \ 7x - 2y - 10z + 3t = 0 \end{cases}$
- $\begin{cases} x + 9y + 2z - 4t = 0 \ 3x - 4y + 5z + 2t = 0 \ 5x - 17y + 8z + 8t = 0 \ 9x - 43y + 14z + 20t = 0 \end{cases}$
- $\begin{cases} -x - y + 3z = 0 \ 4x + 5y + z = 0 \ 2x - 3y + 4z = 0 \end{cases}$