How this calculation works
Reduce the coefficient matrix. Each nonpivot variable is free. Give one free variable the value 1 and the other free variables 0, then solve for the pivot variables. These independent solution vectors span the null space.
A x = 0; nullity = number of columns − rank(A)A worked example
For [1 2 3; 2 4 6], a null-space basis is (−2,1,0), (−3,0,1). The rank is 1 and nullity is 2, totaling three coefficient columns.
Good to know
Exact rational matrices up to 6 × 6. Entries are bounded to ±1,000,000, integer fraction denominators to 1,000,000 and decimal input to 6 places. Intermediate fractions have a size limit. No complex entries, symbolic expressions, measurement uncertainty or numerical-conditioning analysis. Ordered vectors are entered as matrix columns.
Reference: TU Delft: basis and dimension (opens in a new tab) ↗
The null space is a space of inputs
A matrix with m rows and n columns maps n-component inputs to m-component outputs. Its null space consists of inputs that map to the zero output. Null-space basis vectors therefore have n components, even when the matrix has a different number of rows.
For A = [1 2 3; 2 4 6], the second equation duplicates the first. The remaining equation is x₁ + 2x₂ + 3x₃ = 0. Choosing x₂ and x₃ freely gives x = x₂(−2,1,0) + x₃(−3,0,1).
Every scalar combination of the displayed basis vectors is a solution, and every solution has such a combination. The basis vectors are examples of independent directions, not an exhaustive list of solutions.
Keep coefficient columns separate from constants
This is a homogeneous problem A x = 0. Supply only A, without appending a constants column. Appending a zero column would introduce an extra unknown and change the computed null space.
The number of pivot columns is the rank. The number of nonpivot columns is the nullity. Rank plus nullity equals the number of input columns; the output shows that identity as a check on dimensions.
A full-column-rank matrix has no nonzero null-space directions. Its null space still contains the zero vector, but its basis is empty. The zero vector is not a one-element basis because it is not independent.
A few questions, answered
Does a zero-dimensional null space mean there are no solutions?
No. The homogeneous system always has the zero solution. Nullity zero means that zero is the only solution.
Are these eigenvectors?
They are solutions to A x = 0. Nonzero ones are eigenvectors for eigenvalue zero when A is square; the tool does not compute other eigenvalues.
Can I solve A x = b here?
Use the linear-system solver for nonzero constants. This page specifically finds the basis of homogeneous solutions.