# Annihilators, Perpendicular Spaces, A Connection

Whilst writing out revision notes, something struck me!

$\huge X^{\circ} = \varphi(X^{\bot})$

let $V$ be a finite dimensional real inner product space.
let $X \leqslant V$ be a subspace of V.

#### Definition

The annihilator of $X$, denoted $X^{\circ} := \{ f \in V' \mid f(x) = 0 \, \forall x \in X\}$ where $V'$ is the dual space of V.

#### Definition

The perpendicular space of $X$, denoted $X^{\bot} := \{ v \in V \mid \langle x,v \rangle = 0 \, \forall x \in X \}$

It might already be apparent that there is some connection between these two – both involve things going to zero, whenever something is done to everything in $X$.

To make this more explicit, first we need a theorem.

#### Theorem: (weaker version of) Riesz Representation Theorem

$\varphi \colon V \rightarrow V' \colon w \mapsto (v \mapsto \langle v , w \rangle)$ is an isomorphism of vector spaces. Restated, we can associate each linear map uniquely $f \colon V \rightarrow \mathbb{R}$ with a vector $w \in V$ such that $f(x) := \langle x,w \rangle$.

#### Proof:

##### Linearity:

let $u,v \in V, \lambda, \mu \in \mathbb{R}$
$\varphi(\lambda u + \mu v)(x)$
$= \langle x, \lambda u + \mu v \rangle$
$= \lambda \langle x, u \rangle + \mu \langle x , v \rangle$
$= \left ( \lambda \varphi(u) + \mu \varphi(v) \right )(x)$
and hence we see this is a linear map.

##### Injectivity:

Suppose $\langle v , w \rangle = \langle v , w' \rangle \, \forall v \in V$
then $\langle v, w - w' \rangle = 0 \, \forall v \in V$
so specifically $\langle w-w', w - w' \rangle = 0$
thus by positive definiteness, $w = w'$ hence $\varphi$ is injective.

##### Surjectivity:

Since we have that this is an injective linear map between spaces of the same dimension, this follows by Rank Nullity. Alternatively:
let $f \in V'$
take a basis $\varepsilon = \{e_1, \ldots, e_n\}$
then define $w := \sum_{i=1}^n \langle f(e_i),e_i \rangle e_i$
and we can check that this works.

Now to use this theorem to show the connection.

$X^{\circ} := \{ f \in V' \mid f(x) = 0 \, \forall x \in X \}$
but $\forall f \in V', f = \varphi(w)$ for some $w \in V$.
thus $X^{\circ} = \{\varphi(w) \mid \varphi(w)(x) = \langle x , w \rangle = 0 \, \forall x \in X \} = \varphi(X^{\bot})$

So we have the result we wanted:
$\huge X^{\circ} = \varphi(X^{\bot})$

## Applications

This means all of those oh-so-fun identities one derives for annihilator subspace structure can be passed over to perpendicular subspace structure!

#### Example:

Given the identity $X_1^{\circ} + X_2^{\circ} = (X_1 \cap X_2)^{\circ}$
we can deduce $\varphi \left ( (X_1 \cap X_2 )^{\bot} \right ) = (X_1 \cap X_2)^{\circ} = X_1^{\circ} + X_2^{\circ} = \varphi(X_1^{\bot}) + \varphi(X_2^{\bot})$
linearity of $\varphi$ gives us $\varphi(X_1^{\bot}) + \varphi(X_2^{\bot})= \varphi(X_1^{\bot} + X_2^{\bot})$
so by injectivity of $\varphi$ we have $(X_1 \cap X_2)^{\bot} = X_1^{\bot} + X_2^{\bot}$

i.e. compare
$X_1^{\circ} + X_2^{\circ} = (X_1 \cap X_2)^{\circ}$
to
$X_1^{\bot} + X_2^{\bot} = (X_1 \cap X_2)^{\bot}$