4.1 Homogeneous systems and null space

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.

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.

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

The real question is whether there are also nontrivial solutions.

The null space collects all homogeneous solutions

Definition

Null space

If $A$ is a matrix, the null space of $A$ is

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

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

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 $Ax = 0$ by reducing the augmented system $[A \mid 0]$ . 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 = \begin{bmatrix} 1 & 2 & -1 \\ 2 & 4 & -2 \end{bmatrix}.

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

\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

x_1 + 2x_2 - x_3 = 0.

Take $x_2 = s$ and $x_3 = t$ as free variables. Then

x_1 = -2s + t.

Therefore

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}.

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 $Ax = b$ when it is consistent.

Theorem

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

Suppose $x_p$ is one particular solution of $Ax = b$ .

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

x = x_p + v

for some $v \in N(A)$ .

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

Proof

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

A nonhomogeneous example

Worked example

Describe all solutions as a translate of the null space

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

x_p = \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix},

and suppose

N(A) = \operatorname{Span} \left\{ \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} \right\}.

Then every solution has the form

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.

Why free variables force infinitely many homogeneous solutions

The reduced system viewpoint makes one important consequence immediate.

Theorem

A free variable creates infinitely many solutions

If the homogeneous system $Ax = 0$ has at least one free variable, then it has infinitely many solutions.

The reason is not mysterious. A free variable may be assigned any real value, and each different assignment gives a different solution vector unless the parameter disappears completely.

This also gives a short theorem that is worth stating explicitly: if a homogeneous system has more variables than pivot equations, then at least one free variable remains, and the system must therefore have infinitely many solutions.

Worked example

One free variable already produces a whole line of solutions

Suppose row reduction shows that

x_1 - 3x_2 = 0.

If $x_2 = t$ , then $x_1 = 3t$ , so every solution has the form

\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = t \begin{bmatrix} 3 \\ 1 \end{bmatrix}, \qquad t \in R.

Different values of t give different vectors, so the homogeneous system has infinitely many solutions, not just more than one.

Worked example

A trivial null space can also occur

Let

A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.

Then $Ax = 0$ is simply

x_1 = 0, \qquad x_2 = 0.

So the only solution is the zero vector, and therefore

N(A) = \{0\}.

This is the opposite extreme from the earlier example with free variables.

The null space is a subspace

This fact is easy to overlook because we first meet the null space as a solution set. But it is more than a solution set: it is always a subspace of the domain of $A$ .

Theorem

The null space is closed under linear combinations

For any matrix $A$ , the null space N(A) is a subspace. In particular:

$0 \in N(A)$ ,
if $u, v \in N(A)$ , then $u + v \in N(A)$ ,
if $u \in N(A)$ and c is a scalar, then $cu \in N(A)$ .

Proof

Why the null space is a subspace

This matters later because once you know a solution set is a subspace, you may look for a basis, count dimensions, and compare it to the pivot structure of the coefficient matrix.

Nullity counts how many independent directions remain

The previous discussion explains why the null space is not just a pile of solutions. It records how many genuinely independent directions of motion are still left after the pivot equations have imposed all their constraints.

Each free variable contributes one independent parameter. So the dimension of the null space is exactly the number of free variables in the reduced homogeneous system.

In rank language, this becomes

\operatorname{nullity}(A) = n - \operatorname{rank}(A),

but even before that theorem is named formally, you should already read nullity as "the number of independent null-space directions left by the system."

Worked example

Membership in the null space is a direct test

Let

A = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \end{bmatrix}, \qquad x = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}, \qquad z = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}.

Then

Ax = \begin{bmatrix} 0 \\ 0 \end{bmatrix}, \qquad Az = \begin{bmatrix} 1 \\ 0 \end{bmatrix}.

So $x \in N(A)$ but $z \notin N(A)$ . This is the practical meaning of the definition: to test membership in N(A), compute Ax and check whether the result is exactly the zero vector.

How to read a basis for the null space from RREF

In practice, the basis vectors of N(A) come directly from the free-variable description of the reduced system.

The workflow is:

row-reduce $A$ ,
identify pivot and free variables,
set one free variable to 1 and the others to 0,
solve for the pivot variables,
repeat once for each free variable.

The vectors produced this way form a basis candidate for the null space because each one records one independent direction of freedom.

There is also a geometric distinction worth keeping clear:

a homogeneous solution set is always a subspace, so it passes through the origin;
a nonhomogeneous consistent solution set is usually a translate of that subspace by a particular solution.

That difference is exactly why N(A) is the structural core of the system, while $x_p + N(A)$ describes the full solution set of $Ax = b$ .

Homogeneous solutions and column dependence

The null-space equation also explains when the columns of a matrix are dependent.

Let the columns of $A$ be $a_1, a_2, \ldots, a_n$ . Then

Ax = 0

is the same as

x_1 a_1 + x_2 a_2 + \cdots + x_n a_n = 0.

So a nontrivial solution of the homogeneous system is exactly a nontrivial linear relation among the columns.

Worked example

A nontrivial null-space vector gives a dependence relation

Suppose

x = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \in N(A).

Then

Ax = 0

means

1 \cdot a_1 - 2 \cdot a_2 + 1 \cdot a_3 = 0.

This is a nontrivial dependence relation among the columns of $A$ .

What null space says about uniqueness

The structure theorem gives an immediate test.

If $N(A) = \{0\}$ , then a consistent system $Ax = b$ has exactly one solution.
If N(A) contains a nonzero vector, then every consistent system $Ax = b$ 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 $x = 0$ always satisfies $Ax = 0$ .

Common mistake

A particular solution is not the whole solution set

Finding one vector $x_p$ with $Ax_p = b$ is only the start. You still need to add the whole null space to describe every solution.

Quick checks

Quick check

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

Answer in one sentence.

Solution

Answer

Quick check

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

Use the theorem from this note.

Solution

Answer

Quick check

If $Ax = 0$ has a free variable, can the solution set contain only two vectors?

Answer from the parameter form, not from a guess.

Solution

Answer

Exercise

Quick check

Suppose $x_p$ solves $Ax = b$ and $u, v \in N(A)$ . Why do $x_p + u$ and $x_p + v$ both solve $Ax = b$ ?

Write one line using linearity.

Solution

Guided solution

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.

4.1 Homogeneous systems and null space

MATH1030: Linear algebra I

Why homogeneous systems are special

Homogeneous system

The null space collects all homogeneous solutions

Null space

Row reduction tells you the shape of the null space

Solve a homogeneous system and describe the null space

Homogeneous solutions control nonhomogeneous ones

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

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

A nonhomogeneous example

Describe all solutions as a translate of the null space

Why free variables force infinitely many homogeneous solutions

A free variable creates infinitely many solutions

One free variable already produces a whole line of solutions

A trivial null space can also occur

The null space is a subspace

The null space is closed under linear combinations

Why the null space is a subspace

Nullity counts how many independent directions remain

Membership in the null space is a direct test

How to read a basis for the null space from RREF

Homogeneous solutions and column dependence

A nontrivial null-space vector gives a dependence relation

What null space says about uniqueness

Common mistakes

The zero vector always belongs to the null space

A particular solution is not the whole solution set

Quick checks

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

Answer

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

Answer

If Ax=0Ax = 0Ax=0 has a free variable, can the solution set contain only two vectors?

Answer

Exercise

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

Guided solution

Related notes

Section mastery checkpoint

Prerequisites

Key terms in this unit

More notes in this series

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

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

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

If $Ax = 0$ has a free variable, can the solution set contain only two vectors?

Suppose $x_p$ solves $Ax = b$ and $u, v \in N(A)$ . Why do $x_p + u$ and $x_p + v$ both solve $Ax = b$ ?