Linear Algebra · Intermediate · 10 min read

The Rank–Nullity Theorem and the Four Fundamental Subspaces

Every matrix hides four subspaces. Once you know what they are and how their dimensions fit together, a long list of linear algebra facts becomes a single picture.

Meeting the subspaces

For an m×n matrix A, the column space Col(A) is spanned by the columns and lives in ℝ^m. The row space Row(A) is spanned by the rows and lives in ℝⁿ. The null space Null(A) consists of all x with Ax = 0 and lives in ℝⁿ. The left null space, the null space of Aᵀ, lives in ℝ^m.

Dimensions and the theorem

The column space and the row space both have dimension equal to the rank r. The null space has dimension n − r, the number of free variables, and the left null space has dimension m − r. The statement rank + nullity = n is the rank–nullity theorem. It says that the columns of A either contribute a new direction (a pivot) or are redundant (a free variable), and these two cases exhaust all columns.

Finding bases

Reduce A to RREF. The pivot columns of the original matrix are a basis of the column space; the non-zero rows of the RREF are a basis of the row space; and each free variable produces one null space basis vector. Taking pivot columns from the RREF instead of from A is a very common mistake, because row operations change the column space.

Why it matters

Solvability of Ax = b is the question of whether b lies in Col(A). Uniqueness is the question of whether Null(A) is trivial. A square matrix is invertible exactly when its null space is {0}, which is when the rank is full.

Key formulas

rank(A) = dim Col(A) = dim Row(A)
nullity(A) = dim Null(A) = n − rank(A)
dim Null(Aᵀ) = m − rank(A)

Solved example

Subspaces of the 3×4 matrix with rows (1, 2, 1, 3), (2, 4, 0, 2), (3, 6, 1, 5)

Row reduce: R₂ ← R₂ − 2R₁ gives (0, 0, −2, −4); R₃ ← R₃ − 3R₁ gives (0, 0, −2, −4). Then R₃ ← R₃ − R₂ gives 0, and R₂ ← R₂ / (−2) gives (0, 0, 1, 2). Finally R₁ ← R₁ − R₂ gives (1, 2, 0, 1).
The pivots are in columns 1 and 3, so the rank is 2 and the nullity is 4 − 2 = 2.
Column space basis (original columns 1 and 3): (1, 2, 3) and (1, 0, 1).
Row space basis (non-zero RREF rows): (1, 2, 0, 1) and (0, 0, 1, 2).
Null space: the free variables are x₂ and x₄. Setting x₂ = 1 gives (−2, 1, 0, 0). Setting x₄ = 1 gives x₃ = −2 and x₁ = −1, so (−1, 0, −2, 1).
Check: the second vector times row 1 gives −1 + 0 − 2 + 3 = 0, and times row 2 gives −2 + 0 + 0 + 2 = 0; row 3 gives −3 + 0 − 2 + 5 = 0.

Practice questions

Try each question first, then open the answer.

1. A 4×6 matrix has rank 3. What is the dimension of its null space?

6 − 3 = 3.

2. A 5×5 matrix has non-zero determinant. What is its null space?

Only the zero vector, of dimension 0, because the matrix has full rank 5.

3. A 3×5 matrix has rank 3. What is the dimension of its left null space?

3 − 3 = 0, so the left null space is {0}.