Three Structural Properties
Orthogonal complements are defined by a condition involving every vector in a given subspace. Their basic properties follow from how inner products behave under addition and scalar multiplication, together with the positive-definiteness of the inner product. These facts do not require finite dimension: they hold in any real or complex inner product space.
Let \(V\) be an inner product space over \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\). For a subspace \(W\subseteq V\), a vector \(\mathbf{x}\) is orthogonal to \(W\) if \(\langle\mathbf{x},\mathbf{w}\rangle=0\) for every \(\mathbf{w}\in W\). The orthogonal complement collects exactly those vectors. We use the convention that the inner product is linear in its first slot; over \(\mathbb{C}\), it is conjugate-linear in its second slot.
The condition “for every \(\mathbf{w}\in W\)” matters: being orthogonal to a few selected vectors in \(W\) is not by itself enough unless those vectors are known to generate \(W\). When \(W\) is a span, however, it is enough to test the generators, because inner-product linearity carries orthogonality from the generators to each linear combination.
The Orthogonal Complement Is a Subspace
Proof. First, \(\langle\mathbf{0},\mathbf{w}\rangle=0\) for every \(\mathbf{w}\in W\), so \(\mathbf{0}\in W^{\perp}\). Now let \(\mathbf{x},\mathbf{y}\in W^{\perp}\) and let \(a,b\in\mathbb{F}\). For any \(\mathbf{w}\in W\), linearity in the first slot gives \(\langle a\mathbf{x}+b\mathbf{y},\mathbf{w}\rangle=a\langle\mathbf{x},\mathbf{w}\rangle+b\langle\mathbf{y},\mathbf{w}\rangle=a\cdot0+b\cdot0=0\). This holds for every \(\mathbf{w}\in W\), so \(a\mathbf{x}+b\mathbf{y}\in W^{\perp}\). Thus \(W^{\perp}\) contains zero and is closed under linear combinations, which proves it is a subspace. \(\square\)
The argument covers complex scalars as well as real ones: in the convention used here, multiplying the first vector in the inner product by \(a\) multiplies the inner product by \(a\). Even if the scalar were placed in the second slot, the result would still be zero when the original inner product is zero, since the second-slot factor would be \(\overline{a}\).
Worked Example: A Complement in \(\mathbb{R}^3\)
Let \(W=\operatorname{span}\{(1,1,0)^{\top},(0,1,1)^{\top}\}\) in \(\mathbb{R}^3\), with the usual dot product. Find \(W^{\perp}\) and verify that it is a subspace.
Write an arbitrary vector as \(\mathbf{x}=(x,y,z)^{\top}\). It belongs to \(W^{\perp}\) exactly when it is orthogonal to both generators. The resulting equations are
The first equation gives \(x=-y\), and the second gives \(z=-y\). Setting \(y=-t\) yields \(\mathbf{x}=t(1,-1,1)^{\top}\), so \(W^{\perp}=\operatorname{span}\{(1,-1,1)^{\top}\}\). To check the answer, the proposed generator has dot products \(1-1+0=0\) and \(0-1+1=0\) with the two generators of \(W\). By linearity, it is orthogonal to every vector in their span. Every multiple remains orthogonal because its dot products are multiplied by the same scalar. Thus the computed set has exactly the required vectors and is a subspace.
Inclusion Reverses
Suppose two subspaces are nested: \(U\subseteq W\). A vector orthogonal to all of \(W\) is automatically orthogonal to every vector in the smaller subspace \(U\), since each vector of \(U\) is also a vector of \(W\). The resulting inclusion of complements points in the opposite direction.
Proof. Let \(\mathbf{x}\in W^{\perp}\). By definition, \(\langle\mathbf{x},\mathbf{w}\rangle=0\) for every \(\mathbf{w}\in W\). Since \(U\subseteq W\), every \(\mathbf{u}\in U\) also belongs to \(W\), and hence \(\langle\mathbf{x},\mathbf{u}\rangle=0\) for every \(\mathbf{u}\in U\). By definition, \(\mathbf{x}\in U^{\perp}\). This proves \(W^{\perp}\subseteq U^{\perp}\). \(\square\)
This theorem asserts an implication, not an equality: it does not say that the two complements are always equal. A larger subspace imposes more orthogonality conditions, so its complement can be smaller.
Worked Example: Nested Subspaces in \(\mathbb{R}^3\)
Let \(U=\operatorname{span}\{\mathbf{e}_1\}\) and \(W=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\), where \(\mathbf{e}_1=(1,0,0)^{\top}\) and \(\mathbf{e}_2=(0,1,0)^{\top}\). Find both complements and check the reversed inclusion.
For \(\mathbf{x}=(x,y,z)^{\top}\), orthogonality to \(\mathbf{e}_1\) requires \(x=0\). Thus \(U^{\perp}=\{(0,y,z)^{\top}:y,z\in\mathbb{R}\}\). For orthogonality to both generators of \(W\), we require \(x=0\) and \(y=0\), so \(W^{\perp}=\{(0,0,z)^{\top}:z\in\mathbb{R}\}\). Every vector \((0,0,z)^{\top}\) has the form \((0,y,z)^{\top}\) with \(y=0\), proving \(W^{\perp}\subseteq U^{\perp}\). The containment is strict: \(\mathbf{e}_2=(0,1,0)^{\top}\in U^{\perp}\), since \(\mathbf{e}_2\cdot\mathbf{e}_1=0\), but \(\mathbf{e}_2\notin W^{\perp}\), since \(\mathbf{e}_2\cdot\mathbf{e}_2=1\ne0\).
The Intersection Contains Only Zero
A vector in \(W\cap W^{\perp}\) belongs to \(W\) and is orthogonal to every vector in \(W\). In particular, it must be orthogonal to itself. Positive-definiteness says that a vector has inner product zero with itself only when it is the zero vector.
Proof. Since \(W\) is a subspace, \(\mathbf{0}\in W\). Also \(\langle\mathbf{0},\mathbf{w}\rangle=0\) for every \(\mathbf{w}\in W\), so \(\mathbf{0}\in W^{\perp}\). Therefore \(\mathbf{0}\in W\cap W^{\perp}\). Conversely, let \(\mathbf{x}\in W\cap W^{\perp}\). Because \(\mathbf{x}\in W\) and \(\mathbf{x}\in W^{\perp}\), the definition of \(W^{\perp}\), applied to \(\mathbf{w}=\mathbf{x}\), gives \(\langle\mathbf{x},\mathbf{x}\rangle=0\). Positive-definiteness of the inner product then gives \(\mathbf{x}=\mathbf{0}\). Thus the intersection contains zero and no other vector. \(\square\)
Worked Example: A Complement in a Matrix Space
Consider \(V=\mathbb{R}^{2\times2}\) with the inner product \(\langle\mathbf{A},\mathbf{B}\rangle=\operatorname{tr}(\mathbf{A}^{\top}\mathbf{B})\), which is the sum of the products of corresponding entries. Let \(W=\operatorname{span}\{\mathbf{I}_2\}\). Find \(W^{\perp}\) and verify the intersection property.
Write \(\mathbf{X}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\). A matrix is in \(W^{\perp}\) exactly when it is orthogonal to every \(k\mathbf{I}_2\), \(k\in\mathbb{R}\). Compute \(\langle\mathbf{X},k\mathbf{I}_2\rangle=ka+kd=k(a+d)\). This equals zero for every real \(k\) exactly when \(a+d=0\). Therefore \(W^{\perp}=\left\{\begin{bmatrix}a&b\\c&-a\end{bmatrix}:a,b,c\in\mathbb{R}\right\}\). This set contains the zero matrix; adding two matrices of this form adds their \(a\), \(b\), and \(c\) entries and preserves the opposite diagonal entries; scalar multiplication also preserves that form. Hence it is a subspace.
A matrix in \(W\) has the form \(k\mathbf{I}_2=\begin{bmatrix}k&0\\0&k\end{bmatrix}\). To also belong to \(W^{\perp}\), its diagonal entries must sum to zero, so \(k+k=0\), or \(2k=0\). Over \(\mathbb{R}\), this forces \(k=0\). Thus the only matrix in both spaces is \(\mathbf{O}\). As a direct check, \(\langle\mathbf{I}_2,\mathbf{I}_2\rangle=2\ne0\), so no nonzero scalar multiple of \(\mathbf{I}_2\) is orthogonal to \(W\).
Common Pitfall: Reversing the Wrong Inclusion
A common error is to retain the direction of inclusion when taking complements. From \(U\subseteq W\), it is incorrect to conclude \(U^{\perp}\subseteq W^{\perp}\). In the nested-subspace example above, that mistaken conclusion would put \(\mathbf{e}_2\) in \(W^{\perp}\). But \(\mathbf{e}_2\cdot\mathbf{e}_2=1\), so \(\mathbf{e}_2\) is not orthogonal to \(W\). The correct conclusion is \(W^{\perp}\subseteq U^{\perp}\): orthogonality to the larger space requires satisfying at least all the conditions imposed by the smaller one.
These properties are useful before any calculation of a particular complement is attempted. The subspace result guarantees that \(W^{\perp}\) is itself a legitimate subspace; the inclusion rule predicts how complements relate when the original spaces are nested; and the intersection result rules out any nonzero vector being both inside \(W\) and orthogonal to all of \(W\). Each conclusion follows directly from the definition and the inner product axioms.
Check Your Understanding
Use the definitions and the three proved properties to answer the following.
- In \(\mathbb{R}^3\), let \(W=\operatorname{span}\{(1,0,1)^{\top}\}\). Find \(W^{\perp}\) as a set of solutions to an equation.
- If \(U\subseteq W\), which inclusion holds between \(U^{\perp}\) and \(W^{\perp}\)? Explain why.
- Why does \(\mathbf{x}\in W\cap W^{\perp}\) imply \(\langle\mathbf{x},\mathbf{x}\rangle=0\)?
- For \(W=\operatorname{span}\{\mathbf{I}_2\}\) in the matrix space example, determine whether \(\begin{bmatrix}1&2\\3&-1\end{bmatrix}\) belongs to \(W^{\perp}\).
- State the inner product property that rules out a nonzero vector in \(W\cap W^{\perp}\).
