Dimension of Vector Space Tool

Math Linear Algebra • Vector Spaces and Linear Independence

Written by STEM Calculators Team Published February 22, 2026 Updated February 24, 2026

Compute the dimension of a vector space/subspace from vectors or a matrix. In vector mode, \(\dim(\mathrm{span}\{v_i\})=\mathrm{rank}([v_1\cdots v_k])\). If the vectors are independent, they form a basis for their span and \(\dim=k\). In matrix mode, the tool reports \(\mathrm{rank}(A)\), \(\mathrm{nullity}(A)\), and related subspace dimensions.

Mode

Tolerance

Display decimals

Steps

Show row-ops log (RREF) Show 2D plot (vector mode, d=2)

Inputs accept -3.5, 2e-4, fractions like 7/3, and constants pi, e.

Dimension \(d\)

Number of vectors \(k\)

Input orientation

Vectors input

A is 3×3 (columns are vectors)

Pivot columns correspond to a basis subset for the span.

Results

Dimension result

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Rank / pivots

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Independence / basis check (vector mode)

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Nullity (matrix mode)

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Reduced basis subset

In vector mode: pivot vectors form a basis for the span. In matrix mode: row/column space basis is described by pivots.

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Ready

RREF

Pivot columns determine rank; in vector mode they identify a basis subset.

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Enter data and click “Calculate”.

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Theory

What “dimension” means

The dimension of a vector space \(V\), written \(\dim(V)\), is the number of vectors in any basis of \(V\). A basis is a set of vectors that is:

Linearly independent (no redundancy), and
Spanning (every vector in \(V\) can be written as a linear combination of the basis vectors).

A key fact: all bases of the same vector space have the same size. That common size is \(\dim(V)\).

Dimension from vectors: rank and pivots

If you have vectors \(v_1,\dots,v_k \in \mathbb{R}^d\), place them as columns of a matrix \[ A=[v_1\ v_2\ \cdots\ v_k]\in\mathbb{R}^{d\times k}. \] The subspace generated by these vectors is \(\mathrm{span}\{v_1,\dots,v_k\}\), which equals the column space of \(A\).

The dimension of the span is the rank: \[ \dim(\mathrm{span}\{v_i\})=\mathrm{rank}(A). \] When you compute RREF, the number of pivot columns equals \(\mathrm{rank}(A)\). The original vectors corresponding to pivot columns form a basis subset for the span.

Independence check: \[ \{v_i\}\ \text{independent} \iff \mathrm{rank}(A)=k. \] If independent, the set is a basis for its span and \(\dim(\mathrm{span})=k\). If not, the dimension is still \(\mathrm{rank}(A)\), but the set is not a basis.

Spanning \(\mathbb{R}^d\) and the “basis for \(\mathbb{R}^d\)” test

The vectors span all of \(\mathbb{R}^d\) exactly when \(\mathrm{rank}(A)=d\). A set is a basis for \(\mathbb{R}^d\) when it is independent and spans \(\mathbb{R}^d\), which for a finite set often appears as:

\(k=d\) and \(\mathrm{rank}(A)=d\) (square full rank), or
more generally \(\mathrm{rank}(A)=d\) and the set is independent (so necessarily \(k\ge d\)).

Important shortcut: if \(k>d\), the vectors cannot be independent, so they cannot be a basis of \(\mathbb{R}^d\).

Dimension from a matrix: rank–nullity

For a matrix \(A\in\mathbb{R}^{m\times n}\), several subspaces are naturally associated with it:

Column space (span of columns): dimension \(=\mathrm{rank}(A)\).
Row space (span of rows): dimension \(=\mathrm{rank}(A)\).
Null space \(N(A)=\{x:\ Ax=0\}\): dimension \(=\mathrm{nullity}(A)\).
Left null space \(N(A^T)\): dimension \(=m-\mathrm{rank}(A)\).

The rank–nullity theorem states: \[ \mathrm{rank}(A)+\mathrm{nullity}(A)=n. \] This is why the “null space dimension” of \(A\) is computed as \(n-\mathrm{rank}(A)\).

Infinite-dimensional note (university)

Some spaces are infinite-dimensional. For example, the space of all polynomials has an infinite basis \(\{1,x,x^2,x^3,\dots\}\), so its dimension is infinite. In many applied settings (e.g., polynomials up to degree \(n\)), the dimension becomes finite (\(n+1\)).

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