What Happens When We Take the Complement Twice?
For a subspace \(W\) of an inner product space \(V\), the orthogonal complement \(W^\perp\) consists of the vectors in \(V\) orthogonal to every vector in \(W\). Taking a complement imposes orthogonality conditions, so \(W^\perp\) is generally different from \(W\). But if we take the complement of \(W^\perp\), the result is exactly \(W\), provided the space is finite-dimensional. The key is to count how many directions are in each complement, using an orthonormal basis.
Throughout the main result, \(V\) is a finite-dimensional inner product space over \(\mathbb{F}\), where \(\mathbb{F}=\mathbb{R}\) or \(\mathbb{C}\), and \(W\) is a subspace of \(V\). Over \(\mathbb{C}\), the inner product is linear in its first slot and conjugate-linear in its second. In particular, conjugate symmetry means that \(\langle \mathbf{x},\mathbf{y}\rangle=0\) if and only if \(\langle \mathbf{y},\mathbf{x}\rangle=0\).
The space \(V\) matters: both complements are taken inside \(V\), not inside a larger space. We use the standard fact that the orthogonal complement of a subspace is itself a subspace. If \(n=\dim V\) and \(k=\dim W\), the central dimension count will be \(\dim W^\perp=n-k\).
Using an Orthonormal Basis to Count Dimensions
An orthonormal list is a list of unit vectors that are pairwise orthogonal. An orthonormal basis is an orthonormal list that is also a basis. Since \(W\) is finite-dimensional, it has an orthonormal basis; this follows by applying Gram–Schmidt to a basis of \(W\). The basis can be extended to an orthonormal basis of \(V\), by the orthonormal basis extension result.
Proof. Let \(n=\dim V\) and \(k=\dim W\). Choose an orthonormal basis \((\mathbf{u}_1,\dots,\mathbf{u}_k)\) of \(W\) and extend it to an orthonormal basis \((\mathbf{u}_1,\dots,\mathbf{u}_n)\) of \(V\). Because this is an orthonormal basis, every \(\mathbf{x}\in V\) has the expansion \(\mathbf{x}=\sum_{j=1}^n\langle\mathbf{x},\mathbf{u}_j\rangle\mathbf{u}_j\). Indeed, writing \(\mathbf{x}=\sum_{j=1}^n c_j\mathbf{u}_j\) and taking the inner product with \(\mathbf{u}_i\) gives \(\langle\mathbf{x},\mathbf{u}_i\rangle=c_i\).
We claim that \(W^\perp=\operatorname{span}\{\mathbf{u}_{k+1},\dots,\mathbf{u}_n\}\). First, let \(\mathbf{x}\in W^\perp\). For \(1\leq i\leq k\), we have \(\mathbf{u}_i\in W\), so \(\langle\mathbf{x},\mathbf{u}_i\rangle=0\). The expansion of \(\mathbf{x}\) therefore contains only \(\mathbf{u}_{k+1},\dots,\mathbf{u}_n\), proving that \(\mathbf{x}\) lies in their span. Conversely, each \(\mathbf{u}_j\) with \(j>k\) is orthogonal to every \(\mathbf{u}_i\) with \(i\leq k\). Every vector of \(W\) is a linear combination of \(\mathbf{u}_1,\dots,\mathbf{u}_k\), so every vector in the span of \(\mathbf{u}_{k+1},\dots,\mathbf{u}_n\) is orthogonal to every vector of \(W\). This proves the claimed equality. The added vectors are part of a basis, hence are linearly independent, and there are \(n-k\) of them. Thus \(\dim W^\perp=n-k\). If \(k=n\), the added list is empty and its span is \(\{\mathbf{0}\}\); if \(k=0\), the added list is a basis of all of \(V\). The same count applies in both cases. \(\square\)
This proof identifies more than the number of dimensions: it identifies which directions make up the complement. The basis vectors belonging to \(W\) account for \(k\) directions, and the added vectors account for the remaining \(n-k\). The orthonormality is what lets us read off a vector’s coefficients by inner products and recognize exactly when its components along \(W\) vanish.
The Double Orthogonal Complement Theorem
Proof. First, \(W\subseteq(W^\perp)^\perp\). To verify this, take any \(\mathbf{w}\in W\) and any \(\mathbf{z}\in W^\perp\). By the definition of \(W^\perp\), \(\langle\mathbf{z},\mathbf{w}\rangle=0\). Conjugate symmetry gives \(\langle\mathbf{w},\mathbf{z}\rangle=0\), so \(\mathbf{w}\) is orthogonal to every vector in \(W^\perp\), as required.
Next, write \(n=\dim V\) and \(k=\dim W\). The dimension theorem applied first to \(W\) and then to the subspace \(W^\perp\) gives
We have an inclusion of finite-dimensional subspaces \(W\subseteq(W^\perp)^\perp\) with equal dimensions, so they are equal. To justify this last step, if the inclusion were strict, there would be a vector in \((W^\perp)^\perp\) not in \(W\). Adding that vector to a basis of \(W\) would give a linearly independent list in \((W^\perp)^\perp\) with \(k+1\) vectors, contradicting that its dimension is \(k\). Hence \((W^\perp)^\perp=W\). \(\square\)
The inclusion and dimension count have separate roles. Orthogonality gives the inclusion, but inclusion alone does not imply equality. The dimension calculation rules out a larger subspace containing \(W\). Finite dimensionality is essential to this argument because it supplies finite dimensions and the basis extension used to count them.
Worked Examples
Worked Example: A Plane in \(\mathbb{R}^3\)
Let \(V=\mathbb{R}^3\) with the dot product and let \(W=\operatorname{span}\{(1,1,0)^\top,(1,-1,2)^\top\}\). Find \(W^\perp\) and verify the double-complement result.
The two generators are orthogonal because \(1(1)+1(-1)+0(2)=0\). Their squared lengths are \(1^2+1^2+0^2=2\) and \(1^2+(-1)^2+2^2=6\). Thus an orthonormal basis for \(W\) is
The vector \(\mathbf{q}_3=\frac{1}{\sqrt{3}}(1,-1,-1)^\top\) has squared length \((1+1+1)/3=1\). Its dot products with \(\mathbf{q}_1\) and \(\mathbf{q}_2\) are \((1-1)/\sqrt{6}=0\) and \((1+1-2)/\sqrt{18}=0\). Consequently \((\mathbf{q}_1,\mathbf{q}_2,\mathbf{q}_3)\) is an orthonormal basis of \(\mathbb{R}^3\), and \(W^\perp=\operatorname{span}\{\mathbf{q}_3\}\). In particular, \(\dim W=2\) and \(\dim W^\perp=1=3-2\).
A vector \((x,y,z)^\top\) lies in \((W^\perp)^\perp\) exactly when its dot product with \((1,-1,-1)^\top\) is zero, or \(x-y-z=0\). For such a vector, set \(\alpha=(x+y)/2\) and \(\beta=z/2\). The condition gives \(x-y=z\), so \(\alpha+\beta=x\), \(\alpha-\beta=y\), and \(2\beta=z\). Hence
Conversely, both generators of \(W\) are orthogonal to \((1,-1,-1)^\top\), so every vector in \(W\) belongs to \((W^\perp)^\perp\). Thus the two spaces are equal, as the theorem predicts.
Worked Example: A Subspace of a Polynomial Space
Let \(V=P_2(\mathbb{R})\), the space of real polynomials of degree at most two, with inner product defined by coefficient vectors: \(\langle a+bt+ct^2,d+et+ft^2\rangle=ad+be+cf\). Let \(W=\operatorname{span}\{1+t\}\). Find both complements.
The polynomial \(q=(1+t)/\sqrt{2}\) has coefficient vector \((1,1,0)^\top/\sqrt{2}\), whose squared length is \((1+1)/2=1\). Therefore \(q\) is an orthonormal basis of \(W\). A polynomial \(p=a+bt+ct^2\) belongs to \(W^\perp\) exactly when
It follows that \(p=a(1-t)+ct^2\), and hence \(W^\perp=\operatorname{span}\{1-t,t^2\}\), with dimension \(2=3-1\). Now take \(p=a+bt+ct^2\) in \((W^\perp)^\perp\). Orthogonality to \(1-t\) and to \(t^2\) requires \(a-b=0\) and \(c=0\), respectively. Thus \(b=a\), and \(p=a(1+t)\in W\). Every multiple of \(1+t\) satisfies these two conditions, so \((W^\perp)^\perp=W\).
Worked Example: The Zero Subspace
Take \(V=\mathbb{R}^2\) with the dot product and \(W=\{\mathbf{0}\}\). Every vector is orthogonal to \(\mathbf{0}\), since \(\mathbf{x}\cdot\mathbf{0}=0\), so \(W^\perp=\mathbb{R}^2\). A vector \((x,y)^\top\) belongs to \((W^\perp)^\perp\) only if it is orthogonal to both \((1,0)^\top\) and \((0,1)^\top\). Those dot products are \(x\) and \(y\), so both must be zero. Therefore \((W^\perp)^\perp=\{\mathbf{0}\}=W\). The dimensions also agree with the theorem: \(\dim W=0\), \(\dim W^\perp=2\), and \(\dim(W^\perp)^\perp=0\).
A Common Pitfall and Why the Result Matters
Pitfall: thinking the second complement is the first complement again. For \(W=\operatorname{span}\{(1,1)^\top\}\subseteq\mathbb{R}^2\), a student might incorrectly claim that taking the complement twice leaves the line \(\operatorname{span}\{(1,-1)^\top\}\) unchanged. The first complement is indeed that line, since \((x,y)^\top\) is orthogonal to \((1,1)^\top\) precisely when \(x+y=0\). But a vector \((x,y)^\top\) in the second complement must be orthogonal to \((1,-1)^\top\), giving the condition \(x-y=0\), not \(x+y=0\). Thus the second complement is \(\operatorname{span}\{(1,1)^\top\}=W\). The mistake confuses \(W^\perp\) with \((W^\perp)^\perp\); checking the defining dot-product equation corrects it.
The double-complement theorem lets us recover a subspace from the vectors orthogonal to it. It also clarifies the dimensions: if \(W\) occupies \(k\) dimensions of an \(n\)-dimensional space, its orthogonal complement occupies exactly the remaining \(n-k\). The theorem does not say that \(W\) and \(W^\perp\) are the same space, or that they have equal dimensions. It says that complementing twice returns \(W\). In later applications, this relationship helps organize orthogonality conditions and subspaces without losing information about the original subspace.
Check Your Understanding
Use the definitions and dimension count to justify each answer.
- If \(\dim V=7\) and \(\dim W=4\), what are \(\dim W^\perp\) and \(\dim(W^\perp)^\perp\)?
- In \(\mathbb{R}^3\), find the condition on \((x,y,z)^\top\) for it to be orthogonal to \(\operatorname{span}\{(1,0,1)^\top\}\), and state the dimension of that complement.
- Explain why \(W\subseteq(W^\perp)^\perp\), including the role of conjugate symmetry over \(\mathbb{C}\).
- For \(W=V\), determine \(W^\perp\) and \((W^\perp)^\perp\), and check the dimensions.
- In the proof of the dimension theorem, why do the added orthonormal basis vectors span \(W^\perp\) rather than merely lie in it?
