5.2 Annihilators

Section 1 built the dual space $V^{*}$ , the space of linear measurements on $V$ . Now we connect subspaces of $V$ with subspaces of $V^{*}$ . A plane in $\mathbb{R}^3$ can be described by a spanning list or by an equation, and an equation is a functional. This section collects all the equations that a set of vectors satisfies into one subspace of $V^{*}$ , the annihilator, counts its dimension, and shows that it turns sums into intersections and intersections into sums.

5.2.1. The equations a subspace satisfies

Every object comes with its subobjects. The dual space $V^{*}$ has subspaces of its own, and the question is which of them are attached to subspaces of $V$ . The standard example shows the way.

Let $U = \operatorname{span}((1, 0, 1), (0, 1, 1))$ , a plane in $\mathbb{R}^3$ . The functional $\varphi(x, y, z) = x + y - z$ vanishes on both spanning vectors, so by linearity it vanishes on all of $U$ : the plane satisfies the equation $x + y - z = 0$ . So do $2x + 2y - 2z = 0$ and the trivial equation $0 = 0$ . Which functionals vanish on $U$ ? By Theorem 5.1.1 a functional is $\psi(x, y, z) = ax + by + cz$ . It vanishes on the two spanning vectors if and only if $a + c = 0$ and $b + c = 0$ , that is, $\psi = -c(x + y - z) = -c\,\varphi$ . So the equations of the plane form the line $\operatorname{span}(\varphi)$ in $(\mathbb{R}^3)^{*}$ .

For the $z$ -axis $\operatorname{span}((0, 0, 1))$ , the same computation gives $c = 0$ : the equations are $ax + by = 0$ , a plane in $(\mathbb{R}^3)^{*}$ . A bigger subspace satisfies fewer equations, and in both cases the dimensions add up to $3$ . The set of equations deserves a name, and the pattern deserves a theorem.

The annihilator of a set of vectors is the space of all linear equations that every one of those vectors satisfies.

Definition 5.2.1 (Annihilator)

Let $V$ be a vector space over $F$ , and let $S \subseteq V$ be any subset. The annihilator of $S$ is $S^{0} \mathrel{:=} \{ \varphi \in V^{*} : \varphi(\mathbf{s}) = 0 \text{ for all } \mathbf{s} \in S \}.$

In words: $S^{0}$ is a subset of the dual space $V^{*}$ , not of $V$ . A functional belongs to it when it vanishes at every vector of $S$ ; a single non-zero value disqualifies it. The set $S$ needs no structure: it may be a subspace, a finite list of vectors, or even empty. The superscript $0$ recalls the value that the functionals take on $S$ .

Well-definedness. The definition makes sense for every subset. What needs checking is that the result is a subspace, which is what makes its dimension meaningful.

Proposition 5.2.1 (The Annihilator Is a Subspace)

Let $V$ be a vector space over $F$ , and let $S \subseteq V$ . Then $S^{0}$ is a subspace of $V^{*}$ .

Proof. We use the subspace test (Theorem 1.2.1) in $V^{*}$ . The zero functional sends every $\mathbf{s} \in S$ to $0$ , so it lies in $S^{0}$ . Let $\varphi, \psi \in S^{0}$ and $c \in F$ . For every $\mathbf{s} \in S$ , the pointwise operations of Definition 5.1.2 give $(\varphi + \psi)(\mathbf{s}) = \varphi(\mathbf{s}) + \psi(\mathbf{s}) = 0 + 0 = 0, \qquad (c\varphi)(\mathbf{s}) = c\,\varphi(\mathbf{s}) = c \cdot 0 = 0 .$ Hence $\varphi + \psi \in S^{0}$ and $c\varphi \in S^{0}$ , and $S^{0}$ is a subspace of $V^{*}$ .

Example (First Annihilators)

(a) For any vector space $V$ , find $\{\mathbf{0}\}^{0}$ , $\varnothing^{0}$ and $V^{0}$ .
(b) In $M_2(F)$ , find the annihilator of the single matrix $\mathbf{E}_{11}$ .
(c) In $\mathbb{R}[x]_{\le 2}$ , find the annihilator of $S = \{1, x\}$ .

Solution.

(a) Every functional sends $\mathbf{0}$ to $0$ (Theorem 2.1.1), so $\{\mathbf{0}\}^{0} = V^{*}$ . The condition “for all $\mathbf{s} \in \varnothing$ ” holds vacuously, so $\varnothing^{0} = V^{*}$ . A functional in $V^{0}$ takes the value $0$ at every vector, so it is the zero functional, and $V^{0} = \{0\}$ . These degenerate cases fix the direction of everything that follows: the smallest subspace of $V$ has the largest annihilator, and the largest has the smallest.
(b) Let $\varepsilon_{ij}(\mathbf{A}) = a_{ij}$ be the dual basis of the standard basis $(\mathbf{E}_{11}, \mathbf{E}_{12}, \mathbf{E}_{21}, \mathbf{E}_{22})$ . By Theorem 5.1.3 (c), every $\psi \in M_2(F)^{*}$ is $\psi = \sum_{i,j} \psi(\mathbf{E}_{ij})\,\varepsilon_{ij}$ . Such a $\psi$ lies in $\{\mathbf{E}_{11}\}^{0}$ exactly when $\psi(\mathbf{E}_{11}) = 0$ . Hence $\{\mathbf{E}_{11}\}^{0} = \operatorname{span}(\varepsilon_{12}, \varepsilon_{21}, \varepsilon_{22}) ,$ the functionals $\mathbf{A} \mapsto c_{12}a_{12} + c_{21}a_{21} + c_{22}a_{22}$ . It has dimension $3$ , and $3 + 1 = 4 = \dim M_2(F)$ .
(c) Let $(\varphi_0, \varphi_1, \varphi_2)$ be the dual basis of $(1, x, x^2)$ , so $\varphi_k$ reads off the coefficient of $x^k$ . A functional $\psi = \psi(1)\varphi_0 + \psi(x)\varphi_1 + \psi(x^2)\varphi_2$ lies in $S^{0}$ exactly when $\psi(1) = \psi(x) = 0$ , that is, when $\psi = \psi(x^2)\varphi_2$ . So $S^{0} = \operatorname{span}(\varphi_2)$ : the only equations satisfied by both $1$ and $x$ are the multiples of “the coefficient of $x^2$ is $0$ ”.

Non-example by minimal change. Shift the plane of the opening example off the origin: $A = \{ (x, y, z) \in \mathbb{R}^3 : x + y - z = 1 \}$ . The set is still cut out by the same left-hand side, but $\varphi(x, y, z) = x + y - z$ takes the value $1$ , not $0$ , on $A$ , so the defining clause $\varphi(\mathbf{s}) = 0$ fails and $\varphi \notin A^{0}$ . In fact $A^{0} = \{0\}$ : the set $A$ contains $(1, 0, 0)$ , $(0, 1, 0)$ and $(0, 0, -1)$ , and a functional $ax + by + cz$ vanishing at these three vectors has $a = b = c = 0$ . Annihilators record only homogeneous equations.

Why this definition. We allow any subset $S$ , not only subspaces, because subspaces are usually handed to us by spanning lists, and we want to test a functional on the list alone. The theorem at the end of this section shows that this loses nothing: $S^{0} = (\operatorname{span} S)^{0}$ . The name is literal. To annihilate is to reduce to nothing, and the functionals in $S^{0}$ reduce every vector of $S$ to $0$ .

Warning. $U^{0}$ lives in $V^{*}$ , and it is not a complement of $U$ . It is tempting to turn the functional $ax + by$ into the vector $(a, b)$ and to picture $U^{0}$ as a subspace of $F^2$ “perpendicular” to $U$ . Over $\mathbb{F}_2$ this picture fails completely. Let $U = \operatorname{span}((1, 1)) \subseteq \mathbb{F}_2^2$ . A functional $ax + by$ vanishes on $(1, 1)$ exactly when $a + b = 0$ , that is, $a = b$ , so $U^{0} = \{0,\ x + y\}$ . The functional $x + y$ corresponds to the vector $(1, 1)$ , which lies in $U$ . So the “picture” of $U^{0}$ is $U$ itself, not a complement. Perpendicularity needs an inner product, which is the orthogonal complement of Chapter 11; the annihilator needs nothing.

The payoff of the definition is the pattern we saw in the opening example. The equations of a subspace take up exactly the dimensions that the subspace leaves free.

5.2.2. Counting equations

Theorem 5.2.2 (Dimension of the Annihilator)

Let $V$ be a finite-dimensional vector space over $F$ , and let $U$ be a subspace of $V$ . Then $\dim U + \dim U^{0} = \dim V .$ More precisely, if $(\mathbf{u}_1, \dots, \mathbf{u}_m, \mathbf{v}_{m+1}, \dots, \mathbf{v}_n)$ is a basis of $V$ extending a basis $(\mathbf{u}_1, \dots, \mathbf{u}_m)$ of $U$ , with dual basis $(\varphi_1, \dots, \varphi_n)$ , then $(\varphi_{m+1}, \dots, \varphi_n)$ is a basis of $U^{0}$ .

Idea. This is a dimension formula, so we start from the smallest subspace, $U$ , take a basis and extend it to $V$ . The dual basis now splits in two. The last $n - m$ functionals vanish on every $\mathbf{u}_i$ , by the delta, so they vanish on $U$ . The first $m$ do not. To see that nothing else is in $U^{0}$ , expand a functional $\psi \in U^{0}$ in the dual basis: its coefficients are its values on the basis, and the first $m$ values are $\psi(\mathbf{u}_i) = 0$ .

Proof. Since $U$ is a subspace of the finite-dimensional space $V$ , it has a basis $(\mathbf{u}_1, \dots, \mathbf{u}_m)$ (Theorem 1.6.7, Corollary 1.5.5), and by the Basis Extension Theorem (Theorem 1.6.5) it extends to a basis $\mathcal{B} = (\mathbf{u}_1, \dots, \mathbf{u}_m, \mathbf{v}_{m+1}, \dots, \mathbf{v}_n)$ of $V$ , where $n = \dim V$ . Let $(\varphi_1, \dots, \varphi_n)$ be the dual basis of $\mathcal{B}$ (Theorem 5.1.3).

The $\varphi_j$ with $j > m$ lie in $U^{0}$ . Let $j > m$ and $\mathbf{u} \in U$ , say $\mathbf{u} = a_1\mathbf{u}_1 + \dots + a_m\mathbf{u}_m$ . By Theorem 2.1.4, $\varphi_j(\mathbf{u}) = \sum_{i=1}^{m} a_i\varphi_j(\mathbf{u}_i) = 0$ , since $\varphi_j(\mathbf{u}_i) = \delta_{ji} = 0$ for $i \le m < j$ .

They span $U^{0}$ . Let $\psi \in U^{0}$ . By Theorem 5.1.3 (c), $\psi = \sum_{i=1}^{m} \psi(\mathbf{u}_i)\,\varphi_i + \sum_{j=m+1}^{n} \psi(\mathbf{v}_j)\,\varphi_j = \sum_{j=m+1}^{n} \psi(\mathbf{v}_j)\,\varphi_j ,$ because $\psi(\mathbf{u}_i) = 0$ for $\mathbf{u}_i \in U$ . Hence $U^{0} \subseteq \operatorname{span}(\varphi_{m+1}, \dots, \varphi_n)$ , and the reverse inclusion holds because $U^{0}$ is a subspace (Proposition 5.2.1, Theorem 1.3.1).

They are independent, being part of the basis $(\varphi_1, \dots, \varphi_n)$ of $V^{*}$ . Therefore $(\varphi_{m+1}, \dots, \varphi_n)$ is a basis of $U^{0}$ , and $\dim U^{0} = n - m = \dim V - \dim U$ . This proves the theorem.

Alternatively, restrict functionals to $U$ . The map $R \colon V^{*} \to U^{*}$ , $\varphi \mapsto \varphi|_U$ , is linear, and its kernel is $U^{0}$ by definition. It is surjective: given $\xi \in U^{*}$ , Theorem 2.1.5 gives $\varphi \in V^{*}$ with $\varphi(\mathbf{u}_i) = \xi(\mathbf{u}_i)$ and $\varphi(\mathbf{v}_j) = 0$ , and then $\varphi|_U = \xi$ because both agree on a basis of $U$ . By the Rank–Nullity Theorem (Theorem 2.3.1) and Corollary 5.1.4, $\dim V = \dim V^{*} = \dim U^{0} + \dim U^{*} = \dim U^{0} + \dim U$ . This route shows why the formula holds: a functional on $V$ is a functional on $U$ plus a choice of values off $U$ , and $U^{0}$ measures that choice.

The theorem is the second signature move of the chapter: pass to the dual and count. One immediate consequence: if $U$ is a proper subspace of a finite-dimensional $V$ , then $\dim U^{0} > 0$ , so some non-zero functional vanishes on $U$ . Equivalently, a list spans a finite-dimensional $V$ if and only if the only functional that kills every vector of the list is $0$ : apply the previous sentence to the span $U$ of the list, and note that a functional kills every vector of the list exactly when it kills $U$ , by Theorem 2.1.4.

In $F^n$ the whole computation becomes elimination, because functionals are rows.

Example (An Annihilator as a Null Space)

Find a basis of $U^{0}$ for $U = \operatorname{span}((1, 2, 0), (0, 1, 1)) \subseteq \mathbb{R}^3$ , and check the dimension formula.

Solution. By Theorem 5.1.1, every functional is $\varphi_{\mathbf{a}}(\mathbf{x}) = \mathbf{a}^{\top}\mathbf{x}$ for a unique $\mathbf{a} = (a_1, a_2, a_3)$ . By linearity, $\varphi_{\mathbf{a}}$ vanishes on $U$ if and only if it vanishes on the two spanning vectors (a combination of vectors killed by $\varphi_{\mathbf{a}}$ is killed, by Theorem 2.1.4). These two conditions are $\begin{aligned} &a_1 + 2a_2 = 0, \qquad a_2 + a_3 = 0, \\ &\text{that is,} \qquad \mathbf{A}\mathbf{a} = \mathbf{0} \quad \text{with} \quad \mathbf{A} = \begin{pmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \end{pmatrix}, \end{aligned}$ the matrix whose rows are the spanning vectors. So $\varphi_{\mathbf{a}} \in U^{0}$ if and only if $\mathbf{a} \in \operatorname{nul}(\mathbf{A})$ . Subtracting twice row 2 from row 1 gives the RREF $\begin{pmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \end{pmatrix},$ with $a_3$ free. By Theorem 3.5.4, $\operatorname{nul}(\mathbf{A}) = \operatorname{span}((2, -1, 1))$ . Hence $U^{0} = \operatorname{span}(\psi)$ with $\psi(x, y, z) = 2x - y + z .$ Check: $\psi(1, 2, 0) = 2 - 2 + 0 = 0$ and $\psi(0, 1, 1) = -1 + 1 = 0$ . The two spanning vectors are independent (neither is a multiple of the other), so $\dim U = 2$ , and $\dim U + \dim U^{0} = 2 + 1 = 3$ , as Theorem 5.2.2 predicts. As a by-product, $U$ is the plane $2x - y + z = 0$ : $U \subseteq \operatorname{ker}\psi$ , both have dimension $2$ (Proposition 5.1.2), and Theorem 1.6.8 gives equality.

The same recipe works in general. For $U = \operatorname{span}(\mathbf{u}_1, \dots, \mathbf{u}_m) \subseteq F^n$ , let $\mathbf{A}$ have rows $\mathbf{u}_1^{\top}, \dots, \mathbf{u}_m^{\top}$ . Then $U^{0} = \{ \varphi_{\mathbf{a}} : \mathbf{a} \in \operatorname{nul}(\mathbf{A}) \}$ , and the isomorphism of Theorem 5.1.1 carries a basis of $\operatorname{nul}(\mathbf{A})$ to a basis of $U^{0}$ .

Away from $F^n$ , counting often replaces the computation entirely.

Example (The Equations of a Root Condition)

Let $U = \{ p \in \mathbb{R}[x]_{\le 2} : p(1) = 0 \}$ . Find $U^{0}$ .

Solution. Let $\varepsilon_1(p) = p(1)$ , a linear functional (Example (Four Families of Functionals)). By definition $U = \operatorname{ker}\varepsilon_1$ , so $\varepsilon_1 \in U^{0}$ . Since $\varepsilon_1(1) = 1$ , $\varepsilon_1 \neq 0$ , and Proposition 5.1.2 gives $\dim U = 3 - 1 = 2$ . By Theorem 5.2.2, $\dim U^{0} = 3 - 2 = 1$ . Now $\operatorname{span}(\varepsilon_1)$ is a subspace of $U^{0}$ (Proposition 5.2.1) of dimension $1$ , so $U^{0} = \operatorname{span}(\varepsilon_1)$ by Theorem 1.6.8.

We never found a basis of $U$ and never solved an equation. The conclusion is concrete: a functional on $\mathbb{R}[x]_{\le 2}$ vanishes on every polynomial with a root at $1$ only if it is a multiple of $p \mapsto p(1)$ .

Quick check. Let $U = \operatorname{span}((1, 1, 1)) \subseteq \mathbb{R}^3$ . What is $\dim U^{0}$ ? Name a basis of $U^{0}$ without solving a system.

Solution. $\dim U^{0} = 3 - 1 = 2$ by Theorem 5.2.2. The functionals $x - y$ and $y - z$ vanish at $(1, 1, 1)$ , hence on $U$ , and they are independent: in $a(x - y) + b(y - z) = 0$ , the coefficient of $x$ gives $a = 0$ , and then $b = 0$ . Two independent vectors in a $2$ -dimensional space form a basis (Theorem 1.6.4).

5.2.3. Sums, intersections and spans

Taking annihilators reverses inclusions: more vectors satisfy fewer common equations. It also exchanges sums with intersections. In $\mathbb{R}^3$ , let $U$ be the $xy$ -plane and $W$ the $xz$ -plane. Then $U^{0} = \operatorname{span}(z)$ and $W^{0} = \operatorname{span}(y)$ , writing $x, y, z$ for the coordinate functionals. The intersection $U \cap W$ is the $x$ -axis, with annihilator $\operatorname{span}(y, z) = U^{0} + W^{0}$ . The sum $U + W = \mathbb{R}^3$ has annihilator $\{0\} = U^{0} \cap W^{0}$ . The theorem says this always happens.

Theorem 5.2.3 (Properties of Annihilators)

Let $V$ be a vector space over $F$ .

(a) If $S \subseteq T \subseteq V$ , then $T^{0} \subseteq S^{0}$ .
(b) For every subset $S \subseteq V$ , $S^{0} = (\operatorname{span} S)^{0}$ .
(c) For subspaces $U, W$ of $V$ , $(U + W)^{0} = U^{0} \cap W^{0}$ .
(d) If $V$ is finite-dimensional, then for subspaces $U, W$ of $V$ , $(U \cap W)^{0} = U^{0} + W^{0}$ .

Idea. Parts (a)–(c) are unwinding definitions: fix a functional in one side and test it on the vectors of the other. In (d), the inclusion $U^{0} + W^{0} \subseteq (U \cap W)^{0}$ is also direct. The reverse inclusion is the hard half. Given $\varphi$ vanishing on $U \cap W$ , we would have to split it as $\psi + \chi$ with $\psi$ killing $U$ and $\chi$ killing $W$ , and nothing in sight tells us how. So we count instead: one inclusion plus equal dimensions gives equality. The plan is ① compute $\dim(U^{0} + W^{0})$ by the dimension formula for sums inside $V^{*}$ ; ② replace $U^{0} \cap W^{0}$ by $(U + W)^{0}$ using (c); ③ turn every annihilator dimension into a dimension in $V$ with Theorem 5.2.2; ④ recognize the dimension formula in $V$ . Finite dimension is spent in step ③.

Proof.

(a) Let $\varphi \in T^{0}$ . Every $\mathbf{s} \in S$ lies in $T$ , so $\varphi(\mathbf{s}) = 0$ . Hence $\varphi \in S^{0}$ .
(b) Since $S \subseteq \operatorname{span} S$ (Theorem 1.3.1), part (a) gives $(\operatorname{span} S)^{0} \subseteq S^{0}$ . Conversely, let $\varphi \in S^{0}$ and $\mathbf{v} \in \operatorname{span} S$ , say $\mathbf{v} = a_1\mathbf{s}_1 + \dots + a_k\mathbf{s}_k$ with $\mathbf{s}_i \in S$ and $k \ge 0$ . By Theorem 2.1.4, $\varphi(\mathbf{v}) = \sum_i a_i\varphi(\mathbf{s}_i) = 0$ . Hence $\varphi \in (\operatorname{span} S)^{0}$ .
(c) Since $U \subseteq U + W$ and $W \subseteq U + W$ (Theorem 1.7.1), part (a) gives $(U + W)^{0} \subseteq U^{0}$ and $(U + W)^{0} \subseteq W^{0}$ , so $(U + W)^{0} \subseteq U^{0} \cap W^{0}$ . Conversely, let $\varphi \in U^{0} \cap W^{0}$ , and let $\mathbf{u} + \mathbf{w} \in U + W$ with $\mathbf{u} \in U$ , $\mathbf{w} \in W$ . Then $\varphi(\mathbf{u} + \mathbf{w}) = \varphi(\mathbf{u}) + \varphi(\mathbf{w}) = 0 + 0 = 0$ . Hence $\varphi \in (U + W)^{0}$ .
(d) $(\supseteq)$ Let $\varphi = \psi + \chi$ with $\psi \in U^{0}$ and $\chi \in W^{0}$ , and let $\mathbf{v} \in U \cap W$ . Since $\mathbf{v} \in U$ and $\mathbf{v} \in W$ , $\varphi(\mathbf{v}) = \psi(\mathbf{v}) + \chi(\mathbf{v}) = 0 + 0 = 0$ . Hence $U^{0} + W^{0} \subseteq (U \cap W)^{0}$ .

$(\subseteq)$ Let $n = \dim V$ . The subspaces $U$ , $W$ , $U \cap W$ (Theorem 1.2.2) and $U + W$ (Theorem 1.7.1) of $V$ are finite-dimensional by Theorem 1.6.7. The space $V^{*}$ has dimension $n$ by Corollary 5.1.4, so its subspaces $U^{0}$ , $W^{0}$ are finite-dimensional too. By the Dimension Formula for Sums (Theorem 1.7.3) in $V^{*}$ , then (c), $\begin{aligned} \dim(U^{0} + W^{0}) &= \dim U^{0} + \dim W^{0} - \dim(U^{0} \cap W^{0}) \\ &= \dim U^{0} + \dim W^{0} - \dim(U + W)^{0} . \end{aligned}$ By Theorem 5.2.2 applied to $U$ , $W$ and $U + W$ , the right-hand side is $\begin{aligned} &(n - \dim U) + (n - \dim W) - \bigl(n - \dim(U + W)\bigr) \\ &\qquad = n - \bigl(\dim U + \dim W - \dim(U + W)\bigr) \\ &\qquad = n - \dim(U \cap W), \end{aligned}$ where the last equality is Theorem 1.7.3 in $V$ . By Theorem 5.2.2 once more, $n - \dim(U \cap W) = \dim(U \cap W)^{0}$ . So $U^{0} + W^{0}$ is a subspace of $(U \cap W)^{0}$ (by $(\supseteq)$ and Theorem 1.7.1) with the same finite dimension, and Theorem 1.6.8 gives $U^{0} + W^{0} = (U \cap W)^{0}$ . This proves the theorem.

Part (b) is what made the computations of this section legitimate: to find the annihilator of a subspace, test functionals on a spanning list only. Parts (c) and (d) form a dictionary. A subspace given as a sum has equations obtained by intersecting equation spaces; a subspace given as an intersection has equations obtained by adding them. Later in this chapter this dictionary becomes a practical method for computing sums and intersections of subspaces.

Remark. Parts (a)–(c) and the inclusion $(\supseteq)$ in (d) hold in every vector space; only the counting step used finite dimension. The equality (d) is in fact true for every $V$ , but the proof must then construct the splitting $\varphi = \psi + \chi$ by hand; Exercise C1 below does this. In finite dimension, (a) also becomes strict: if $U \subsetneq W$ , then $\dim U < \dim W$ (Theorem 1.6.8), so $\dim W^{0} < \dim U^{0}$ and $W^{0} \subsetneq U^{0}$ .

5.2.4. Exercises

A. Check your understanding

Exercise (A1)

(a) Define the annihilator $S^{0}$ of a subset $S$ of a vector space $V$ .
(b) True or false: $U^{0}$ is a subspace of $V$ . Justify your answer.
(c) Let $U$ be a $2$ -dimensional subspace of $\mathbb{R}^5$ . What is $\dim U^{0}$ ?
(d) For a finite-dimensional $V$ , what are $\{\mathbf{0}\}^{0}$ and $V^{0}$ ?
(e) True or false: if $U \subsetneq W$ are subspaces of a finite-dimensional $V$ , then $W^{0} \subsetneq U^{0}$ . Justify your answer.
(f) To compute $U^{0}$ for $U = \operatorname{span}(\mathbf{u}_1, \dots, \mathbf{u}_m) \subseteq F^n$ , which matrix do you row reduce, and why?

Solution.

(a) $S^{0} = \{ \varphi \in V^{*} : \varphi(\mathbf{s}) = 0 \text{ for all } \mathbf{s} \in S \}$ (Definition 5.2.1).
(b) False. $U^{0}$ is a subspace of the dual space $V^{*}$ (Proposition 5.2.1); its elements are functionals, not vectors of $V$ .
(c) $\dim U^{0} = 5 - 2 = 3$ , by Theorem 5.2.2.
(d) $\{\mathbf{0}\}^{0} = V^{*}$ and $V^{0} = \{0\}$ , by Example (First Annihilators) (a).
(e) True. By Theorem 5.2.3 (a), $W^{0} \subseteq U^{0}$ . Since $U \subsetneq W$ , $\dim U < \dim W$ by Theorem 1.6.8, so $\dim W^{0} = \dim V - \dim W < \dim V - \dim U = \dim U^{0}$ . Hence the inclusion is strict.
(f) The matrix $\mathbf{A}$ whose rows are $\mathbf{u}_1^{\top}, \dots, \mathbf{u}_m^{\top}$ . By Theorem 5.1.1 a functional is $\mathbf{x} \mapsto \mathbf{a}^{\top}\mathbf{x}$ , and it kills every $\mathbf{u}_i$ exactly when $\mathbf{A}\mathbf{a} = \mathbf{0}$ ; so $U^{0}$ corresponds to $\operatorname{nul}(\mathbf{A})$ (Example (An Annihilator as a Null Space)).

B. Practice

Exercise (B1: Computing annihilators)

(a) Let $U = \operatorname{span}((1, 2, 1, 0), (2, 4, 3, 1), (1, 2, 2, 1)) \subseteq \mathbb{R}^4$ . Find a basis of $U^{0}$ , and verify that $\dim U + \dim U^{0} = 4$ .
(b) Let $U = \{ p \in \mathbb{R}[x]_{\le 3} : p(1) = p(-1) = 0 \}$ . Show that $U^{0} = \operatorname{span}(\varepsilon_1, \varepsilon_{-1})$ , where $\varepsilon_c(p) = p(c)$ .

Solution.

(a) Let $\mathbf{A}$ have the three spanning vectors as rows. As in Example (An Annihilator as a Null Space), $\varphi_{\mathbf{a}} \in U^{0}$ if and only if $\mathbf{A}\mathbf{a} = \mathbf{0}$ . Subtracting $2$ times row 1 from row 2 and row 1 from row 3 gives rows $(1, 2, 1, 0)$ , $(0, 0, 1, 1)$ , $(0, 0, 1, 1)$ ; subtracting row 2 from row 3 and then row 2 from row 1 gives the RREF $\begin{pmatrix} 1 & 2 & 0 & -1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$ The free variables are $a_2$ and $a_4$ , and Theorem 3.5.4 gives $\operatorname{nul}(\mathbf{A}) = \operatorname{span}((-2, 1, 0, 0), (1, 0, -1, 1))$ . Hence $U^{0}$ has the basis $\psi_1(x_1, x_2, x_3, x_4) = -2x_1 + x_2, \qquad \psi_2(x_1, x_2, x_3, x_4) = x_1 - x_3 + x_4 .$ Check on the first two spanning vectors: $\psi_1(1, 2, 1, 0) = 0$ , $\psi_2(1, 2, 1, 0) = 0$ , $\psi_1(2, 4, 3, 1) = 0$ , $\psi_2(2, 4, 3, 1) = 2 - 3 + 1 = 0$ ; the third vector is their difference, $(2, 4, 3, 1) - (1, 2, 1, 0)$ . The RREF has two pivots, so $\dim U = \dim\operatorname{row}(\mathbf{A}) = 2$ (Theorem 3.5.5), and $2 + 2 = 4$ .
(b) By definition $\varepsilon_1, \varepsilon_{-1} \in U^{0}$ . Independent: if $a\varepsilon_1 + b\varepsilon_{-1} = 0$ , applying it to $x + 1$ gives $2a = 0$ and applying it to $x - 1$ gives $-2b = 0$ . Dimension of $U$ : the map $T \colon \mathbb{R}[x]_{\le 3} \to \mathbb{R}^2$ , $p \mapsto (p(1), p(-1))$ , is linear with kernel $U$ , and its image contains $T(x + 1) = (2, 0)$ and $T(x - 1) = (0, -2)$ , which span $\mathbb{R}^2$ . So $\operatorname{rank} T = 2$ and $\dim U = 4 - 2 = 2$ by Theorem 2.3.1. By Theorem 5.2.2, $\dim U^{0} = 2$ . Hence the $2$ -dimensional subspace $\operatorname{span}(\varepsilon_1, \varepsilon_{-1})$ of $U^{0}$ is all of $U^{0}$ (Theorem 1.6.8).

Exercise (B2: Equations of the symmetric matrices)

Let $U = \{ \mathbf{A} \in M_2(\mathbb{R}) : \mathbf{A}^{\top} = \mathbf{A} \}$ . Find $U^{0}$ , and verify the dimension formula.

Solution. Every symmetric matrix is $a\mathbf{E}_{11} + b(\mathbf{E}_{12} + \mathbf{E}_{21}) + c\mathbf{E}_{22}$ , so $U = \operatorname{span}(\mathbf{E}_{11}, \mathbf{E}_{12} + \mathbf{E}_{21}, \mathbf{E}_{22})$ . Let $(\varepsilon_{ij})$ be the dual basis of the standard basis, so that every $\psi \in M_2(\mathbb{R})^{*}$ is $\psi = \sum_{i,j} c_{ij}\varepsilon_{ij}$ with $c_{ij} = \psi(\mathbf{E}_{ij})$ (Theorem 5.1.3 (c)). By Theorem 5.2.3 (b), $\psi \in U^{0}$ if and only if $\psi$ kills the three spanning matrices: $c_{11} = 0, \qquad c_{12} + c_{21} = 0, \qquad c_{22} = 0 .$ So $\psi = c_{12}(\varepsilon_{12} - \varepsilon_{21})$ , and $U^{0} = \operatorname{span}(\varepsilon_{12} - \varepsilon_{21})$ : the only equation of the symmetric matrices, up to scaling, is $a_{12} - a_{21} = 0$ . The three spanning matrices are independent (they involve disjoint sets of entries), so $\dim U = 3$ , and $\dim U + \dim U^{0} = 3 + 1 = 4 = \dim M_2(\mathbb{R})$ .

Exercise (B3: The equations of a hyperplane)

Let $V$ be a finite-dimensional vector space over $F$ , and let $\varphi \in V^{*}$ be non-zero. Prove that $(\operatorname{ker}\varphi)^{0} = \operatorname{span}(\varphi)$ .

Solution. By definition of the kernel, $\varphi$ vanishes on $\operatorname{ker}\varphi$ , so $\varphi \in (\operatorname{ker}\varphi)^{0}$ , and $\operatorname{span}(\varphi) \subseteq (\operatorname{ker}\varphi)^{0}$ by Proposition 5.2.1. Since $\varphi \neq 0$ , $\dim\operatorname{span}(\varphi) = 1$ , and Proposition 5.1.2 gives $\dim\operatorname{ker}\varphi = \dim V - 1$ . By Theorem 5.2.2, $\dim(\operatorname{ker}\varphi)^{0} = \dim V - (\dim V - 1) = 1$ . A $1$ -dimensional subspace of a $1$ -dimensional space is the whole space (Theorem 1.6.8), so $(\operatorname{ker}\varphi)^{0} = \operatorname{span}(\varphi)$ .

C. Going deeper

Exercise (C1: The intersection formula without counting)

Let $U, W$ be subspaces of a vector space $V$ over $F$ , and let $\varphi \in (U \cap W)^{0}$ .

(a) Suppose $V$ is finite-dimensional. Construct $\psi \in U^{0}$ and $\chi \in W^{0}$ with $\varphi = \psi + \chi$ , without using any dimension formula.
(b) Explain how to modify the construction when $V$ is infinite-dimensional. Deduce that $(U \cap W)^{0} = U^{0} + W^{0}$ holds in every vector space.

Hint: extend a basis of $U \cap W$ to a basis of $U$ and, separately, to a basis of $W$ .

Solution.

(a) Let $(\mathbf{x}_1, \dots, \mathbf{x}_k)$ be a basis of $U \cap W$ (Theorem 1.2.2, Theorem 1.6.7). Extend it to a basis $(\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{u}_1, \dots, \mathbf{u}_p)$ of $U$ and to a basis $(\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{w}_1, \dots, \mathbf{w}_q)$ of $W$ (Theorem 1.6.5).
The combined list $(\mathbf{x}_1, \dots, \mathbf{x}_k, \mathbf{u}_1, \dots, \mathbf{u}_p, \mathbf{w}_1, \dots, \mathbf{w}_q)$ is independent. Let $\sum a_i\mathbf{x}_i + \sum b_i\mathbf{u}_i + \sum c_j\mathbf{w}_j = \mathbf{0}$ . Then $\sum c_j\mathbf{w}_j = -\sum a_i\mathbf{x}_i - \sum b_i\mathbf{u}_i$ lies in $W$ and in $U$ , hence in $U \cap W$ , so $\sum c_j\mathbf{w}_j = \sum d_i\mathbf{x}_i$ for some $d_i \in F$ . Since $(\mathbf{x}, \mathbf{w})$ is a basis of $W$ , all $c_j$ and $d_i$ are $0$ . Then $\sum a_i\mathbf{x}_i + \sum b_i\mathbf{u}_i = \mathbf{0}$ , and since $(\mathbf{x}, \mathbf{u})$ is a basis of $U$ , all $a_i$ and $b_i$ are $0$ .

Extend the combined list to a basis of $V$ by vectors $\mathbf{y}_1, \dots, \mathbf{y}_r$ (Theorem 1.6.5). By Theorem 2.1.5 there is $\psi \in V^{*}$ with $\psi(\mathbf{x}_i) = 0, \qquad \psi(\mathbf{u}_i) = 0, \qquad \psi(\mathbf{w}_j) = \varphi(\mathbf{w}_j), \qquad \psi(\mathbf{y}_l) = 0 .$ Then $\psi$ kills a basis of $U$ , so $\psi \in U^{0}$ (Theorem 5.2.3 (b)). Put $\chi \mathrel{:=} \varphi - \psi$ . Then $\chi(\mathbf{x}_i) = \varphi(\mathbf{x}_i) - 0 = 0$ , because $\mathbf{x}_i \in U \cap W$ and $\varphi \in (U \cap W)^{0}$ ; and $\chi(\mathbf{w}_j) = \varphi(\mathbf{w}_j) - \varphi(\mathbf{w}_j) = 0$ . So $\chi$ kills a basis of $W$ , and $\chi \in W^{0}$ . Hence $\varphi = \psi + \chi \in U^{0} + W^{0}$ .
(b) Replace the finite lists by sets. By Theorem 1.8.3, $U \cap W$ has a basis $X$ , and by Theorem 1.8.2 it extends to a basis $X \cup P$ of $U$ and to a basis $X \cup Q$ of $W$ , with $P \cap X = Q \cap X = \varnothing$ . First, $P \cap Q = \varnothing$ : a vector of $P \cap Q$ lies in $U \cap W = \operatorname{span} X$ , which contradicts the independence of $X \cup P$ . So every vector of $X \cup P \cup Q$ lies in exactly one of $X$ , $P$ , $Q$ . A linear relation among finitely many distinct vectors of $X \cup P \cup Q$ splits into an $X$ -part, a $P$ -part and a $Q$ -part, and the independence argument of (a) applies word for word; so $X \cup P \cup Q$ is independent. By Theorem 1.8.2 it extends to a basis $B$ of $V$ . Every $\mathbf{v} \in V$ is a finite combination $\mathbf{v} = \sum_{\mathbf{b}} c_{\mathbf{b}}\mathbf{b}$ of distinct vectors of $B$ , with unique coefficients because $B$ is independent. Define $\psi(\mathbf{v}) \mathrel{:=} \sum_{\mathbf{b} \in Q} c_{\mathbf{b}}\,\varphi(\mathbf{b}),$ a finite sum over the vectors of $Q$ that occur. Adding or scaling expansions gives expansions, so by uniqueness the coefficients add and scale, and $\psi$ is linear. It vanishes on $X \cup P$ and agrees with $\varphi$ on $Q$ , so the rest of (a) is unchanged: $\psi \in U^{0}$ and $\chi = \varphi - \psi$ kills the basis $X \cup Q$ of $W$ . Together with the inclusion $(\supseteq)$ of Theorem 5.2.3 (d), which used no finiteness, this gives $(U \cap W)^{0} = U^{0} + W^{0}$ in every vector space.

Exercise (C2: Equations of a double root)

Let $U = \{ p \in \mathbb{R}[x]_{\le 3} : p(1) = 0 \text{ and } p'(1) = 0 \}$ , and let $\varepsilon_1(p) = p(1)$ and $\eta_1(p) = p'(1)$ . Prove that $U^{0} = \operatorname{span}(\varepsilon_1, \eta_1)$ . Deduce that $\psi(p) = p(0) - p(2) + 2p'(1)$ does not vanish on $U$ .

Solution. Both $\varepsilon_1$ and $\eta_1$ are linear (evaluation, and differentiation followed by evaluation), and they vanish on $U$ by definition. Independent: if $a\varepsilon_1 + b\eta_1 = 0$ , applying it to $1$ gives $a = 0$ , and applying it to $x - 1$ gives $b = 0$ . Dimension of $U$ : the linear map $T \colon \mathbb{R}[x]_{\le 3} \to \mathbb{R}^2$ , $p \mapsto (p(1), p'(1))$ , has kernel $U$ , and $T(1) = (1, 0)$ , $T(x - 1) = (0, 1)$ , so $T$ is surjective. By Theorem 2.3.1, $\dim U = 4 - 2 = 2$ , and by Theorem 5.2.2, $\dim U^{0} = 2$ . The $2$ -dimensional subspace $\operatorname{span}(\varepsilon_1, \eta_1)$ of $U^{0}$ is therefore all of $U^{0}$ (Theorem 1.6.8).

For the deduction, suppose $\psi \in U^{0}$ . Then $\psi = a\varepsilon_1 + b\eta_1$ . Applying both sides to $1$ gives $1 - 1 + 0 = a$ , so $a = 0$ ; applying them to $x - 1$ gives $-1 - 1 + 2 = b$ , so $b = 0$ . Then $\psi = 0$ . But $\psi(x^2) = 0 - 4 + 4 = 0$ and $\psi(x^3) = 0 - 8 + 6 = -2 \neq 0$ , a contradiction. Hence $\psi \notin U^{0}$ . Check with an explicit witness: $p = x(x - 1)^2 \in U$ has $\psi(p) = 0 - 2 + 0 = -2 \neq 0$ .

Exercise (C3: Dual of a direct sum decomposition)

(a) Let $V$ be finite-dimensional with $V = U \oplus W$ . Prove that $V^{*} = U^{0} \oplus W^{0}$ .
(b) In $\mathbb{R}^2$ , let $U = \operatorname{span}((1, 0))$ and $W = \operatorname{span}((1, 1))$ . Write $\psi(x, y) = x$ as $\psi_U + \psi_W$ with $\psi_U \in U^{0}$ and $\psi_W \in W^{0}$ .

Solution.

(a) Since the sum is direct, $U \cap W = \{\mathbf{0}\}$ (Theorem 1.7.4), and $U + W = V$ . By Theorem 5.2.3 (c) and Example (First Annihilators) (a), $U^{0} \cap W^{0} = (U + W)^{0} = V^{0} = \{0\}$ . By Theorem 5.2.3 (d), $U^{0} + W^{0} = (U \cap W)^{0} = \{\mathbf{0}\}^{0} = V^{*}$ . So $V^{*} = U^{0} + W^{0}$ , and the sum is direct by Theorem 1.7.4.
(b) A functional $ax + by$ vanishes on $(1, 0)$ if and only if $a = 0$ , so $U^{0} = \operatorname{span}(y)$ . It vanishes on $(1, 1)$ if and only if $b = -a$ , so $W^{0} = \operatorname{span}(x - y)$ . Solving $x = s\,y + t(x - y)$ gives $t = 1$ and $s - t = 0$ , so $s = 1$ : $\psi = \underbrace{y}_{\in U^{0}} + \underbrace{(x - y)}_{\in W^{0}} .$ By (a) this decomposition is unique.