Rank

The dimension of a matrix's column space — measuring how many linearly independent directions its transformation covers.

Rank counts how many independent directions the columns span

Rank 1 matrix

Matrix A has c₂ = 2c₁, so it has one independent direction.

Rank 2 matrix

Matrix B has two independent columns, so its column space fills the plane.

Definition

The rank of a matrix $A$ is the dimension of its column space — the number of linearly independent columns.

Equivalently, the rank equals:

For an $m \times n$ matrix: $\text{rank}(A) \leq \min(m, n)$ .

Full column rank: $\text{rank}(A) = n$ (all columns independent; $A\mathbf{x}=\mathbf{0}$ has only trivial solution)
Full row rank: $\text{rank}(A) = m$ (all rows independent; $A\mathbf{x}=\mathbf{b}$ always has a solution)
Full rank: $\text{rank}(A) = \min(m,n)$

Key properties

Row rank always equals column rank, even though rows and columns live in different spaces
$\text{rank}(A) \leq \min(m,n)$ — rank can never exceed either dimension
$\text{rank}(A^T) = \text{rank}(A)$ — transposing doesn't change the rank
Rank is invariant under row reduction (elementary row operations preserve rank)

Common mistakes

Assuming a "big" matrix has high rank: a matrix can be enormous in size yet have very low rank if its rows/columns are mostly redundant combinations of a few directions
Forgetting rank can drop under multiplication: $\text{rank}(AB) \leq \min(\text{rank}(A), \text{rank}(B))$ — multiplying by a low-rank matrix can only shrink rank, never grow it

Finding rank by row reduction

$A = \begin{pmatrix}1&2&3\\2&4&6\\0&1&1\end{pmatrix}$

Row 2 = $2\times$ Row 1: subtract to get zeros. After reduction:

$\begin{pmatrix}1&2&3\\0&0&0\\0&1&1\end{pmatrix} \to \begin{pmatrix}1&2&3\\0&1&1\\0&0&0\end{pmatrix}$

Two pivot positions → $\text{rank}(A) = 2$ .

Try it

A $4\times 5$ matrix $A$ has $\text{rank}(A) = 3$ . What are the dimensions of the null space and column space of $A$ ?

Solution

Column space (image): $\dim = \text{rank}(A) = 3$
Null space (kernel): by the rank-nullity theorem, $\text{rank}(A) + \text{nullity}(A) = n = 5$ , so $\text{nullity}(A) = 5 - 3 = 2$ . There are 2 free variables; the null space is 2-dimensional.

The null space is a subspace of $\mathbb{R}^5$ (domain); the column space is a subspace of $\mathbb{R}^4$ (codomain).

Related concepts

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