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}