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4.1 Homogeneous systems and null space

Use homogeneous systems and null spaces to describe all solutions systematically, not just one solution at a time.

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MATH1030: Linear algebra I

Rigorous linear algebra notes on systems, matrices, structure, and proof, with interaction used only where it clarifies the mathematics.

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Chapter 1

Systems of equations

Learn to read equations as full solution sets.

1.1

1.1 Equations and solution sets

Chapter 2

Matrices and elimination

Build matrix intuition and use row reduction with purpose.

2.1

2.1 Matrix basics

2.2

2.2 Augmented matrices and row operations

2.3

2.3 Gaussian elimination and RREF

2.4

2.4 Solution-set types

Chapter 3

Matrix algebra

Matrix multiplication, transpose, and structural matrix notation.

3.1

3.1 Matrix multiplication and identity matrices

3.2

3.2 Transpose and special matrices

Chapter 4

Solution structure

Homogeneous systems, null spaces, and the shape of full solution sets.

4.1

4.1 Homogeneous systems and null space

Chapter 5

Invertibility

Understand when a matrix can be undone and why that matters.

5.1

5.1 Invertible matrices

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Chapter 6

Vector spaces

Move from matrix procedures to the structure of spaces, span, independence, and basis.

6.1

6.1 Vector spaces

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6.2

6.2 Subspaces

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6.3

6.3 Linear combinations and span

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6.4

6.4 Linear dependence and independence

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6.5

6.5 Basis and dimension

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6.6

6.6 Column space, row space, and rank

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By the time you can row-reduce a system, the next question is no longer "Can I solve this example?" but rather "What is the structure of every solution?" Homogeneous systems are the cleanest place to ask that question, and the null space is the language that answers it.

Why homogeneous systems are special

A homogeneous linear system is a system whose constant terms are all 0. In matrix form, it looks like

Ax=0.Ax = 0.

This situation is special for one immediate reason: the zero vector always solves it.

Definition

Homogeneous system

A homogeneous linear system is a linear system of the form

Ax=0.Ax = 0.

Its trivial solution is the zero vector x=0x = 0.

The real question is whether there are also nontrivial solutions.

The null space collects all homogeneous solutions

Definition

Null space

If AA is a matrix, the null space of AA is

N(A)={x:Ax=0}.N(A) = \{x : Ax = 0\}.

So N(A) is exactly the solution set of the homogeneous system Ax=0Ax = 0.

This definition turns a list of solutions into a mathematical object. Instead of saying "here are some vectors that work," you can describe the whole set at once.

Row reduction tells you the shape of the null space

To find N(A), you solve Ax=0Ax = 0 by reducing the augmented system [A∣0][A \mid 0]. The pivots tell you which variables are determined; the free variables tell you how many directions of freedom remain.

Worked example

Solve a homogeneous system and describe the null space

Let

A=[12−124−2].A = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{bmatrix}.

To solve Ax=0Ax = 0, row-reduce:

[12−1024−20]∼[12−100000].\left[ \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 2 & 4 & -2 & 0 \end{array} \right] \sim \left[ \begin{array}{ccc|c} 1 & 2 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right].

So the equation is

x1+2x2−x3=0.x_1 + 2x_2 - x_3 = 0.

Take x2=sx_2 = s and x3=tx_3 = t as free variables. Then

x1=−2s+t.x_1 = -2s + t.

Therefore

x=[−2s+tst]=s[−210]+t[101].x = \begin{bmatrix} -2s + t \\ s \\ t \end{bmatrix} = s \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix} + t \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.

So

N(A)=Span⁡{[−210],[101]}.N(A) = \operatorname{Span} \left\{ \begin{bmatrix} -2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\}.

This example shows why null-space descriptions are powerful. They tell you not only whether solutions exist, but how every solution is built.

Homogeneous solutions control nonhomogeneous ones

The same idea explains the structure of a system Ax=bAx = b when it is consistent.

Theorem

Every solution is a particular solution plus a null-space vector

Suppose xpx_p is one particular solution of Ax=bAx = b.

Then a vector x solves Ax=bAx = b if and only if

x=xp+vx = x_p + v

for some v∈N(A)v \in N(A).

This is the key structural theorem behind free-variable formulas.

Proof

Why the full solution set has the form xp+N(A)x_p + N(A)

A nonhomogeneous example

Worked example

Describe all solutions as a translate of the null space

Suppose the system Ax=bAx = b has one particular solution

xp=[301],x_p = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix},

and suppose

N(A)=Span⁡{[1−10]}.N(A) = \operatorname{Span} \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \right\}.

Then every solution has the form

x=[301]+t[1−10]=[3+t−t1],t∈R.x = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} + t \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 + t \\ -t \\ 1 \end{bmatrix}, \qquad t \in R.

The null space gives the direction of freedom; the particular solution tells you where that family of solutions sits.

What null space says about uniqueness

The structure theorem gives an immediate test.

  • If N(A)={0}N(A) = \{0\}, then a consistent system Ax=bAx = b has exactly one solution.
  • If N(A) contains a nonzero vector, then every consistent system Ax=bAx = b has infinitely many solutions, because you can add scalar multiples of that vector to a particular solution.

So null space measures the hidden freedom in the system.

Common mistakes

Common mistake

The zero vector always belongs to the null space

Students sometimes think a homogeneous system can have no solution. That is impossible, because x=0x = 0 always satisfies Ax=0Ax = 0.

Common mistake

A particular solution is not the whole solution set

Finding one vector xpx_p with Axp=bAx_p = b is only the start. You still need to add the whole null space to describe every solution.

Quick checks

Quick check

Why does Ax=0Ax = 0 always have at least one solution?

Answer in one sentence.

Solution

Answer

Quick check

If N(A)={0}N(A) = \{0\} and Ax=bAx = b is consistent, how many solutions does it have?

Use the theorem from this note.

Solution

Answer

Exercise

Quick check

Suppose xpx_p solves Ax=bAx = b and u,v∈N(A)u, v \in N(A). Why do xp+ux_p + u and xp+vx_p + v both solve Ax=bAx = b?

Write one line using linearity.

Solution

Guided solution

Related notes

This note builds on 2.3 Gaussian elimination and RREF and 2.4 Solution-set types. It prepares the way for 5.1 Invertible matrices and connects naturally with 6.2 Subspaces.

Prerequisites

2.4

2.4 Solution-set types

Key terms in this unit

More notes in this series

Previous section

3.2 Transpose and special matrices

Next section

5.1 Invertible matrices