Chapter 29. Annihilators

An annihilator is a subspace of the dual space consisting of all linear functionals that vanish on a given set of vectors. If (V) is a vector space over a field (F), and (S\subseteq V), then the annihilator of (S) is

$$ S^0={\varphi\in V^*:\varphi(s)=0 \text{ for all } s\in S}. $$

The elements of (S^0) are linear measurements that give zero on every vector in (S). When (S) is a subspace, its annihilator records all linear equations that are satisfied by every vector in (S). The annihilator is always a subspace of (V^*).

29.1 Definition

Let (V) be a vector space over (F). Let (S\subseteq V). The annihilator of (S) is

$$ S^0={\varphi\in V^*:\varphi(s)=0 \text{ for every } s\in S}. $$

The notation (S^0) is common. Some books use (S^\circ) or (\operatorname{Ann}(S)).

If (U\subseteq V) is a subspace, then (U^0) is the set of all linear functionals on (V) that vanish on (U).

Thus

$$ U^0 \subseteq V^*. $$

The annihilator lives in the dual space, not in the original vector space.

29.2 First Example

Let

$$ V=\mathbb{R}^2 $$

and let

$$ U=\operatorname{span} \left( \begin{bmatrix} 1\ 0 \end{bmatrix} \right). $$

Then (U) is the (x)-axis.

A linear functional on (\mathbb{R}^2) has the form

$$ \varphi(x,y)=ax+by. $$

For (\varphi) to vanish on (U), we need

$$ \varphi(t,0)=0 $$

for every (t\in\mathbb{R}). But

$$ \varphi(t,0)=at. $$

This is zero for every (t) exactly when

$$ a=0. $$

Therefore

$$ U^0={\varphi(x,y)=by:b\in\mathbb{R}}. $$

So the annihilator is one-dimensional. It consists of all scalar multiples of the functional that reads the second coordinate.

29.3 Annihilators Are Subspaces

For every subset (S\subseteq V), the annihilator (S^0) is a subspace of (V^*).

First, the zero functional belongs to (S^0), because

$$ 0(s)=0 $$

for every (s\in S).

Next, let

$$ \varphi,\psi\in S^0. $$

Then for every (s\in S),

$$ (\varphi+\psi)(s)=\varphi(s)+\psi(s)=0+0=0. $$

Thus

$$ \varphi+\psi\in S^0. $$

For a scalar (c),

$$ (c\varphi)(s)=c\varphi(s)=c0=0. $$

Thus

$$ c\varphi\in S^0. $$

Therefore (S^0) is a subspace of (V^*).

29.4 Extreme Cases

There are two useful extreme cases.

First,

$$ {0}^0=V^*. $$

Every linear functional sends the zero vector to zero, so every functional vanishes on ({0}).

Second,

$$ V^0={0}. $$

The only linear functional that vanishes on every vector of (V) is the zero functional.

These identities show the reversal built into annihilators: a smaller subset has a larger annihilator, and a larger subset has a smaller annihilator.

29.5 Inclusion Reversal

If

$$ S\subseteq T\subseteq V, $$

then

$$ T^0\subseteq S^0. $$

Indeed, if a functional vanishes on all of (T), then it certainly vanishes on all of (S). Thus it belongs to (S^0).

This reversal is important. Annihilators turn containment around.

A large subspace imposes many conditions on a functional, so fewer functionals vanish on it. A small subspace imposes fewer conditions, so more functionals vanish on it.

29.6 Annihilator of a Span

The annihilator of a set equals the annihilator of its span:

$$ S^0=(\operatorname{span} S)^0. $$

If a functional vanishes on every vector of (S), then by linearity it vanishes on every linear combination of vectors from (S). Hence it vanishes on (\operatorname{span} S).

Conversely, if it vanishes on (\operatorname{span} S), then it vanishes on (S), since

$$ S\subseteq \operatorname{span} S. $$

Therefore the annihilator depends only on the subspace generated by the set.

29.7 Computing an Annihilator in Coordinates

Let (U\subseteq \mathbb{R}^n) be spanned by columns of a matrix

$$ A= \begin{bmatrix} |&|&&|\ u_1&u_2&\cdots&u_k\ |&|&&| \end{bmatrix}. $$

A functional (\varphi\in(\mathbb{R}^n)^*) can be represented by a row vector

$$ y^T. $$

The condition (\varphi(u_i)=0) for every (i) becomes

$$ y^Tu_i=0. $$

Equivalently,

$$ y^TA=0. $$

Transposing,

$$ A^Ty=0. $$

Thus the annihilator of the column space of (A) corresponds to the null space of (A^T):

$$ (\operatorname{Col}(A))^0 \cong \operatorname{Null}(A^T). $$

This is the coordinate form of the left null space.

29.8 Example in (\mathbb{R}^3)

Let

$$ U=\operatorname{span} \left( \begin{bmatrix} 1\ 1\ 0 \end{bmatrix}, \begin{bmatrix} 0\ 1\ 1 \end{bmatrix} \right) \subseteq \mathbb{R}^3. $$

A linear functional has the form

$$ \varphi(x,y,z)=ax+by+cz. $$

We require

$$ \varphi(1,1,0)=0 $$

and

$$ \varphi(0,1,1)=0. $$

These give

$$ a+b=0, $$

$$ b+c=0. $$

Hence

$$ a=-b, \qquad c=-b. $$

Let

$$ b=t. $$

Then

$$ (a,b,c)=(-t,t,-t)=t(-1,1,-1). $$

Therefore

$$ U^0= \operatorname{span} {\varphi(x,y,z)=-x+y-z}. $$

The annihilator is one-dimensional. This agrees with the dimension formula:

$$ \dim U^0=3-2=1. $$

29.9 Dimension Formula

If (V) is finite-dimensional and (U\subseteq V) is a subspace, then

$$ \dim U^0=\dim V-\dim U. $$

The number on the right is the codimension of (U) in (V). Thus

$$ \dim U^0=\operatorname{codim} U. $$

An annihilator contains one independent linear condition for each direction missing from (U). Equivalently, it measures how many independent functionals can vanish on (U).

29.10 Proof of the Dimension Formula

Let

$$ \dim V=n, \qquad \dim U=k. $$

Choose a basis of (U):

$$ u_1,\ldots,u_k. $$

Extend it to a basis of (V):

$$ u_1,\ldots,u_k,v_{k+1},\ldots,v_n. $$

Let the dual basis be

$$ u^1,\ldots,u^k,v^{k+1},\ldots,v^n. $$

A functional in (U^0) must vanish on

$$ u_1,\ldots,u_k. $$

The dual basis functionals

$$ v^{k+1},\ldots,v^n $$

vanish on all of (U), and they form a basis of (U^0). There are (n-k) of them.

Therefore

$$ \dim U^0=n-k=\dim V-\dim U. $$

29.11 Annihilator of a Sum

If (U) and (W) are subspaces of (V), then

$$ (U+W)^0=U^0\cap W^0. $$

A functional vanishes on (U+W) exactly when it vanishes on every vector of (U) and every vector of (W).

Indeed, if

$$ \varphi\in (U+W)^0, $$

then (U\subseteq U+W) and (W\subseteq U+W), so

$$ \varphi\in U^0\cap W^0. $$

Conversely, if

$$ \varphi\in U^0\cap W^0, $$

then for any

$$ u+w\in U+W, $$

we have

$$ \varphi(u+w)=\varphi(u)+\varphi(w)=0. $$

Thus

$$ \varphi\in (U+W)^0. $$

29.12 Annihilator of an Intersection

For finite-dimensional spaces,

$$ (U\cap W)^0=U^0+W^0. $$

This identity is dual to the previous one. The annihilator changes intersections into sums and sums into intersections.

The inclusion

$$ U^0+W^0\subseteq (U\cap W)^0 $$

is immediate: if a functional vanishes on (U), and another vanishes on (W), then their sum vanishes on every vector belonging to both.

The reverse inclusion is deeper and follows from the dimension formula.

29.13 Double Annihilator

If (V) is finite-dimensional and (U\subseteq V), then

$$ U^{00}=U, $$

after identifying (V) with its double dual (V^{**}).

Here

$$ U^{00} $$

means the annihilator of (U^0) inside (V^{**}). Under the natural identification

$$ V\cong V^{**}, $$

it returns the original subspace (U).

This says that in finite dimensions, a subspace is completely determined by the linear functionals that vanish on it.

29.14 Annihilators and Quotient Spaces

There is a natural isomorphism

$$ (V/U)^*\cong U^0. $$

A functional on the quotient (V/U) is the same as a functional on (V) that vanishes on (U).

To see this, let

$$ \lambda:V/U\to F $$

be linear. Define

$$ \varphi(v)=\lambda(v+U). $$

Then (\varphi) is a linear functional on (V). If (u\in U), then

$$ u+U=U, $$

so

$$ \varphi(u)=\lambda(U)=0. $$

Thus

$$ \varphi\in U^0. $$

Conversely, if (\varphi\in U^0), define

$$ \lambda(v+U)=\varphi(v). $$

This is well-defined because (\varphi) vanishes on (U).

29.15 Annihilators and Linear Equations

A subspace can often be described as the common zero set of linear functionals.

For example, in (\mathbb{R}^3),

$$ U={(x,y,z):x+y+z=0} $$

is the kernel of the functional

$$ \varphi(x,y,z)=x+y+z. $$

Thus

$$ U=\ker \varphi. $$

The annihilator (U^0) is the set of all linear functionals that vanish on this plane. Since the plane has dimension (2), its annihilator has dimension (1). Hence

$$ U^0=\operatorname{span}(\varphi). $$

This shows that annihilators encode systems of homogeneous linear equations.

29.16 Orthogonal Complements and Annihilators

In a finite-dimensional inner product space, a vector (a\in V) defines a functional

$$ \varphi_a(v)=\langle v,a\rangle. $$

Under this identification, the annihilator of a subspace corresponds to its orthogonal complement:

$$ U^0 \cong U^\perp. $$

The annihilator is more general because it does not require an inner product. Orthogonal complements depend on a chosen inner product. Annihilators depend only on the vector space and its dual.

29.17 Row Space and Null Space

For a matrix (A), the null space is the annihilator of the row space.

Let the rows of (A) be

$$ r_1,\ldots,r_m. $$

Then (x\in \operatorname{Null}(A)) means

$$ r_i x=0 $$

for every row (r_i).

Thus (x) is annihilated by every row functional. In dual language,

$$ \operatorname{Null}(A) = \operatorname{Row}(A)^0, $$

after identifying row vectors with linear functionals on (F^n).

This is one of the fundamental relationships between the four matrix subspaces.

29.18 Column Space and Left Null Space

Similarly, the left null space of (A) is the annihilator of the column space.

The left null space is

$$ \operatorname{Null}(A^T). $$

A vector (y\in F^m) lies in (\operatorname{Null}(A^T)) exactly when

$$ A^Ty=0. $$

Equivalently,

$$ y^Ta=0 $$

for every column (a) of (A).

Thus (y) defines a functional that vanishes on (\operatorname{Col}(A)). Hence

$$ \operatorname{Null}(A^T) \cong \operatorname{Col}(A)^0. $$

29.19 Practical Computation

To compute (U^0) for a subspace (U\subseteq F^n):

Step	Operation
1	Put spanning vectors of (U) as columns of a matrix (A)
2	Write a general functional as (y^T)
3	Solve (y^TA=0), equivalently (A^Ty=0)
4	Convert the solution vectors (y) into functionals (y^Tx)
5	The resulting functionals form a basis for (U^0)

This procedure reduces annihilator computation to a null space computation.

29.20 Summary

An annihilator is the set of all linear functionals that vanish on a given set or subspace. It is a subspace of the dual space.

The key ideas are:

Concept	Meaning
Annihilator	(S^0={\varphi:\varphi(s)=0\text{ for all }s\in S})
Ambient space	(S^0\subseteq V^*)
Inclusion reversal	(S\subseteq T) implies (T^0\subseteq S^0)
Dimension formula	(\dim U^0=\dim V-\dim U)
Sum rule	((U+W)^0=U^0\cap W^0)
Intersection rule	((U\cap W)^0=U^0+W^0) in finite dimensions
Double annihilator	(U^{00}=U) in finite dimensions
Quotient relation	((V/U)^*\cong U^0)
Matrix computation	(U^0) is found by solving (A^Ty=0)

Annihilators express subspaces through the linear functionals that vanish on them. They provide a coordinate-free way to describe linear equations, quotient duals, orthogonal complements, and the relationships among row spaces, column spaces, and null spaces.