Null Space & Column Space of a Matrix

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# Null Space & Column Space of a Matrix ## 🎯 Learning Objectives After this topic you will be able to: - Find a basis for the null space of a matrix - Find a basis for the column space of a matrix - Relate nullity and rank to the geometry of $A\mathbf{x} = \mathbf{b}$ - Understand the relationship between the four...

Null Space & Column Space of a Matrix

🎯 Learning Objectives

After this topic you will be able to:

Find a basis for the null space of a matrix
Find a basis for the column space of a matrix
Relate nullity and rank to the geometry of $A\mathbf{x} = \mathbf{b}$
Understand the relationship between the four fundamental subspaces
Determine whether a given vector is in the null space or column space

📋 Prerequisites

Rank & Dimension (Week 4) — rank, nullity, rank-nullity theorem
Gaussian Elimination (Week 2) — solving homogeneous systems
Subspaces (Week 3) — definition of subspaces

1. The Four Fundamental Subspaces

Every

m \times n

matrix

A

defines four subspaces:

Subspace	Notation	Definition	Lives in
Null space	$N(A)$	$\{\mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0}\}$	$\mathbb{R}^n$
Column space	$C(A)$	$\{A\mathbf{x} \in \mathbb{R}^m \mid \mathbf{x} \in \mathbb{R}^n\}$	$\mathbb{R}^m$
Row space	$R(A)$	Span of rows of $A$	$\mathbb{R}^n$
Left null space	$N(A^T)$	$\{\mathbf{y} \in \mathbb{R}^m \mid A^T\mathbf{y} = \mathbf{0}\}$	$\mathbb{R}^m$

(Diagram)

2. Null Space (Kernel)

2.1 Definition and Geometry

Definition (Null Space). The null space of an $m \times n$ matrix $A$ is:

> N(A) = \{\mathbf{x} \in \mathbb{R}^n \mid A\mathbf{x} = \mathbf{0}\} >

The null space contains all vectors that

A

annihilates (maps to zero). It is always a subspace of

\mathbb{R}^n

. Geometric intuition: If

A

projects

\mathbb{R}^3

onto the

xy

-plane, the

z

-axis becomes the null space — every vector

(0,0,z)

maps to

(0,0,0)

2.2 Finding a Basis for the Null Space

Algorithm:

Row reduce $A$ to RREF
Identify pivot columns (basic variables) and non-pivot columns (free variables)
Express basic variables in terms of free variables
Write the solution as a linear combination of basis vectors, one per free variable

Example 1: Null space basis

> A = \begin{bmatrix} 1 & 2 & 0 & 1 & -1 \\ 0 & 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} >

(already in RREF)

Step 1: Identify pivots. Pivot columns: 1, 3. Free variables: $x_2, x_4, x_5$ .

Step 2: Write equations:

> x_1 + 2x_2 + x_4 - x_5 = 0 \Rightarrow x_1 = -2x_2 - x_4 + x_5 >

> x_3 + 2x_4 + x_5 = 0 \Rightarrow x_3 = -2x_4 - x_5 >

Step 3: Parametric form:

> \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = x_2\begin{bmatrix} -2 \\ 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} + x_4\begin{bmatrix} -1 \\ 0 \\ -2 \\ 1 \\ 0 \end{bmatrix} + x_5\begin{bmatrix} 1 \\ 0 \\ -1 \\ 0 \\ 1 \end{bmatrix} >

Basis for $N(A)$ : $\{(-2,1,0,0,0), (-1,0,-2,1,0), (1,0,-1,0,1)\}$ .

$\text{nullity}(A) = 3$ .

2.3 Checking Membership

To check if

\mathbf{x} \in N(A)

: compute

A\mathbf{x}

. If

A\mathbf{x} = \mathbf{0}

, then

\mathbf{x}

is in the null space.

Example 2: Membership test

Is $(2,-1,0,0,0)$ in $N(A)$ from Example 1?

> A\mathbf{x} = \begin{bmatrix} 1(2) + 2(-1) + 0 + 0 + 0 \\ 0 + 0 + 0 + 0 + 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} >

✓

Yes, it's in $N(A)$ . Indeed, it's $-\frac{1}{2}$ times the first basis vector.

3. Column Space (Range)

3.1 Definition

Definition (Column Space). The column space of $A$ is:

> C(A) = \{\mathbf{y} \in \mathbb{R}^m \mid \mathbf{y} = A\mathbf{x} \text{ for some } \mathbf{x} \in \mathbb{R}^n\} >

Equivalently, $C(A) = \text{span}\{\text{columns of } A\}$ . The column space captures all possible outputs of $A$ .

3.2 Finding a Basis for the Column Space

Algorithm:

Row reduce $A$ to RREF
Identify pivot columns
The original columns of $A$ at the pivot positions form a basis for $C(A)$

Example 3: Column space basis

> A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 2 & 4 & 1 & 3 \\ 1 & 2 & -1 & 0 \end{bmatrix} >

Row reduce:

> \xrightarrow{R_2-2R_1, R_3-R_1} \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 \end{bmatrix} \xrightarrow{R_3+R_2} \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} >

Pivot columns: 1 and 3.

Basis for $C(A)$ :

> \left\{\begin{bmatrix}1\\2\\1\end{bmatrix}, \begin{bmatrix}0\\1\\-1\end{bmatrix}\right\} >

.

$\text{rank}(A) = 2$ .

3.3 Checking Membership

To check if

\mathbf{b} \in C(A)

: solve

A\mathbf{x} = \mathbf{b}

. If a solution exists,

\mathbf{b}

is in the column space.

Example 4: Column space membership

Is $\mathbf{b} = (3,5,1)$ in $C(A)$ from Example 3?

Solve $A\mathbf{x} = \mathbf{b}$ :

> \left[\begin{array}{cccc|c} 1 & 2 & 0 & 1 & 3 \\ 2 & 4 & 1 & 3 & 5 \\ 1 & 2 & -1 & 0 & 1 \end{array}\right] >

After row reduction, check consistency. If consistent, $\mathbf{b} \in C(A)$ .

4. Column Space vs. Row Space

Aspect	Column Space $C(A)$	Row Space $R(A)$
Subspace of	$\mathbb{R}^m$	$\mathbb{R}^n$
Basis from	Pivot columns of original $A$	Non-zero rows of RREF
Dimension	$\text{rank}(A)$	$\text{rank}(A)$
Related to	Outputs of $A$	Linear relations among columns

Example 5: Row space basis

Using $A$ from Example 3, RREF is

> \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix} >

.

Basis for $R(A)$ : $\{(1,2,0,1), (0,0,1,1)\}$ .

5. The Fundamental Theorem of Linear Algebra

Theorem (Fundamental Theorem of Linear Algebra, Part I). For an $m \times n$ matrix $A$ :

$N(A)$ and $R(A)$ are orthogonal complements in $\mathbb{R}^n$

$N(A^T)$ and $C(A)$ are orthogonal complements in $\mathbb{R}^m$

$\dim(N(A)) + \dim(R(A)) = n$

$\dim(N(A^T)) + \dim(C(A)) = m$ Orthogonal complements means every vector in one subspace is orthogonal to every vector in the other, and together they span the whole space.

6. Edge Cases & Gotchas

Situation	What Happens
Zero matrix $0_{m \times n}$	$N(A) = \mathbb{R}^n$ (whole domain), $C(A) = \{\mathbf{0}\}$
Identity $I_n$	$N(A) = \{\mathbf{0}\}$ , $C(A) = \mathbb{R}^n$
Square invertible	$N(A) = \{\mathbf{0}\}$ , $C(A) = \mathbb{R}^n$
One row	$N(A)$ is $(n-1)$ -dimensional; $C(A)$ is a line

7. Common Pitfalls

❌ Pitfall 1: Thinking $N(A)$ and $C(A)$ live in the same space

N(A) \subseteq \mathbb{R}^n

(domain),

C(A) \subseteq \mathbb{R}^m

(codomain). For

m \neq n

, they are different spaces entirely.

❌ Pitfall 2: Using RREF columns for column space basis

The column space basis must use original columns from

A

, not the RREF columns. The RREF columns are linear combinations of original columns and may live in a different coordinate system.

❌ Pitfall 3: Forgetting to check $\mathbf{0}$ for null space membership

\mathbf{0}

is always in

N(A)

. The question is whether there are non-zero vectors as well.

8. Formula Summary Table

Concept	Basis	Dimension
Null space $N(A)$	Parametric solution vectors of $A\mathbf{x}=\mathbf{0}$	$\text{nullity}(A) = n - r$
Column space $C(A)$	Original pivot columns of $A$	$\text{rank}(A) = r$
Row space $R(A)$	Non-zero rows of RREF	$\text{rank}(A) = r$
Left null space $N(A^T)$	Solve $A^T\mathbf{y} = \mathbf{0}$	$m - r$

9. 📝 Practice Questions

Q1: Null space of a 2×2 matrix

Find a basis for $N(A)$ where

> A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

.

Solution: Solve $A\mathbf{x} = \mathbf{0}$ : $x + 2y = 0$ , $2x + 4y = 0$ ⇒ $x = -2y$ , $y$ free. $N(A) = \text{span}\{(-2,1)\}$ . Dimension = 1. Q2: Column space of a 2×2 matrix

Find a basis for $C(A)$ where

> A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

.

Solution: Columns are $(1,2)$ and $(2,4)$ . Column 2 = 2(column 1), so they are dependent. $C(A) = \text{span}\{(1,2)\}$ . Dimension = 1. Q3: Three independent columns

Find a basis for $C(A)$ where

> A = \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & 2 & 3 \end{bmatrix} >

.

Solution: Row reduce:

> \begin{bmatrix} 1 & 0 & -1 \\ 2 & 1 & 0 \\ 1 & 2 & 3 \end{bmatrix} \xrightarrow{R_2-2R_1, R_3-R_1} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \end{bmatrix} \xrightarrow{R_3-2R_2} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix} >

Pivot columns: 1 and 2. Basis for $C(A)$ : columns 1 and 2 of original $A$ : $\{(1,2,1), (0,1,2)\}$ . Q4: Complete solution to Ax = b

Find the complete solution to $A\mathbf{x} = \mathbf{b}$ where

> A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

,

> \mathbf{b} = \begin{bmatrix} 3 \\ 6 \end{bmatrix} >

.

Solution: Particular solution: Solve $x + 2y = 3$ . Set $y = 0$ , then $x = 3$ . $\mathbf{x}_p = (3,0)$ .

Homogeneous solution: $N(A) = \text{span}\{(-2,1)\}$ .

Complete solution:

> \mathbf{x} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} + t\begin{bmatrix} -2 \\ 1 \end{bmatrix} >

, $t \in \mathbb{R}$ . Q5: Column space membership test

Is $\mathbf{b} = (1, 2, 3)$ in $C(A)$ for

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

?

Solution: Solve $A\mathbf{x} = \mathbf{b}$ :

> \begin{cases} x = 1 \\ y = 2 \\ x + y = 3 \end{cases} \Rightarrow 1 + 2 = 3 \text{ ✓} >

Yes, $\mathbf{b} \in C(A)$ with $\mathbf{x} = (1,2)$ . Q6: Null space of a 3×3 singular matrix

Find $N(A)$ for

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

.

Solution: Solve $A\mathbf{x} = \mathbf{0}$ : $x + y + z = 0$ , $y + z = 0$ ⇒ $y = -z$ , $x = -y - z = z - z = 0$ . So $x = 0$ , $y = -z$ , $z$ free.

$N(A) = \text{span}\{(0,-1,1)\}$ . Nullity = 1. Q7: Finding left null space

Find a basis for $N(A^T)$ where

> A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

.

Solution:

> A^T = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} >

(same here). Solve $A^T\mathbf{y} = \mathbf{0}$ : $y_1 + 2y_2 = 0$ , $2y_1 + 4y_2 = 0$ ⇒ $y_1 = -2y_2$ , $y_2$ free.

$N(A^T) = \text{span}\{(-2,1)\}$ . Dimension = $m - r = 2 - 1 = 1$ . Q8: Rank from nullity

If $A$ is $7 \times 5$ with nullity 3, what is the rank?

Solution: $\text{rank}(A) = 5 - 3 = 2$ . Q9: When does N(A) = {0}?

For what type of matrix does $N(A) = \{\mathbf{0}\}$ ?

Solution: When $A$ has full column rank, i.e., $\text{rank}(A) = n$ (number of columns). Then nullity = $n - n = 0$ . Q10: Four subspaces dimensions

For $A_{4 \times 6}$ with rank 2, give dimensions of all four fundamental subspaces.

Solution:

$\dim(N(A)) = 6 - 2 = 4$

$\dim(C(A)) = 2$

$\dim(R(A)) = 2$

$\dim(N(A^T)) = 4 - 2 = 2$

🔗 Cross-References

Next topic: Rank-Nullity Theorem (review)
Next: Linear Transformations Introduction
Week 6 (Kernel & Image): Generalises null space and column space
BSMA1001 (Maths 1): Solving systems of equations
BSCS2004 (ML Foundations): Underdetermined and overdetermined systems Join Discord Previous4.2 Rank & DimensionNext5.2 Linear Transformations

What this note is really saying

Null Space & Column Space of a Matrix

🎯 Learning Objectives

📋 Prerequisites

1. The Four Fundamental Subspaces

2. Null Space (Kernel)

2.1 Definition and Geometry

2.2 Finding a Basis for the Null Space

2.3 Checking Membership

3. Column Space (Range)

3.1 Definition

3.2 Finding a Basis for the Column Space

3.3 Checking Membership

4. Column Space vs. Row Space

5. The Fundamental Theorem of Linear Algebra

6. Edge Cases & Gotchas

7. Common Pitfalls

❌ Pitfall 1: Thinking N(A)N(A)N(A) and C(A)C(A)C(A) live in the same space

❌ Pitfall 2: Using RREF columns for column space basis

❌ Pitfall 3: Forgetting to check 0\mathbf{0}0 for null space membership

8. Formula Summary Table

9. 📝 Practice Questions

🔗 Cross-References

Intelligence Hub

❌ Pitfall 1: Thinking $N(A)$ and $C(A)$ live in the same space

❌ Pitfall 3: Forgetting to check $\mathbf{0}$ for null space membership