Rank-Nullity
Theorem
Discover the fundamental relationship between the rank and nullity of a linear transformation: how the dimensions of image and kernel always add up to the dimension of the domain.
What Is a Linear Map?
A linear transformation (or linear map) is a function \( T : V \to W \) between vector spaces that preserves addition and scalar multiplication:
Think of \( T \) as a machine: you feed in vectors from \( V \), and it outputs vectors in \( W \). The Rank-Nullity Theorem tells us exactly how the input space splits between what gets mapped to zero and what produces distinct outputs.
The Kernel (Null Space)
The kernel of \( T \), written \( \ker(T) \), is the set of all vectors in \( V \) that \( T \) maps to the zero vector:
The nullity of \( T \) is the dimension of the kernel: \( \text{nullity}(T) = \dim(\ker(T)) \). A larger kernel means more vectors "collapse" to zero: more information is lost.
The Image (Range)
The image (or range) of \( T \), written \( \text{im}(T) \), is the set of all vectors in \( W \) that are actually "hit" by \( T \):
The rank of \( T \) is the dimension of the image: \( \text{rank}(T) = \dim(\text{im}(T)) \). A larger image means \( T \) produces more "distinct directions" in the output space.
How V Splits Into Two Pieces
The Rank-Nullity Theorem
Equivalently, for a matrix \( A \) of size \( m \times n \):
Why Columns, Not Rows?
For an \( m \times n \) matrix \( A \), the map is \( T(\mathbf{x}) = A\mathbf{x} \). The input \( \mathbf{x} \) lives in \( \mathbb{R}^n \), so \( \dim(V) = n \). The theorem partitions the domain (columns), not the codomain (rows).
Rank vs Nullity
| Rank | Nullity | |
|---|---|---|
| Measures | Dimension of image | Dimension of kernel |
| Interpretation | Distinct output directions | Directions crushed to zero |
| Matrix view | Number of pivot columns | Number of free variables |
| If maximized | \( T \) is surjective (if \( m \le n \)) | \( T \) maps everything to \( \mathbf{0} \) |
| If zero | \( T \) is the zero map | \( T \) is injective (one-to-one) |
How to Apply the Theorem
Example 1: A \( 2 \times 3 \) Matrix
Find the rank and nullity of \( A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{pmatrix} \).
Interpretation: row 2 is a multiple of row 1, so the map collapses a 2-dimensional subspace to zero, and only a 1-dimensional line is actually reached in the output.
Example 2: A \( 3 \times 3 \) Matrix
Find the rank and nullity of \( B = \begin{pmatrix} 1 & 0 & 2 \\ 0 & 1 & -1 \\ 1 & 1 & 1 \end{pmatrix} \).
The null space is spanned by one vector (the free-variable solution). The image is a 2D plane inside \( \mathbb{R}^3 \).
Example 3: Full Rank
Let \( C = \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \). Find its rank and nullity.
Important Consequences
Common Mistakes
Summary Cheat Sheet
| Concept | Definition | Matrix View |
|---|---|---|
| Rank | \( \dim(\text{im}(T)) \) | # of pivot columns |
| Nullity | \( \dim(\ker(T)) \) | # of free variables |
| Theorem | \( \dim(V) = \text{rank} + \text{nullity} \) | \( n = \text{pivots} + \text{free vars} \) |
| Injective | \( \ker(T) = \{\mathbf{0}\} \) | nullity = 0, all columns pivot |
| Surjective | \( \text{im}(T) = W \) | rank = m, every row has pivot |
Question 1 of 5
A \( 4 \times 6 \) matrix \( A \) has rank 3. What is the nullity of \( A \)?
Question 2 of 5
If a linear map \( T : \mathbb{R}^5 \to \mathbb{R}^3 \) is surjective (onto), what is the nullity of \( T \)?
Question 3 of 5
Which quantity does the Rank-Nullity Theorem equate to \( \text{rank}(A) + \text{nullity}(A) \)?
Question 4 of 5
A \( 3 \times 3 \) matrix \( A \) has \( \text{nullity}(A) = 0 \). Which of the following must be true?
Question 5 of 5
A matrix \( A \) has 5 columns, 2 pivot columns, and 3 free variables. Which statement is correct?