Subspace Membership Checker

Math Linear Algebra • Vector Spaces and Linear Independence

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

Test whether a vector \(v\in\mathbb{R}^d\) belongs to the subspace \(S=\mathrm{span}\{b_1,\dots,b_k\}\). This is done by solving \(Bc=v\) (with \(B=[b_1\ \cdots\ b_k]\)). The tool checks consistency via ranks (RREF) and, when possible, returns coefficients \(c\).

Dimension \(d\)

# basis/generators \(k\)

Basis orientation

Tolerance

Display decimals

Steps

Show row-ops log (RREF) Show parametric form (if infinite solutions) Simplify coefficients to integers (when possible) If no: show closest projection (least squares) Show 2D plot when \(d=2\)

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

Basis/generators \(b_1,\dots,b_k\)

B is 2×2

We test \(v\in \mathrm{span}\{b_i\}\) by solving \(Bc=v\).

Vector \(v\)

Enter the target vector to test for membership.

Results

Membership

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rank(B)

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rank([B|v])

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Residual \(\|Bc-v\|\)

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One coefficient solution \(c\) (if member)

If there are free variables, this tool returns one solution by setting free parameters to 0.

—

Ready

RREF panels

Consistency check uses \(\\mathrm{rank}(B)\) and \(\\mathrm{rank}([B|v])\). If the ranks match, \(v\in\\mathrm{Col}(B)=\\mathrm{span}\\{b_i\\}\).

RREF(B)

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RREF([B | v])

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Enter basis vectors and \(v\), then click “Calculate”.

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Theory

What subspace membership means

Let \(S=\mathrm{span}\{b_1,\dots,b_k\}\subseteq \mathbb{R}^d\). A vector \(v\) is in \(S\) exactly when it can be written as a linear combination: \[ v = c_1 b_1 + \cdots + c_k b_k. \] If we place the spanning vectors as columns of a matrix \[ B=[b_1\ \cdots\ b_k]\in\mathbb{R}^{d\times k}, \] then membership becomes the linear system \[ Bc=v. \]

Consistency via ranks (RREF test)

Create the augmented matrix \([B|v]\). The system \(Bc=v\) is consistent if and only if \[ \mathrm{rank}(B)=\mathrm{rank}([B|v]). \] This is exactly what RREF reveals: if a row becomes \([0\ 0\ \cdots\ 0\ |\ \beta]\) with \(\beta\neq 0\), then the system is inconsistent, so \(v\notin\mathrm{span}(B)\).

Finding the coefficients

If the system is consistent, there is at least one coefficient vector \(c\). If \(B\) has full column rank (rank \(=k\)), the solution is unique. If rank \(infinitely many solutions: some variables are free parameters.

A standard approach from RREF is:

Identify pivot columns (pivot variables) and free columns (free variables).
Set one free variable to \(1\) (others \(0\)) to build special solutions, or set all free variables to \(0\) to get one simple solution.

If not in the span: closest vector (projection)

When \(v\notin\mathrm{span}(B)\), there is no exact solution to \(Bc=v\). A common “best alternative” is the closest vector in the span: \[ v_{\mathrm{proj}}=\arg\min_{x\in\mathrm{span}(B)} \|v-x\|. \] This is the least-squares projection of \(v\) onto the column space of \(B\). The distance \(\|v-v_{\mathrm{proj}}\|\) measures how far \(v\) is from the subspace.

2D intuition

In \(\mathbb{R}^2\), a subspace can only be \(\{0\}\), a line through the origin, or all of \(\mathbb{R}^2\). So membership is easy to visualize: either \(v\) lies on that line (or is the zero vector), or it doesn’t.

University note: affine spaces

Subspaces pass through the origin. Many problems instead ask whether a point lies in an affine subspace \(p+S=\{p+s:\ s\in S\}\). In that case you test membership by checking whether \(v-p\in S\), i.e. solve \(Bc=v-p\).

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