What is the null space of this matrix?

How it works

The null space of an m × n matrix A is the set of all vectors x with Ax = 0.

1. RREF. Reduce A to its reduced row echelon form R by Gauss-Jordan elimination, as the RREF calculator does: for each column from the left, the first remaining row with a nonzero entry becomes the pivot row; it is swapped up, scaled so the pivot is 1, and subtracted from every other row to clear the column. Ax = 0 and Rx = 0 have the same solutions.
2. Pivot and free variables. A column of R with a leading 1 is a pivot column; each other column j gives a free variable x_j.
3. Special solutions. For each free column j: set x_j = 1 and every other free variable to 0. Row r of R holds the pivot of some column p, and gives x_p = −R[r][j]. The vector is the special solution for x_j.
4. Basis. The special solutions, in the order of their free columns, are a basis of the null space (Interactive Linear Algebra §2.7). Their number is the nullity, n − rank (the Rank Theorem).

When every column has a pivot, the null space is only the zero vector and the nullity is 0.

Rules

• The matrix has 1 to 10 rows and 1 to 10 columns. Each cell is an integer, a decimal or a fraction. A cell typed as a fraction (1/3 or 3/2) is read as that exact fraction, and a decimal as its exact value (0.25 is 1/4), so all the arithmetic is exact.

Output format

• Every entry is an exact fraction in lowest terms, written as an integer or as p/q, with the true minus sign (−). A vector is written in square brackets with its entries separated by commas: [−2, 1, 0, 0]. Matrices are written row by row in square brackets, rows separated by semicolons: [1, 2, 0, −2; 0, 0, 1, 2; 0, 0, 0, 0].
• Basis: the special solutions, separated by commas, in the order of their free columns; with no free variables, the zero vector only, [0, 0, …] with n zeros.
• All solutions of Ax = 0: x = x₂·[…] + x₄·[…], each free variable (named x with its column number as subscript digits) times its special solution; with no free variables, x = [0, 0, …].
• Free variables: x₂, x₄, or none. Pivot columns: 1, 3, or none.
• Steps, separated by semicolons: RREF: R; Pivot columns: …; free variables: …; for each free variable Set x_j = 1: v, with the words "and the other free variables to 0" (after a space) before the colon when there are two or more free variables; then Nullity = n columns − rank r = k. Example: Set x₂ = 1 and the other free variables to 0: [−2, 1, 0, 0].
• The answer sentence reads The null space has dimension k, with basis B., with the nullity and the basis as above.

Worked examples by hand

A 3 × 4 matrix. A = [1, 2, 2, 2; 2, 4, 6, 8; 3, 6, 8, 10].

1. R2 → R2 − 2 × R1 gives [0, 0, 2, 4]; R3 → R3 − 3 × R1 gives [0, 0, 2, 4].
2. Column 2 has no nonzero entry below row 1, so it is free. In column 3, R2 → ½ × R2 gives [0, 0, 1, 2]; R1 → R1 − 2 × R2 gives [1, 2, 0, −2]; R3 → R3 − 2 × R2 gives [0, 0, 0, 0].

R = [1, 2, 0, −2; 0, 0, 1, 2; 0, 0, 0, 0]: pivot columns 1 and 3, free variables x₂ and x₄, rank 2.

• x₂ = 1, x₄ = 0: x₁ = −2, x₃ = 0, so [−2, 1, 0, 0].
• x₄ = 1, x₂ = 0: x₁ = −(−2) = 2, x₃ = −2, so [2, 0, −2, 1].

Nullity = 4 − 2 = 2. Every solution is x = x₂·[−2, 1, 0, 0] + x₄·[2, 0, −2, 1]. Check: A × [−2, 1, 0, 0] = [−2 + 2, −4 + 4, −6 + 6] = 0.

A 3 × 3 matrix of rank 2. [1, 2, 3; 4, 5, 6; 7, 8, 9] reduces to [1, 0, −1; 0, 1, 2; 0, 0, 0]. x₃ is free: x₃ = 1 gives x₁ = 1 and x₂ = −2, so the basis is [1, −2, 1] and the nullity is 1.

An invertible matrix. [2, 1; 1, 3] has a pivot in both columns, so the null space is the zero vector only, [0, 0], and the nullity is 0.

One equation. [1, 1, 1] is already in RREF. x₂ and x₃ are free: [−1, 1, 0] and [−1, 0, 1], nullity 2.

Decimals. [0.5, 0.25; 1, 0.5] is [1/2, 1/4; 1, 1/2]. R1 → 2 × R1 gives [1, 1/2]; R2 → R2 − R1 gives [0, 0]. R = [1, 1/2; 0, 0]; x₂ = 1 gives x₁ = −1/2: [−1/2, 1].