← Back to linear algebra 1

Homogeneous Linear Systems

A homogeneous linear system is one where all constants are zero. These systems have special properties that make them fundamental in linear algebra.

Definition

Definition (Homogeneous System):
A system of linear equations is homogeneous if it can be written as: Ax⃗=0⃗A\vec{x} = \vec{0} where AA is the coefficient matrix and 0⃗\vec{0} is the zero vector.

In equation form, every equation has 0 on the right side: {a11x1+a12x2+⋯+a1nxn=0a21x1+a22x2+⋯+a2nxn=0⋮am1x1+am2x2+⋯+amnxn=0\begin{cases} a_{11}x_1 + a_{12}x_2 + \cdots + a_{1n}x_n = 0 \\ a_{21}x_1 + a_{22}x_2 + \cdots + a_{2n}x_n = 0 \\ \vdots \\ a_{m1}x_1 + a_{m2}x_2 + \cdots + a_{mn}x_n = 0 \end{cases}

Example 1: Homogeneous System

{x+2y−z=03x−y+2z=0\begin{cases} x + 2y - z = 0 \\ 3x - y + 2z = 0 \end{cases}

Matrix form: [12−13−12][xyz]=[00]\begin{bmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

The Trivial Solution

Theorem (Existence of Trivial Solution):
Every homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} has at least one solution: the trivial solution x⃗=0⃗\vec{x} = \vec{0}.

Proof: A0⃗=0⃗A\vec{0} = \vec{0} by the properties of matrix-vector multiplication. ∎

This means homogeneous systems are always consistent. The question is: are there other solutions?

The solution x⃗=0⃗\vec{x} = \vec{0} is called "trivial" because it's obvious and automatic—it works for any homogeneous system without calculation.

What's interesting are the nontrivial solutions (non-zero solutions), which reveal the structure of the system.

Nontrivial Solutions

Theorem (Nontrivial Solutions):
A homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0} has nontrivial solutions (non-zero solutions) if and only if there are free variables in the row-reduced form.

Equivalently: If AA is m×nm \times n with m<nm < n (more variables than equations), or if not all columns are pivot columns, then nontrivial solutions exist.

Example 2: System with Nontrivial Solutions

Solve the homogeneous system: {x+2y−z=02x+4y−2z=0\begin{cases} x + 2y - z = 0 \\ 2x + 4y - 2z = 0 \end{cases}

Row Reduction

Step 0 of 1

The row-reduced form shows:

General solution: {x=−2y+zy is freez is free\begin{cases} x = -2y + z \\ y \text{ is free} \\ z \text{ is free} \end{cases}

Or in vector form: x⃗=[xyz]=[−2y+zyz]=y[−210]+z[101]\vec{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -2y + z \\ y \\ z \end{bmatrix} = y\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + z\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}

Infinitely many solutions parametrized by yy and zz.

Let's verify some nontrivial solutions:

With y=1,z=0y = 1, z = 0: x⃗=[−210]\vec{x} = \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}

Check:

  • First equation: (−2)+2(1)−0=0(-2) + 2(1) - 0 = 0 ✓
  • Second equation: 2(−2)+4(1)−2(0)=02(-2) + 4(1) - 2(0) = 0 ✓

With y=0,z=1y = 0, z = 1: x⃗=[101]\vec{x} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}

Check:

  • First equation: 1+2(0)−1=01 + 2(0) - 1 = 0 ✓
  • Second equation: 2(1)+4(0)−2(1)=02(1) + 4(0) - 2(1) = 0 ✓

Example 3: Only Trivial Solution

Solve: {x+y=0x−y=0\begin{cases} x + y = 0 \\ x - y = 0 \end{cases}

Row Reduction to RREF

Step 0 of 3

The RREF shows both columns are pivot columns (no free variables).

Solution: x=0,y=0x = 0, y = 0 (trivial solution only).

Structure of Solution Sets

Homogeneous systems have a special structure:

Theorem (Solution Set Structure):
If x⃗1\vec{x}_1 and x⃗2\vec{x}_2 are solutions to Ax⃗=0⃗A\vec{x} = \vec{0}, then:

  1. x⃗1+x⃗2\vec{x}_1 + \vec{x}_2 is also a solution
  2. cx⃗1c\vec{x}_1 is also a solution for any scalar cc

Proof:

  1. A(x⃗1+x⃗2)=Ax⃗1+Ax⃗2=0⃗+0⃗=0⃗A(\vec{x}_1 + \vec{x}_2) = A\vec{x}_1 + A\vec{x}_2 = \vec{0} + \vec{0} = \vec{0}
  2. A(cx⃗1)=c(Ax⃗1)=c0⃗=0⃗A(c\vec{x}_1) = c(A\vec{x}_1) = c\vec{0} = \vec{0}

Both use linearity of matrix-vector multiplication. ∎

The solution set of Ax⃗=0⃗A\vec{x} = \vec{0} is called the null space (or kernel) of AA.

Key properties:

  • Always contains 0⃗\vec{0} (trivial solution)
  • Closed under addition: sum of solutions is a solution
  • Closed under scalar multiplication: scalar multiple of a solution is a solution

This makes the solution set a subspace of Rn\mathbb{R}^n.

Geometric interpretation:

  • If only trivial solution: Just the origin (0-dimensional)
  • If 1 free variable: A line through the origin (1-dimensional)
  • If 2 free variables: A plane through the origin (2-dimensional)
  • And so on...

Example 4: Combining Solutions

From Example 2, we found two solutions: v⃗1=[−210],v⃗2=[101]\vec{v}_1 = \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}

Then 3v⃗1−2v⃗23\vec{v}_1 - 2\vec{v}_2 is also a solution: 3v⃗1−2v⃗2=3[−210]−2[101]=[−630]+[−20−2]=[−83−2]3\vec{v}_1 - 2\vec{v}_2 = 3\begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} - 2\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -6 \\ 3 \\ 0 \end{bmatrix} + \begin{bmatrix} -2 \\ 0 \\ -2 \end{bmatrix} = \begin{bmatrix} -8 \\ 3 \\ -2 \end{bmatrix}

Verify: (−8)+2(3)−(−2)=−8+6+2=0(-8) + 2(3) - (-2) = -8 + 6 + 2 = 0 ✓

Relationship to Non-Homogeneous Systems

If Ax⃗=b⃗A\vec{x} = \vec{b} is a non-homogeneous system (where b⃗≠0⃗\vec{b} \neq \vec{0}), its solution set relates to the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}:

Theorem:
If x⃗p\vec{x}_p is a particular solution to Ax⃗=b⃗A\vec{x} = \vec{b}, then every solution has the form: x⃗=x⃗p+x⃗h\vec{x} = \vec{x}_p + \vec{x}_h where x⃗h\vec{x}_h is a solution to the homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}.

Geometric interpretation: The solution set of Ax⃗=b⃗A\vec{x} = \vec{b} is the solution set of Ax⃗=0⃗A\vec{x} = \vec{0} translated by x⃗p\vec{x}_p.

Proof: Suppose x⃗\vec{x} is any solution to Ax⃗=b⃗A\vec{x} = \vec{b}. Let x⃗h=x⃗−x⃗p\vec{x}_h = \vec{x} - \vec{x}_p. Then: Ax⃗h=A(x⃗−x⃗p)=Ax⃗−Ax⃗p=b⃗−b⃗=0⃗A\vec{x}_h = A(\vec{x} - \vec{x}_p) = A\vec{x} - A\vec{x}_p = \vec{b} - \vec{b} = \vec{0}

So x⃗h\vec{x}_h solves the homogeneous system. Thus x⃗=x⃗p+x⃗h\vec{x} = \vec{x}_p + \vec{x}_h. ∎

Visualization:

  • Homogeneous system Ax⃗=0⃗A\vec{x} = \vec{0}: Solution set passes through origin (line, plane, etc.)
  • Non-homogeneous system Ax⃗=b⃗A\vec{x} = \vec{b}: Solution set is parallel but shifted (doesn't pass through origin)

For example:

  • Ax⃗=0⃗A\vec{x} = \vec{0} might have solutions forming a line through origin
  • Ax⃗=b⃗A\vec{x} = \vec{b} has solutions forming a parallel line, but shifted away from origin

Practice Problems

Consider the homogeneous system: [213101112][xyz]=[000]\begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}

Verify that x⃗=0⃗\vec{x} = \vec{0} is a solution.

Just compute A0⃗A\vec{0} directly.

A0⃗=[213101112][000]=0[211]+0[101]+0[312]=[000]A\vec{0} = \begin{bmatrix} 2 & 1 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & 2 \end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = 0\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} + 0\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} + 0\begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} ✓

This works for any matrix AA—the trivial solution is always a solution.

Determine if the following homogeneous system has nontrivial solutions: {x−2y+z=02x−4y+2z=0\begin{cases} x - 2y + z = 0 \\ 2x - 4y + 2z = 0 \end{cases}

If so, find the general solution.

Row reduce and look for free variables.

Augmented matrix (augmented with zero column): [1−2102−420]\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 2 & -4 & 2 & 0 \end{array}\right]

Row reduce: [1−2102−420]→R2−2R1[1−2100000]\left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 2 & -4 & 2 & 0 \end{array}\right] \xrightarrow{R_2 - 2R_1} \left[\begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

Only one pivot column (first column). Variables yy and zz are free.

From x−2y+z=0x - 2y + z = 0: x=2y−zx = 2y - z

General solution: x⃗=[2y−zyz]=y[210]+z[−101]\vec{x} = \begin{bmatrix} 2y - z \\ y \\ z \end{bmatrix} = y\begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} + z\begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}

Answer: Yes, infinitely many nontrivial solutions.

If v⃗1=[12−1]\vec{v}_1 = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} and v⃗2=[211]\vec{v}_2 = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} are both solutions to Ax⃗=0⃗A\vec{x} = \vec{0}, is v⃗3=[551]\vec{v}_3 = \begin{bmatrix} 5 \\ 5 \\ 1 \end{bmatrix} also a solution?

Check if v⃗3\vec{v}_3 can be written as a linear combination of v⃗1\vec{v}_1 and v⃗2\vec{v}_2.

If v⃗3=c1v⃗1+c2v⃗2\vec{v}_3 = c_1\vec{v}_1 + c_2\vec{v}_2 for some c1,c2c_1, c_2, then yes (by the solution set structure theorem).

Try to solve: c1[12−1]+c2[211]=[551]c_1\begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix} + c_2\begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 5 \\ 1 \end{bmatrix}

System: {c1+2c2=52c1+c2=5−c1+c2=1\begin{cases} c_1 + 2c_2 = 5 \\ 2c_1 + c_2 = 5 \\ -c_1 + c_2 = 1 \end{cases}

From first equation: c1=5−2c2c_1 = 5 - 2c_2

Substitute into third: −(5−2c2)+c2=1-(5 - 2c_2) + c_2 = 1 −5+2c2+c2=1-5 + 2c_2 + c_2 = 1 3c2=63c_2 = 6 c2=2c_2 = 2

Then c1=5−2(2)=1c_1 = 5 - 2(2) = 1

Check with second equation: 2(1)+2=4≠52(1) + 2 = 4 \neq 5 ✗

Answer: No, v⃗3\vec{v}_3 is not a solution.

Summary

Homogeneous systems are fundamental for understanding concepts like linear independence, basis, and dimension in the next notes.