Functionals That Vanish on a Subspace
A subspace can be described not only by the vectors it contains, but also by the linear functionals that vanish on every one of those vectors. The collection of such functionals is its annihilator. In a finite-dimensional vector space, its dimension is determined exactly by how many dimensions the subspace leaves outside itself. The key is to extend a basis of the subspace to a basis of the whole space: the dual basis elements associated with the added vectors form a basis for the annihilator.
Let \(V\) be a finite-dimensional vector space over a field \(\mathbb{F}\), where \(\mathbb{F}\) is \(\mathbb{R}\) or \(\mathbb{C}\), and let \(U\) be a subspace of \(V\). A linear functional on \(V\) is a linear map \(f:V\to\mathbb{F}\). The dual space \(V^*\) is the vector space of all such functionals. A functional vanishes on \(U\) if \(f(\mathbf{u})=0\) for every \(\mathbf{u}\in U\).
To see that \(U^0\) is a subspace, the zero functional belongs to it. If \(f,g\in U^0\) and \(a,b\in\mathbb{F}\), then for every \(\mathbf{u}\in U\), \((af+bg)(\mathbf{u})=af(\mathbf{u})+bg(\mathbf{u})=0\). Thus every linear combination of members of \(U^0\) also belongs to \(U^0\). This confirms that its dimension is well-defined.
The dimension formula counts independent ways to assign functional values in directions outside \(U\), while requiring the functional to have value zero on \(U\). It applies over either stated field; no inner product is needed.
The Dimension Formula
Proof. Write \(\dim V=n\) and \(\dim U=k\). Choose a basis \((\mathbf{u}_1,\dots,\mathbf{u}_k)\) for \(U\). By the basis extension theorem for finite-dimensional vector spaces, extend it to a basis \(\mathcal{B}=(\mathbf{u}_1,\dots,\mathbf{u}_k,\mathbf{v}_{k+1},\dots,\mathbf{v}_n)\) for \(V\). Let \((\phi_1,\dots,\phi_n)\) be the dual basis: \(\phi_i\) is the functional that takes value \(1\) on the \(i\)th vector of \(\mathcal{B}\) and value \(0\) on every other vector of \(\mathcal{B}\). In particular, \(\phi_i(\mathbf{u}_j)=1\) if \(i=j\), and \(0\) otherwise, for \(1\leq i,j\leq k\).
We show that \(\phi_{k+1},\dots,\phi_n\) form a basis for \(U^0\). Each of them vanishes on every basis vector \(\mathbf{u}_1,\dots,\mathbf{u}_k\) of \(U\), so by linearity each vanishes on all of \(U\). Therefore these functionals belong to \(U^0\). They are linearly independent because they are members of a basis of \(V^*\).
It remains to show that they span \(U^0\). Every \(f\in V^*\) has the dual-basis expansion \(f=\sum_{i=1}^n f(\mathbf{b}_i)\phi_i\), where \(\mathbf{b}_i\) denotes the \(i\)th vector of \(\mathcal{B}\). If \(f\in U^0\), then \(f(\mathbf{u}_i)=0\) for \(1\leq i\leq k\). Hence its first \(k\) coefficients in this expansion are zero, and \(f=\sum_{i=k+1}^n f(\mathbf{v}_i)\phi_i\). Thus the listed functionals span \(U^0\), so they are a basis for it. There are \(n-k\) of them, giving \(\dim U^0=n-k=\dim V-\dim U\). If \(k=0\) or \(k=n\), the same argument uses an empty initial or final part of the basis, respectively, and gives the same formula. \(\square\)
The proof identifies more than the dimension. Once a basis of \(U\) is extended to a basis of \(V\), the dual basis elements corresponding to the added vectors are a basis of \(U^0\). This provides a direct way to describe the annihilator, not just count its dimensions.
Calculating an Annihilator in Coordinates
Worked Example: A Two-Dimensional Subspace of \(\mathbb{R}^4\)
Let \(U=\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2\}\subseteq\mathbb{R}^4\), where \(\mathbf{u}_1=(1,1,0,0)^{\top}\) and \(\mathbf{u}_2=(0,1,1,0)^{\top}\). Find \(U^0\) and its dimension.
First, the two vectors are independent: if \(r\mathbf{u}_1+s\mathbf{u}_2=\mathbf{0}\), then \((r,r+s,s,0)^{\top}=\mathbf{0}\), so \(r=0\) and \(s=0\). Therefore \(\dim U=2\). A general linear functional on \(\mathbb{R}^4\) has the form \(f(\mathbf{x})=a x_1+b x_2+c x_3+d x_4\), where \(\mathbf{x}=(x_1,x_2,x_3,x_4)^{\top}\).
For \(f\) to lie in \(U^0\), it must vanish on both basis vectors of \(U\). Substitution gives \(f(\mathbf{u}_1)=a+b=0\) and \(f(\mathbf{u}_2)=b+c=0\). Thus \(a=-b\) and \(c=-b\), while \(d\) is unrestricted. Consequently, \(f(\mathbf{x})=b(-x_1+x_2-x_3)+d x_4\). A basis for \(U^0\) is therefore the pair of functionals \(g(\mathbf{x})=-x_1+x_2-x_3\) and \(h(\mathbf{x})=x_4\).
Check: \(g(\mathbf{u}_1)=-1+1=0\), \(g(\mathbf{u}_2)=1-1=0\), and \(h(\mathbf{u}_1)=h(\mathbf{u}_2)=0\). They are independent: if \(r g+s h\) is the zero functional, evaluating at \((1,0,0,0)^{\top}\) gives \(-r=0\), and evaluating at \((0,0,0,1)^{\top}\) gives \(s=0\). The dimension formula also gives \(\dim U^0=4-2=2\), as required.
In coordinates, a functional is specified by its values on a basis, and requiring it to vanish on \(U\) imposes conditions on those values. The theorem guarantees the number of independent functional directions left after these conditions are imposed. The following example carries out the same idea in a space whose vectors are polynomials.
Worked Example: An Annihilator in a Polynomial Space
Let \(V=P_2(\mathbb{R})\), the space of real polynomials of degree at most \(2\), and let \(U=\operatorname{span}\{x^2-x\}\). Find a basis and dimension for \(U^0\).
The polynomial \(x^2-x\) is nonzero, so it is a basis for \(U\) and \(\dim U=1\). The coefficient functionals \(L_0,L_1,L_2\) are defined by \(L_0(a+bx+cx^2)=a\), \(L_1(a+bx+cx^2)=b\), and \(L_2(a+bx+cx^2)=c\). Because \((1,x,x^2)\) is a basis for \(V\), every functional \(f\in V^*\) has a unique expression \(f=\alpha L_0+\beta L_1+\gamma L_2\).
For \(f\) to annihilate \(U\), it is enough and necessary that it vanish on the generator \(x^2-x\). Evaluating, \(f(x^2-x)=\gamma-\beta\), so the condition is \(\gamma=\beta\). Thus \(f=\alpha L_0+\beta(L_1+L_2)\), and \(L_0,L_1+L_2\) form a basis for \(U^0\). They are independent because evaluating a relation \(\alpha L_0+\beta(L_1+L_2)=0\) first at \(1\) gives \(\alpha=0\), then at \(x\) gives \(\beta=0\).
Check: \(L_0(x^2-x)=0\), and \((L_1+L_2)(x^2-x)=-1+1=0\). The two functionals are independent, so \(\dim U^0=2\). The formula agrees: \(\dim V=3\) and \(\dim U=1\), hence \(\dim U^0=3-1=2\).
One Annihilating Direction for a Plane
Worked Example: A Plane in \(\mathbb{R}^3\)
Let \(U=\operatorname{span}\{\mathbf{u}_1,\mathbf{u}_2\}\subseteq\mathbb{R}^3\), where \(\mathbf{u}_1=(1,0,1)^{\top}\) and \(\mathbf{u}_2=(0,1,1)^{\top}\). Determine \(U^0\).
The vectors are independent: \(r\mathbf{u}_1+s\mathbf{u}_2=(r,s,r+s)^{\top}=\mathbf{0}\) implies \(r=s=0\). Thus \(\dim U=2\). Write a general functional as \(f(\mathbf{x})=a x_1+b x_2+c x_3\). Vanishing on \(\mathbf{u}_1\) and \(\mathbf{u}_2\) requires \(a+c=0\) and \(b+c=0\). Therefore \(a=b=-c\), and \(f(\mathbf{x})=c(-x_1-x_2+x_3)\).
So \(U^0\) is spanned by the nonzero functional \(g(\mathbf{x})=-x_1-x_2+x_3\), and has dimension \(1\). Check: \(g(\mathbf{u}_1)=-1+1=0\) and \(g(\mathbf{u}_2)=-1+1=0\). The theorem gives \(\dim U^0=3-2=1\), agreeing with the direct calculation.
Endpoint Cases and a Common Mistake
The formula includes the boundary cases. If \(U=\{\mathbf{0}\}\), then every functional vanishes on \(U\), so \(U^0=V^*\) and \(\dim U^0=\dim V\). If \(U=V\), a functional in \(U^0\) must vanish on every vector of \(V\), so it is the zero functional; hence \(\dim U^0=0\). These conclusions agree with the formula when \(\dim U=0\) and when \(\dim U=\dim V\), respectively.
Common mistake: using \(\dim U\) instead of the complementary dimension. Suppose \(V=\mathbb{R}^4\) and \(U\) is the two-dimensional subspace in the first worked example. A wrong calculation might say \(\dim U^0=\dim U=2\), treating this as a general rule. It happens to give the right number in this particular case, but it is not the formula. For a one-dimensional subspace of \(\mathbb{R}^4\), the same incorrect rule predicts an annihilator of dimension \(1\), whereas the theorem gives \(4-1=3\). The distinction matters: it is the dimension of the whole space minus the dimension of the subspace.
Another useful distinction is that \(U^0\) consists of functionals, so it lies in \(V^*\). Although an inner product may sometimes be used to represent functionals by vectors, that identification requires additional structure and is not part of the definition or proof here. The dimension formula follows from linearity and basis extension alone.
Check Your Understanding
For each question, distinguish the subspace \(U\) from its annihilator \(U^0\subseteq V^*\).
- If \(\dim V=6\) and \(\dim U=4\), what is \(\dim U^0\)?
- In \(\mathbb{R}^3\), let \(U=\operatorname{span}\{(1,0,0)^{\top}\}\). Find a basis for \(U^0\) by describing the functionals that vanish on the generator.
- In \(P_2(\mathbb{R})\), use the coefficient functionals \(L_0,L_1,L_2\) to find a basis for the annihilator of \(\operatorname{span}\{1+x\}\).
- Explain why \(U^0=V^*\) when \(U=\{\mathbf{0}\}\).
- In the basis-extension proof, why must the first \(k\) coefficients in the dual-basis expansion of a member of \(U^0\) be zero?
