Lesson 4: The Rank-Nullity Theorem
Estimated time: 35-45 minutes
Learning Objectives
- State the Rank-Nullity Theorem
- Understand the proof idea and why it works
- Apply the theorem to determine rank, nullity, or domain dimension
- Connect rank and nullity to one-to-one and onto properties
Statement of the Theorem
Rank-Nullity Theorem: Let T : R^n → R^m be a linear transformation with matrix A. Then:
where rank(A) = dim(range(T)) = number of pivot columns, and nullity(A) = dim(ker(T)) = number of free variables.
In plain language: the number of dimensions in the output (rank) plus the number of dimensions crushed to zero (nullity) always adds up to the number of input dimensions (n).
Why It Works (Proof Sketch)
When you row reduce an n-column matrix A, every column is either a pivot column or a free column. There is no third option.
- Pivot columns correspond to independent directions in the range.
- Free columns correspond to independent directions in the kernel (each free variable gives one basis vector of the null space).
Since every column is one or the other: (number of pivots) + (number of free variables) = n. That is rank + nullity = n.
Applying the Theorem
Worked Example 1: Finding Nullity from Rank
A is a 4x6 matrix with rank 3. What is the nullity?
rank + nullity = n = 6 (number of columns).
3 + nullity = 6, so nullity = 3.
The kernel is 3-dimensional: three free variables, so Ax = 0 has a 3-parameter family of solutions.
Worked Example 2: A Concrete Matrix
A = [1 2 1 0; 0 1 1 1; 1 3 2 1]. Row reduce:
[0 1 1 1] → [0 1 1 1]
[1 3 2 1] → [0 0 0 0]
Pivots in columns 1, 2. Free variables: x3, x4. So rank = 2, nullity = 2. Check: 2 + 2 = 4 = n. ✓
Worked Example 3: Using the Theorem Strategically
T : R^5 → R^3 is linear. Can T be one-to-one?
If T were one-to-one, nullity = 0, so rank = 5. But rank ≤ min(m, n) = min(3, 5) = 3.
Since rank cannot be 5, T cannot be one-to-one. The kernel must have dimension at least 2.
Connections to One-to-One and Onto
For T : R^n → R^m:
T is one-to-one ⇔ nullity = 0 ⇔ rank = n ⇔ A has a pivot in every column. Requires n ≤ m.
T is onto ⇔ rank = m ⇔ A has a pivot in every row. Requires m ≤ n.
T is both (invertible) ⇔ rank = n = m ⇔ A is a square invertible matrix.
Worked Example 4
T : R^4 → R^4 with rank 3. What can we conclude?
nullity = 4 - 3 = 1. The kernel is a line (1-dimensional).
T is NOT one-to-one (nullity ≠ 0) and NOT onto (rank ≠ 4 = m).
For square matrices, failing one means failing both.
A Dimension-Counting Tool
The Rank-Nullity Theorem is essentially a dimension-counting principle. The domain splits into two complementary pieces: one piece maps faithfully into the range, and the other collapses to zero. Their dimensions always add up to the total.
Analogy
Think of a projector casting a 3D scene onto a 2D screen. One dimension of depth is "lost" (crushed to the screen). rank = 2 (the screen) + nullity = 1 (the depth direction) = 3 (the 3D space).
Check Your Understanding
1. A 3x5 matrix has nullity 2. What is the rank?
2. Can a linear transformation T : R^3 → R^5 be onto?
3. T : R^7 → R^4 is onto. What is the dimension of the kernel?
4. A square matrix A has nullity 0. Is A invertible?
Key Takeaways
- Rank-Nullity Theorem: rank(A) + nullity(A) = n (number of columns)
- rank = number of pivot columns = dim(range)
- nullity = number of free variables = dim(kernel)
- A map R^n → R^m cannot be one-to-one if n > m, and cannot be onto if n < m
- For square matrices: one-to-one, onto, and invertible are all equivalent