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}

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Annihilators, Perpendicular Spaces, A Connection

2 thoughts on “Annihilators, Perpendicular Spaces, A Connection

  1. Yes, I swear a similar feeling struck me once, but I never cared to look into it, good job!
    For some reason the curriculum is obsessed with infinite dimensional vector spaces. Does this correspondence hold for those wretched things?

    1. jebavarde says:

      I did worry about infinite things. Some of these identities do break for infinite dimensional ‘stuff’, so maybe the correspondence does go boom. I don’t really know much about infinite dimensional vector spaces. The words analytic basis always seem to crop up.

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