Functionals That Vanish on a Subspace
A subspace can be described by the vectors it contains, but it can also be described by the linear functionals that vanish on all of those vectors. Such functionals record constraints on a vector: if a functional vanishes throughout a subspace, then every vector in that subspace satisfies the corresponding linear equation. The collection of those functionals is called the annihilator. It belongs to the dual space, not to the original vector space.
Throughout, \(V\) is a finite-dimensional vector space over \(\mathbb{R}\), \(V^*\) is its dual space, and \(U\) is a subspace of \(V\). A linear functional is a linear map \(f:V\to\mathbb{R}\). In particular, \(f(u+v)=f(u)+f(v)\) and \(f(cu)=cf(u)\) for vectors \(u,v\in V\) and scalars \(c\in\mathbb{R}\).
The zero functional belongs to \(U^0\), since it takes the value zero on every vector. More generally, if \(f\) and \(g\) vanish on \(U\), then their sum and every scalar multiple also vanish on \(U\). This closure is the reason the annihilator is itself a subspace of \(V^*\).
Proof. The zero functional \(0:V\to\mathbb{R}\) satisfies \(0(u)=0\) for every \(u\in U\), so \(0\in U^0\). Now take \(f,g\in U^0\) and \(a,b\in\mathbb{R}\). For every \(u\in U\), the definition of \(U^0\) gives \(f(u)=0\) and \(g(u)=0\). Therefore \((af+bg)(u)=af(u)+bg(u)=a\cdot0+b\cdot0=0\). Since \(af+bg\) is a linear functional, it belongs to \(V^*\), and since it vanishes on every \(u\in U\), it belongs to \(U^0\). Thus \(U^0\) contains zero and is closed under linear combinations, so it is a subspace of \(V^*\). \(\square\)
Testing on Generators
To check whether a functional belongs to \(U^0\), it may seem necessary to test every vector in \(U\). When \(U\) is given as a span, linearity reduces the work to a finite list. A list \(\mathbf{u}_1,\dots,\mathbf{u}_k\) spans \(U\) when every vector in \(U\) can be written as a linear combination of those vectors.
Proof. First suppose \(f\in U^0\). Each generator \(\mathbf{u}_i\) belongs to \(U\), so the definition gives \(f(\mathbf{u}_i)=0\). Conversely, suppose \(f(\mathbf{u}_i)=0\) for every \(i\). An arbitrary \(\mathbf{u}\in U\) has the form \(\mathbf{u}=c_1\mathbf{u}_1+\cdots+c_k\mathbf{u}_k\) for real scalars \(c_i\). By linearity, \(f(\mathbf{u})=c_1f(\mathbf{u}_1)+\cdots+c_kf(\mathbf{u}_k)=0\). This holds for every \(\mathbf{u}\in U\), so \(f\in U^0\). \(\square\)
In \(\mathbb{R}^n\), every linear functional can be written in coordinates. If \(\mathbf{e}_1,\dots,\mathbf{e}_n\) are the standard basis vectors, then a vector \(\mathbf{x}=(x_1,\dots,x_n)^\top\) equals \(\sum_{j=1}^n x_j\mathbf{e}_j\). Consequently, for any \(f\in(\mathbb{R}^n)^*\), linearity gives \(f(\mathbf{x})=\sum_{j=1}^n x_j f(\mathbf{e}_j)\). Writing \(a_j=f(\mathbf{e}_j)\), the functional has the coordinate expression \(f(\mathbf{x})=a_1x_1+\cdots+a_nx_n\). The coefficients \(a_j\) are determined by \(f\), since \(a_j=f(\mathbf{e}_j)\). This is a coordinate description of a functional; it does not require an inner product.
Lines and Planes in Coordinate Space
Worked Example: The Annihilator of a Line
Let \(U=\operatorname{span}\{(1,2,-1)^\top\}\subseteq\mathbb{R}^3\). Find every functional in \(U^0\).
Write a general functional on \(\mathbb{R}^3\) as \(f(x_1,x_2,x_3)=ax_1+bx_2+cx_3\), where \(a,b,c\in\mathbb{R}\). By the spanning-set test, \(f\) belongs to \(U^0\) exactly when it vanishes on the generator:
Choose \(b=s\) and \(c=t\) freely. The equation forces \(a=-2s+t\). Therefore all the functionals in the annihilator are
Check: On the generator, \(f_{s,t}(1,2,-1)=(-2s+t)+2s-t=0\). Every vector in the line is a scalar multiple of that generator, so linearity makes \(f_{s,t}\) vanish on the whole line. The equation \(a+2b-c=0\) also shows that no other coefficient choices have been omitted.
Worked Example: The Annihilator of a Plane
Let \(U=\operatorname{span}\{(1,0,1)^\top,(0,1,1)^\top\}\subseteq\mathbb{R}^3\). Find \(U^0\).
Take \(f(x_1,x_2,x_3)=ax_1+bx_2+cx_3\). Vanishing on the first generator requires \(a+c=0\); vanishing on the second requires \(b+c=0\). Solving these equations gives \(a=-c\) and \(b=-c\). Set \(c=-t\), so \(a=b=t\). Thus the only possible functionals are
Check: \(f_t(1,0,1)=t(1+0-1)=0\), and \(f_t(0,1,1)=t(0+1-1)=0\). Every vector in \(U\) is a linear combination of these two generators, so \(f_t\) vanishes throughout \(U\). Conversely, the two generator equations force the coefficients to have exactly this form. The two generators are independent: if \(\alpha(1,0,1)+\beta(0,1,1)=(0,0,0)\), the first coordinate gives \(\alpha=0\), and the second gives \(\beta=0\).
These examples illustrate the same procedure: write a general functional, impose one vanishing equation for each generator, solve for its coefficients, and verify that the resulting functionals vanish on every generator. The final check on the generators suffices because of linearity.
Coordinate Subspaces
Worked Example: A Coordinate Subspace in \(\mathbb{R}^4\)
Let \(U=\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_3\}\subseteq\mathbb{R}^4\). Determine its annihilator.
A general functional on \(\mathbb{R}^4\) has the form \(f(x_1,x_2,x_3,x_4)=a_1x_1+a_2x_2+a_3x_3+a_4x_4\). Its values on the two generators are \(f(\mathbf{e}_1)=a_1\) and \(f(\mathbf{e}_3)=a_3\). Both must be zero, so \(a_1=a_3=0\), while \(a_2\) and \(a_4\) are unrestricted. Therefore
Check: A vector of \(U\) has the form \((r,0,s,0)^\top\). For every displayed functional, \(f(r,0,s,0)=a_2\cdot0+a_4\cdot0=0\). Conversely, vanishing at \(\mathbf{e}_1\) and \(\mathbf{e}_3\) forces exactly the two coefficient conditions used above.
More generally, if a coordinate subspace is spanned by selected standard basis vectors, vanishing on those basis vectors sets the corresponding functional coefficients to zero. Coefficients attached to the other coordinates remain unrestricted. This follows directly from evaluating the functional on each selected basis vector; it is not a rule about orthogonality.
Annihilators in Polynomial Spaces
Functionals can also be described using the coefficients of a polynomial. Let \(P_2(\mathbb{R})\) denote the vector space of real polynomials of degree at most two, including the zero polynomial. Each \(p\in P_2(\mathbb{R})\) has a unique expression \(p(x)=p_0+p_1x+p_2x^2\). A general linear functional on this space is determined by its values on \(1,x,x^2\), so it has the form \(f(p)=ap_0+bp_1+cp_2\). Indeed, expanding \(p\) in those three basis polynomials and applying linearity gives this expression.
Worked Example: A Subspace of \(P_2(\mathbb{R})\)
Let \(U=\operatorname{span}\{1+x,\ x+x^2\}\). Find all functionals in \(U^0\).
For \(f(p)=ap_0+bp_1+cp_2\), vanishing on \(1+x\) requires \(a+b=0\), because its coefficients are \((1,1,0)\). Vanishing on \(x+x^2\), whose coefficients are \((0,1,1)\), requires \(b+c=0\). Hence \(a=-b\) and \(c=-b\). Setting \(a=t\) gives \(b=-t\) and \(c=t\), so
Check: \(f_t(1+x)=t(1-1+0)=0\), and \(f_t(x+x^2)=t(0-1+1)=0\). Each polynomial in \(U\) is a linear combination of the two generators, so each \(f_t\) vanishes on all of \(U\). The two equations \(a+b=0\) and \(b+c=0\) also show that every functional in the annihilator is included in the displayed family.
The coefficient coordinates here refer to the polynomial basis \((1,x,x^2)\). A functional is not itself a polynomial: it takes polynomials as inputs and returns real numbers. Keeping the input space and the dual space distinct prevents a common confusion when moving between vector coordinates and functional coordinates.
Common Mistake: Checking the Wrong Condition
Suppose \(U=\operatorname{span}\{(1,1)^\top\}\subseteq\mathbb{R}^2\). A tempting but incorrect answer is that \(f(x_1,x_2)=x_1+x_2\) belongs to \(U^0\), perhaps because its coefficients match the entries of the generator. But evaluating the functional gives \(f(1,1)=1+1=2\ne0\), so it does not vanish on \(U\).
The correction is to impose the defining condition on the generator. For a general functional \(g(x_1,x_2)=ax_1+bx_2\), the required equation is \(g(1,1)=a+b=0\). Thus \(b=-a\), and all annihilating functionals are \(g_t(x_1,x_2)=t(x_1-x_2)\), where \(t\in\mathbb{R}\). Substitution confirms \(g_t(1,1)=t-t=0\). The sign matters: adding the coordinate terms produced a functional that does not vanish, while subtracting them produces one that does.
An annihilator gives a collection of linear constraints that every vector in the original subspace satisfies. To compute one from a spanning list, no extra structure such as an inner product is needed: specify a general functional and solve the equations obtained by evaluating it on the generators. An orthogonal complement, by contrast, is defined using an inner product on the vector space. The two notions may have related coordinate formulas in particular settings, but their definitions place them in different spaces and should not be conflated.
Check Your Understanding
For each calculation, begin with a general functional and test it on a spanning list.
- Find the annihilator of \(\operatorname{span}\{(2,1)^\top\}\) in \(\mathbb{R}^2\).
- Find the annihilator of \(\operatorname{span}\{(1,0,0)^\top,(0,0,1)^\top\}\) in \(\mathbb{R}^3\).
- For \(U=\operatorname{span}\{1-x,\ x^2\}\subseteq P_2(\mathbb{R})\), determine all functionals \(f(p)=ap_0+bp_1+cp_2\) in \(U^0\).
- Explain why checking a functional on every vector of a spanning list is sufficient to prove it belongs to the annihilator.
- Give a functional that does not belong to the annihilator of \(\operatorname{span}\{(1,1)^\top\}\), and verify the failure by evaluation.
