Orthogonal Complement Finder

Math Linear Algebra • Vector Spaces and Linear Independence

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

Find a basis for the orthogonal complement \[ V^\perp=\{w\in\mathbb{R}^n:\ w\cdot v=0\ \text{for all } v\in V\}. \] If \(V=\mathrm{span}\{v_1,\dots,v_k\}\) and \(B=[v_1\ \cdots\ v_k]\), then \[ V^\perp = \mathrm{Null}(B^T). \] This tool computes \(\mathrm{rref}(B)\), \(\mathrm{rref}(B^T)\), outputs a basis for \(V^\perp\), and verifies dot products.

Ambient dimension \(n\)

# vectors spanning \(V\) (\(k\))

Input orientation

Tolerance

Display decimals

Steps

Show row-ops log (for \(B^T\)) Simplify basis to integers (when possible) Orthonormalize \(V^\perp\) basis (Gram–Schmidt) Show dot-product checks Decompose test vector \(x=x_V+x_\perp\) Show 2D plot when \(n=2\)

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

Vectors spanning \(V\)

B is 3×1

We build \(B=[v_1\ \cdots\ v_k]\) and compute \(V^\perp=\mathrm{Null}(B^T)\).

Results

rank(B) = dim(V)

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dim(\(V^\perp\)) = \(n-\mathrm{rank}(B)\)

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Basis for \(V^\perp\)

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Optional orthonormal basis

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Ready

RREF

We compute RREF of \(B\) (for rank) and RREF of \(B^T\) (to build \(V^\perp=\mathrm{Null}(B^T)\)).

RREF(B)

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RREF(\(B^T\))

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Enter vectors spanning \(V\) and click “Calculate”.

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Theory

Orthogonal complement \(V^\perp\)

Let \(V\subseteq \mathbb{R}^n\) be a subspace. The orthogonal complement of \(V\) is \[ V^\perp = \{w\in\mathbb{R}^n : w\cdot v = 0 \ \text{for all } v\in V\}. \] Intuitively, \(V^\perp\) contains all directions “perpendicular to every direction in \(V\)”. It is always a subspace (closed under addition and scalar multiplication).

Matrix formulation: \(V^\perp=\mathrm{Null}(B^T)\)

Suppose \(V=\mathrm{span}\{v_1,\dots,v_k\}\) and form the matrix \[ B=[v_1\ v_2\ \cdots\ v_k]\in\mathbb{R}^{n\times k}. \] A vector \(w\in\mathbb{R}^n\) is orthogonal to every \(v_i\) exactly when \[ v_i\cdot w = 0 \quad \text{for all } i. \] Stack these dot products as a linear system: since \(B^T w\) is the vector whose entries are \(v_i\cdot w\), we get \[ w\in V^\perp \iff B^T w = 0 \iff w\in \mathrm{Null}(B^T). \] So finding a basis for \(V^\perp\) is a null-space problem.

Dimensions: \(\dim(V)+\dim(V^\perp)=n\)

In \(\mathbb{R}^n\), the orthogonal complement satisfies the dimension identity \[ \dim(V) + \dim(V^\perp) = n. \] With the matrix \(B\), \(\dim(V)=\mathrm{rank}(B)\). By rank–nullity applied to \(B^T\in\mathbb{R}^{k\times n}\), \[ \dim(\mathrm{Null}(B^T)) = n-\mathrm{rank}(B^T). \] Since \(\mathrm{rank}(B^T)=\mathrm{rank}(B)\), this gives \(\dim(V^\perp)=n-\mathrm{rank}(B)\).

How RREF produces a basis for \(V^\perp\)

To find \(\mathrm{Null}(B^T)\), compute the reduced row echelon form \(R=\mathrm{rref}(B^T)\). Then:

Pivot columns correspond to pivot variables.
Free columns correspond to free variables.

For each free variable, set it to \(1\) (others to \(0\)) and solve for pivot variables from \(R\). Each choice yields one null-space vector. These vectors form a basis for \(V^\perp\).

Geometric intuition (especially in 2D and 3D)

In \(\mathbb{R}^2\), any subspace is either \(\{0\}\), a line through the origin, or all of \(\mathbb{R}^2\). The orthogonal complement swaps lines with perpendicular lines:

If \(V\) is a line, \(V^\perp\) is the perpendicular line.
If \(V=\mathbb{R}^2\), then \(V^\perp=\{0\}\).
If \(V=\{0\}\), then \(V^\perp=\mathbb{R}^2\).

In \(\mathbb{R}^3\), a 1D subspace (a line) has a 2D orthogonal complement (a plane through the origin), and a 2D subspace (a plane) has a 1D orthogonal complement (a line normal to the plane).

University note: Hilbert spaces

Orthogonal complements generalize beyond \(\mathbb{R}^n\) to inner-product spaces, including Hilbert spaces. Many ideas remain the same: \(V^\perp\) is defined using the inner product, and decompositions like \(x=x_V+x_\perp\) become foundational for projections, Fourier series, and least-squares theory.

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