Orthogonal Complement¶

Definition¶

Let $V$ be a real vector space with an inner product, written $\boldsymbol{x}\cdot\boldsymbol{y}$ (in $\mathbb{R}^n$: the dot product), and let $W\subseteq V$$V$. The orthogonal complement of $W$ is

the set of vectors orthogonal to every vector of $W$. It is read "$W$ perp". Usually $W$ is a subspace.

Interpretation¶

$W^\perp$ collects all directions perpendicular to $W$. In $\mathbb{R}^3$ the complement of a plane through the origin is its normal line, and the complement of a line through the origin is the plane perpendicular to it.

The figure draws the first example below: $W=\operatorname{span}\{(1,1,1)^{\top}\}$ is a line, and $W^\perp$ is the plane $x_1+x_2+x_3=0$ with basis $(-1,1,0)^{\top},(-1,0,1)^{\top}$. Every vector of the plane is orthogonal to $(1,1,1)^{\top}$, the two meet only in $\boldsymbol{0}$, and the dimensions add up to $1+2=3$$3$.

"Complement" does not mean set complement: $V\setminus W$ is never a subspace, as it misses $\boldsymbol{0}$. Nor is $W^\perp$ just some subspace $U$ with $V=W\oplus U$$U$: such complements are many, but the orthogonal one is unique. It depends on the inner product; with a different inner product on $\mathbb{R}^n$ the orthogonal complement can change.

To compute it, it suffices to test a spanning set: if $W=\operatorname{span}\{\boldsymbol{w}_1,\ldots,\boldsymbol{w}_k\}$$\mathrm{span}\{{\mathbf{w}}_1,\dots ,{\mathbf{w}}_k\}$, then $\boldsymbol{v}\in W^\perp$ if and only if $\boldsymbol{v}\cdot\boldsymbol{w}_i=0$ for $i=1,\ldots,k$$1,\dots ,k$. In $\mathbb{R}^n$ this is the homogeneous system $M^{\top}\boldsymbol{v}=\boldsymbol{0}$ with $M=\begin{pmatrix}\boldsymbol{w}_1&\cdots&\boldsymbol{w}_k\end{pmatrix}$$\left(\begin{array}{ccc}{\mathbf{w}}_1& \cdots & {\mathbf{w}}_k\end{array}\right)$, so $W^\perp$ is the null space (kernel) of $M^{\top}$.

Key Results¶

• $W^\perp$ is a subspace for every subset $W$, and $W^\perp=(\operatorname{span}W)^\perp$$(\mathrm{span}W{)}^{\perp }$.
• $W\cap W^\perp=\{\boldsymbol{0}\}$ for a subspace $W$, since $\boldsymbol{w}\cdot\boldsymbol{w}=0$ forces $\boldsymbol{w}=\boldsymbol{0}$$0$.
• If $V$ is finite-dimensional and $W$ a subspace: $V=W\oplus W^\perp$ (a direct sum), $\dim W+\dim W^\perp=\dim V$$\dim V$, and $(W^\perp)^\perp=W$$W$.
• Inclusions reverse: $W_1\subseteq W_2$ implies $W_2^\perp\subseteq W_1^\perp$$W_1^{\perp }$. Also $(W_1+W_2)^\perp=W_1^\perp\cap W_2^\perp$$W_2^{\perp }$.
• $\{\boldsymbol{0}\}^\perp=V$ and $V^\perp=\{\boldsymbol{0}\}$$\{0\}$.
• For a real matrix $A$, $\operatorname{Null}(A)=\operatorname{Row}(A)^\perp$ and $\operatorname{Null}(A^{\top})=\operatorname{Col}(A)^\perp$ (Fundamental Theorem of Linear Algebra).

Examples¶

A line in $\mathbb{R}^3$. For $W=\operatorname{span}\{(1,1,1)^{\top}\}$$\mathrm{span}\{(1,1,1{)}^{\top }\}$, $W^\perp=\{\boldsymbol{x}\colon x_1+x_2+x_3=0\}$$0\}$, with basis $(-1,1,0)^{\top},(-1,0,1)^{\top}$. Dimensions: $1+2=3$$3$.

A plane in $\mathbb{R}^3$. For $W=\{\boldsymbol{x}\colon x_1-2x_2+x_3=0\}$$0\}$, $W^\perp=\operatorname{span}\{(1,-2,1)^{\top}\}$$\mathrm{span}\{(1,-2,1{)}^{\top }\}$: the coefficient vector is the normal vector of the plane.

A plane in $\mathbb{R}^4$. For $W=\operatorname{span}\{(1,0,1,0)^{\top},(0,1,0,1)^{\top}\}$$\mathrm{span}\{(1,0,1,0{)}^{\top },(0,1,0,1{)}^{\top }\}$, the conditions $x_1+x_3=0$ and $x_2+x_4=0$ give

and $2+2=4$$4$. Note that $(1,0,0,0)^{\top}$ is not in $W$, yet it is not in $W^\perp$ either: most vectors lie in neither.

Prerequisites¶

• Inner Product
• Orthogonal Vectors
• Subspace
• Direct Sum

Related Concepts¶

• Orthogonal Projection
• Fundamental Theorem of Linear Algebra
• Gram-Schmidt Algorithm
• Normal Vector
• Hyperplane
• Orthonormal Basis

Sources¶

• Math V: Advanced Linear Algebra · Week 5

Aliases: perp, orthocomplement, perpendicular complement Keywords: W^⊥, W perp, (W^⊥)^⊥ = W, dim W + dim W^⊥ = n, R^n = W ⊕ W^⊥, normal line Symbols: element of, intersection, real numbers, subset

Last updated: 22 October 2026 19:04