Annihilators and the Dual Space
A linear functional can detect whether a vector lies outside a subspace: if a functional vanishes on the subspace but not on the vector, that vector cannot belong to it. Annihilators collect all functionals that vanish on a given subspace. In finite dimensions, this collection contains enough information to recover the subspace exactly. It also reverses inclusion: a larger subspace imposes more vanishing conditions, so it has a smaller annihilator.
Let \(V\) be a finite-dimensional vector space over \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\). Its dual space \(V^*\) is the space of linear functionals \(f:V\to\mathbb{F}\). For a subspace \(U\subseteq V\), its annihilator is a subspace of \(V^*\). To take an annihilator of a subspace of \(V^*\), we use the dual space \(V^{**}=(V^*)^*\).
The double annihilator is initially a subspace of \(V^{**}\), not literally of \(V\). The identification needed to compare it with \(U\) is the evaluation map \(J:V\to V^{**}\), defined by \(J(\mathbf{v})(f)=f(\mathbf{v})\). For finite-dimensional \(V\), this map is an isomorphism. We use it to view \(V\) as \(V^{**}\); more precisely, the double-annihilator result below is \(J(U)=(U^0)^0\).
Dimension Facts Needed for the Proofs
We will use two dimension calculations. The first says that an annihilator has the complementary dimension. Here is a basis argument that also applies when the space being annihilated is \(V^*\).
Proof. Write \(k=\dim S\) and \(n=\dim X\). Choose a basis \((\mathbf{x}_1,\dots,\mathbf{x}_k)\) of \(S\), and extend it to a basis \((\mathbf{x}_1,\dots,\mathbf{x}_n)\) of \(X\). Let \((f_1,\dots,f_n)\) be the corresponding dual basis, so \(f_i(\mathbf{x}_j)=1\) if \(i=j\) and \(0\) otherwise. A functional \(f=\sum_{i=1}^n c_i f_i\) vanishes on \(S\) exactly when \(f(\mathbf{x}_j)=c_j=0\) for \(1\le j\le k\). Thus \(S^0=\operatorname{span}\{f_{k+1},\dots,f_n\}\), which has dimension \(n-k\). This also covers \(k=0\) and \(k=n\), with the indicated span having dimension \(n\) and \(0\), respectively. \(\square\)
The second calculation concerns the sum and intersection of two subspaces. Its basis proof makes explicit the dimension relationship used later.
Proof. Choose a basis \((\mathbf{z}_1,\dots,\mathbf{z}_r)\) of \(U\cap W\). Extend it to a basis of \(U\) by adjoining \(\mathbf{u}_1,\dots,\mathbf{u}_a\), and extend it to a basis of \(W\) by adjoining \(\mathbf{w}_1,\dots,\mathbf{w}_b\). The list \((\mathbf{z}_1,\dots,\mathbf{z}_r,\mathbf{u}_1,\dots,\mathbf{u}_a,\mathbf{w}_1,\dots,\mathbf{w}_b)\) spans \(U+W\), because the chosen lists span \(U\) and \(W\). It is independent as well: in a zero combination, the combination of the \(\mathbf{w}_j\) belongs to \(U\) because the other terms do, and it also belongs to \(W\); hence it belongs to \(U\cap W\). Since the \(\mathbf{w}_j\) extend a basis of \(U\cap W\) to a basis of \(W\), their combination can belong to \(U\cap W\) only if all its coefficients are zero. The remaining combination of the \(\mathbf{z}_i\) and \(\mathbf{u}_j\) is then zero, so all its coefficients are zero by the basis property in \(U\). The displayed list is therefore a basis of \(U+W\), giving \(\dim(U+W)=r+a+b\). Since \(\dim U=r+a\), \(\dim W=r+b\), and \(\dim(U\cap W)=r\), this is the asserted formula. \(\square\)
The Double-Annihilator Theorem
Proof. First, \(J\) is injective. If \(\mathbf{v}\ne\mathbf{0}\), place \(\mathbf{v}\) in a basis of \(V\) and take the corresponding coordinate functional \(f\) with \(f(\mathbf{v})=1\). Then \(J(\mathbf{v})(f)=1\), so \(J(\mathbf{v})\ne0\). Since \(\dim V^{**}=\dim V^*=\dim V\), an injective linear map \(J:V\to V^{**}\) is also surjective, and hence is an isomorphism. For every \(\mathbf{u}\in U\) and \(f\in U^0\), \(J(\mathbf{u})(f)=f(\mathbf{u})=0\). Therefore \(J(U)\subseteq(U^0)^0\). To compare dimensions, the annihilator dimension lemma gives \(\dim U^0=\dim V-\dim U\). Applying it in \(X=V^*\) to the subspace \(U^0\) gives $$ \dim (U^0)^0=\dim V^*-\dim U^0 =\dim V-(\dim V-\dim U)=\dim U. $$ Because \(J\) is injective, \(\dim J(U)=\dim U\). The inclusion \(J(U)\subseteq(U^0)^0\) is thus an inclusion between finite-dimensional subspaces of equal dimension, so the subspaces are equal. \(\square\)
The theorem includes the boundary cases. If \(U=\{\mathbf{0}\}\), then \(U^0=V^*\), whose annihilator in \(V^{**}\) is \(\{\mathbf{0}\}\). If \(U=V\), then \(U^0=\{0\}\), whose annihilator is all of \(V^{**}\), namely \(J(V)\). In between these extremes, the theorem says that knowing precisely which functionals vanish on \(U\) is enough to recover every vector of \(U\).
Worked Example: Recovering a Plane from Its Annihilator
Let \(V=\mathbb{R}^3\), and let \(U=\operatorname{span}\{(1,0,1),(0,1,1)\}\). Compute \(U^0\), then identify its double annihilator with a subspace of \(\mathbb{R}^3\).
Write a functional as \(f(x_1,x_2,x_3)=a x_1+b x_2+c x_3\). For \(f\) to vanish on \(U\), it must vanish on both spanning vectors: $$ a+c=0,\qquad b+c=0. $$ Thus \(a=b=-c\). Taking \(c=t\), the functionals in \(U^0\) are exactly \(f_t(x_1,x_2,x_3)=t(-x_1-x_2+x_3)\), so \(U^0\) is spanned by the functional with coefficient triple \((-1,-1,1)\).
A vector \(\mathbf{x}=(x_1,x_2,x_3)\) belongs to the identified double annihilator exactly when every functional in \(U^0\) vanishes on it. It is enough to test the spanning functional, giving \(-x_1-x_2+x_3=0\), or \(x_3=x_1+x_2\). Therefore $$ (U^0)^0=\{(s,t,s+t):s,t\in\mathbb{R}\}. $$ Every such vector is \(s(1,0,1)+t(0,1,1)\), so this is precisely \(U\).
Check: The two generators of \(U\) satisfy \(-x_1-x_2+x_3=0\): their respective left sides are \(-1+1=0\) and \(-1+1=0\). Conversely, every vector satisfying the equation has the displayed linear-combination form. Thus the computed double annihilator agrees with \(U\).
Inclusion Reverses Under Annihilators
If \(U\subseteq W\), then a functional that vanishes on all of \(W\) certainly vanishes on \(U\). This gives \(W^0\subseteq U^0\). In finite dimensions the converse also holds, because we can take annihilators again and apply the double-annihilator theorem.
Proof. Suppose \(U\subseteq W\). If \(f\in W^0\), then \(f\) vanishes on every vector of \(W\), and therefore on every vector of \(U\). Hence \(f\in U^0\), proving \(W^0\subseteq U^0\). Conversely, suppose \(W^0\subseteq U^0\). Taking annihilators in \(V^{**}\) reverses this inclusion, so \((U^0)^0\subseteq(W^0)^0\). By the double-annihilator theorem, these spaces are \(J(U)\) and \(J(W)\). Since \(J\) is an isomorphism, \(J(U)\subseteq J(W)\) implies \(U\subseteq W\). Both directions hold, as required. \(\square\)
Worked Example: Inclusion and Its Reversal
In \(V=\mathbb{R}^3\), let \(U=\operatorname{span}\{\mathbf{e}_1\}\) and \(W=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\), where \(\mathbf{e}_1=(1,0,0)\) and \(\mathbf{e}_2=(0,1,0)\). We have \(U\subseteq W\). Find the annihilators to see the inclusion reverse.
For \(f(x_1,x_2,x_3)=a x_1+b x_2+c x_3\), vanishing on \(U\) requires \(a=0\). Thus \(U^0\) consists of the functionals \(b x_2+c x_3\), and is spanned by the coordinate functionals \(f_2(x)=x_2\) and \(f_3(x)=x_3\). Vanishing on \(W\) requires both \(a=0\) and \(b=0\), so \(W^0\) is spanned by \(f_3\). Consequently \(W^0\subseteq U^0\).
Check: Every multiple of \(f_3\) is a linear combination of \(f_2\) and \(f_3\), so the computed annihilators have the asserted inclusion. In the other direction, \(J(U)\) is the double annihilator of \(U\), and \(J(W)\) is the double annihilator of \(W\); taking annihilators of \(W^0\subseteq U^0\) gives \(J(U)\subseteq J(W)\), recovering the original inclusion.
Sums and Intersections
The annihilator of a sum is the intersection of the annihilators: a functional vanishes on \(U+W\) exactly when it vanishes on both \(U\) and \(W\). The formula for an intersection goes in the other direction. In finite dimensions, the annihilator of \(U\cap W\) is the sum \(U^0+W^0\), not merely the intersection of those annihilators.
Proof. A functional \(f\) lies in \((U+W)^0\) exactly when it vanishes on every vector \(\mathbf{u}+\mathbf{w}\), with \(\mathbf{u}\in U\) and \(\mathbf{w}\in W\). If \(f\) vanishes on \(U+W\), it vanishes on \(U\) and \(W\), since each is contained in \(U+W\). Conversely, if it vanishes on both, then \(f(\mathbf{u}+\mathbf{w})=f(\mathbf{u})+f(\mathbf{w})=0+0=0\). This proves \((U+W)^0=U^0\cap W^0\). For the second identity, if \(f\in U^0\) and \(g\in W^0\), then \(f+g\) vanishes on \(U\cap W\), so \(U^0+W^0\subseteq(U\cap W)^0\). To compare dimensions, the first identity and the annihilator dimension lemma give $$ \dim(U^0+W^0) =\dim U^0+\dim W^0-\dim(U^0\cap W^0) $$ $$ =(\dim V-\dim U)+(\dim V-\dim W)-(\dim V-\dim(U+W)) =\dim V-\dim(U\cap W). $$ The last equality follows from the dimension formula for \(U+W\). By the annihilator dimension lemma, \(\dim (U\cap W)^0=\dim V-\dim(U\cap W)\). Thus \(U^0+W^0\) is contained in \((U\cap W)^0\) and has the same dimension, so they are equal. \(\square\)
Worked Example: Sums and Intersections of Coordinate Planes
Let \(U=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2\}\) and \(W=\operatorname{span}\{\mathbf{e}_2,\mathbf{e}_3\}\) in \(\mathbb{R}^3\). Compute both annihilators and check the sum-and-intersection identities.
A functional \(f(x_1,x_2,x_3)=a x_1+b x_2+c x_3\) vanishes on \(U\) precisely when \(a=b=0\), so \(U^0\) is spanned by \(f_3(x)=x_3\). It vanishes on \(W\) precisely when \(b=c=0\), so \(W^0\) is spanned by \(f_1(x)=x_1\).
The subspaces share exactly the vectors \((0,t,0)=t\mathbf{e}_2\), so \(U\cap W=\operatorname{span}\{\mathbf{e}_2\}\). Its annihilator consists of functionals with \(b=0\), hence is spanned by \(f_1\) and \(f_3\). This agrees with \(U^0+W^0=\operatorname{span}\{f_3\}+\operatorname{span}\{f_1\}\). Also \(U+W=\mathbb{R}^3\), so \((U+W)^0=\{0\}\); this agrees with \(U^0\cap W^0=\{0\}\), since a multiple of \(f_1\) can equal a multiple of \(f_3\) only when both multiples are zero.
Check: A functional \(a x_1+c x_3\) vanishes on \(\mathbf{e}_2\), and every functional that vanishes on \(\mathbf{e}_2\) has \(b=0\). A functional that vanishes on all three standard basis vectors has \(a=b=c=0\). These checks confirm both identities directly.
Common Pitfall: Reversing the Wrong Relationship
A common mistake is to write \((U\cap W)^0=U^0\cap W^0\), by analogy with \((U+W)^0=U^0\cap W^0\). In the coordinate-plane example, this incorrect equation would say that \((U\cap W)^0=\{0\}\), because \(U^0\cap W^0=\{0\}\). But \(U\cap W=\operatorname{span}\{\mathbf{e}_2\}\), and the nonzero functional \(f_1(x)=x_1\) vanishes on \(\mathbf{e}_2\), so it belongs to \((U\cap W)^0\). The correct calculation gives \((U\cap W)^0=U^0+W^0=\operatorname{span}\{f_1,f_3\}\). The sum/intersection pairs are therefore crossed: the annihilator of a sum is an intersection, while the annihilator of an intersection is a sum.
Finite dimensionality matters in the converses and equalities proved here. For example, the dimension argument for \((U\cap W)^0=U^0+W^0\) uses finite-dimensional dimension formulas, and the identification of a double annihilator with the original space uses the evaluation isomorphism. In finite-dimensional linear algebra, the practical summary is:
| Subspace operation | Annihilator identity |
|---|---|
| Inclusion \(U\subseteq W\) | Reverse inclusion \(W^0\subseteq U^0\), and conversely |
| Sum \(U+W\) | \((U+W)^0=U^0\cap W^0\) |
| Intersection \(U\cap W\) | \((U\cap W)^0=U^0+W^0\) |
| Double annihilator of \(U\) | \((U^0)^0=J(U)\), identified with \(U\) |
These identities are useful because they translate questions about how subspaces fit together into questions about which functionals vanish on them. The double-annihilator theorem guarantees that, in finite dimensions, this translation loses no information: the annihilator determines the original subspace.
Check Your Understanding
Use the definitions and the finite-dimensional results above. When a problem asks for an identity, indicate which inclusion or dimension argument justifies it.
- In \(\mathbb{R}^3\), find the annihilator of \(\operatorname{span}\{(1,1,0)\}\), writing a general functional as \(f(x_1,x_2,x_3)=a x_1+b x_2+c x_3\).
- If \(U\subseteq W\subseteq V\), state the corresponding inclusion between \(U^0\) and \(W^0\), and explain why the direction reverses.
- For \(U=\operatorname{span}\{\mathbf{e}_1\}\) and \(W=\operatorname{span}\{\mathbf{e}_3\}\) in \(\mathbb{R}^3\), compute \((U+W)^0\) and compare it with \(U^0\cap W^0\).
- Explain why the inclusion \(U^0+W^0\subseteq(U\cap W)^0\) holds directly from the definition.
- Under the evaluation identification \(V\cong V^{**}\), what does the double-annihilator theorem say when \(U=V\)?
