Rank–Nullity Theorem

The theorem

For any $m \times n$ matrix $A$: $\operatorname{rank}(A) + \operatorname{nullity}(A) = n$

where $n$ is the number of columns, $\operatorname{rank}(A) = \dim \operatorname{Col}(A)$, and $\operatorname{nullity}(A) = \dim \operatorname{Null}(A)$.

In words: every column is either a pivot column (counted in rank) or a free-variable column (counted in nullity). There's no third category.

Step-by-step: given 3×4, rank = 2

Setup: $m = 3$, $n = 4$, $\operatorname{rank}(A) = 2$.

Apply the theorem: $2 + \operatorname{nullity}(A) = 4$ $\operatorname{nullity}(A) = 4 - 2 = \boxed{2}$

So the null space has two free parameters; it's a 2-dimensional subspace of $\mathbb{R}^4$.

Explicit example

Consider $A = \begin{pmatrix} 1 & 2 & 0 & 3 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \quad \text{(already in RREF)}$

• Pivot columns: 1 and 3. Rank = 2. ✓
• Free-variable columns: 2 and 4. Let $x_2 = s, x_4 = t$.

From RREF:

• $x_1 + 2s + 3t = 0 \implies x_1 = -2s - 3t$
• $x_3 - t = 0 \implies x_3 = t$

General null-space vector: $\mathbf{x} = \begin{pmatrix} -2s - 3t \\ s \\ t \\ t \end{pmatrix} = s \begin{pmatrix} -2 \\ 1 \\ 0 \\ 0 \end{pmatrix} + t \begin{pmatrix} -3 \\ 0 \\ 1 \\ 1 \end{pmatrix}$

The null space is spanned by those 2 vectors — confirming $\operatorname{nullity}(A) = 2$. ✓

Why the theorem is true

Row reduce $A$ to RREF. Every column is classified:

• A pivot column contributes 1 to the rank.
• A free column contributes one basis vector to the null space.

Every column is in exactly one category. Total columns = rank + nullity. $\square$

Implications

1. Solution structure. The solution set of $A \mathbf{x} = \mathbf{b}$ (when consistent) is an affine subspace of dimension $\operatorname{nullity}(A) = n - \operatorname{rank}(A)$. It's a single point iff $\operatorname{nullity}(A) = 0$.

2. Injectivity. The linear map $\mathbf{x} \mapsto A \mathbf{x}$ is one-to-one ⟺ $\operatorname{nullity}(A) = 0$ ⟺ $\operatorname{rank}(A) = n$ (full column rank).

3. Surjectivity. The map is onto $\mathbb{R}^m$ ⟺ $\operatorname{rank}(A) = m$ (full row rank).

4. Invertibility (square case). $A$ is invertible ⟺ rank = $n$ ⟺ nullity = $0$.

Dimension counting in applications

• Network flow: number of independent currents = number of nodes − 1 (rank-nullity in a graph's incidence matrix).
• Least squares: pseudo-inverse exists because rank tells us how much data we can fit.
• PCA: rank limits the dimensionality of the data's principal components.

Common mistakes

• Using $m$ instead of $n$. The theorem says rank + nullity = # columns, not # rows.
• Confusing null space with the zero vector. Null space is the set of vectors that $A$ sends to zero; it always contains $\mathbf{0}$ (trivially) but may contain more.
• Thinking nullity can be negative. It can't; it's a dimension.

Try it in the visualization

A slider controls $n$ (columns) and rank independently. A bar splits into "pivots" and "free" with lengths equal to rank and nullity. Identity $\operatorname{rank} + \operatorname{nullity} = n$ is shown algebraically and as bar lengths summing.