About

The column space of a matrix A is the set of all possible linear combinations of its column vectors. It forms a subspace of R^m where m is the number of rows. Computing it incorrectly leads to wrong conclusions about system solvability: a vector b lies in the column space of A if and only if Ax = b has a solution. Misidentifying the basis means misidentifying which systems are consistent. This tool applies Gaussian elimination with partial pivoting to compute the Reduced Row Echelon Form, identifies pivot columns, and extracts the corresponding columns from the original matrix as the basis for Col(A).

The calculator also computes the rank (dimension of the column space) and nullity via the Rank-Nullity Theorem: rank(A) + nullity(A) = n. Results assume exact rational-like arithmetic with floating-point tolerance 1e-10. For matrices with entries near that threshold, results may be unreliable. Supports real-valued matrices up to 10 × 10.

Formulas

The column space basis is found by reducing A to its Reduced Row Echelon Form and identifying pivot columns. The corresponding columns of the original matrix form the basis.

RREF(A) → pivot columns at positions j₁, j₂, …, j_r

Col(A) = Span{a_j₁, a_j₂, …, a_{j_r}}

The Rank-Nullity Theorem relates the fundamental dimensions:

rank(A) + nullity(A) = n

Where A is an m × n matrix, rank(A) = r is the number of pivot columns, and n is the number of columns. Gaussian elimination proceeds by: (1) find the leftmost non-zero column, (2) swap rows to place the largest absolute value at the pivot position (partial pivoting), (3) scale the pivot row so the leading entry is 1, (4) eliminate all other entries in that column via row replacement.

Reference Data

Property	Symbol / Notation	Definition	Dimension Formula
Column Space	Col(A) or Im(A)	Span of column vectors of A	rank(A)
Row Space	Row(A)	Span of row vectors of A	rank(A)
Null Space	Null(A) or Ker(A)	Solutions to Ax = 0	n − rank(A)
Left Null Space	Null(A^T)	Solutions to A^Ty = 0	m − rank(A)
Rank	rank(A)	Number of pivot positions in RREF	≤ min(m, n)
Nullity	nullity(A)	Dimension of null space	n − rank(A)
Full Column Rank	rank = n	All columns are linearly independent	Nullity = 0
Full Row Rank	rank = m	All rows are linearly independent	Left nullity = 0
Rank-Nullity Theorem	-	Fundamental dimension relationship	rank + nullity = n
Invertible Matrix	A⁻¹ exists	Square matrix with full rank	rank = m = n
Pivot Column	-	Column containing a leading 1 in RREF	Count = rank
Free Variable	-	Non-pivot column variable	Count = nullity
Spanning Set	-	Set whose span equals the subspace	Size ≥ dimension
Linear Independence	-	No vector is a combination of others	Max independent set = rank
Orthogonal Complement	Col(A)^⊥	Null(A^T)	m − rank(A)

Frequently Asked Questions

The column space of the original matrix A and its RREF are generally different subspaces. Row operations preserve the row space and null space, but not the column space. However, the RREF reveals which columns are pivot columns. Those same column positions in the original matrix A form a basis for Col(A). This is why the calculator extracts columns from the original matrix, not from the reduced form.

This calculator uses an epsilon tolerance of 1e-10 when determining if a value is zero during Gaussian elimination. If your matrix has entries that are very close to zero (within that threshold), the algorithm may incorrectly classify a pivot as zero or vice versa, producing a wrong rank. For matrices derived from measurement data with inherent noise, consider rounding entries before input. Singular Value Decomposition (SVD) provides more numerically stable rank determination but is beyond this tool's scope.

Yes. If rank(A) = m (full row rank), then the column space spans all of R^m. This requires at least m columns (n ≥ m) and that those columns contain m linearly independent vectors. In practical terms, the system Ax = b is consistent for every b.

This is a fundamental theorem in linear algebra. The number of pivot positions in the RREF equals both the number of linearly independent rows and the number of linearly independent columns. Intuitively, each pivot simultaneously determines an independent row direction and an independent column direction. Therefore dim(Col(A)) = dim(Row(A)) = rank(A).

If every column is a scalar multiple of one column (or the zero vector), the rank is at most 1. The column space is then either the zero subspace {0} (rank 0, all-zero matrix) or a line through the origin in R^m (rank 1). The nullity would be n − 1, meaning almost all variables are free.

Two checks: (1) The number of basis vectors must equal the rank. (2) The basis vectors must be linearly independent, which you can verify by forming a matrix from them and checking that its RREF has no zero rows. Additionally, every column of the original matrix must be expressible as a linear combination of the basis vectors. This calculator shows the RREF so you can cross-check pivot positions yourself.