Orthogonality: Difference between revisions

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== Overview ==
Orthogonality generalises to an [[inner product space]].


mathematical relationship
== Mathematical definition ==
 
Let [[vector space]] $V$ be an [[inner product space]].
 
Two vectors $u,v\in V$ are '''orthogonal''' if $\langle u, v \rangle=0$.
 
$u\bot v$ denotes this orthogonality.
 
== Properties ==
 
=== $\vec 0$ is orthogonal to every vector ===
 
$$
\forall v \in V, \vec 0 \bot v
$$
 
'''Proof''':
 
$$
Recall [[Vector_space#Mathematical_definition|the properties of a vector space by definition]]. By additive identity, $\vec 0 +\vec 0 = \vec 0$
 
Therefore,
 
$$
\langle \vec 0, u \rangle = \langle \vec 0+\vec 0, u\rangle
$$
 
Recall the [[inner product|properties of the inner product by definition]], by linearity in the 1st slot,
 
$$
\langle \vec 0+\vec 0, u\rangle = \langle\vec 0, u \rangle+\langle \vec 0, u\rangle
$$
 
Subtracting $\langle \vec 0, u\rangle$ from both sides gives
 
$$
\langle \vec 0, u \rangle=0.
$$

Revision as of 14:36, 29 September 2026

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Orthogonality generalises to an inner product space.

Mathematical definition

Let vector space $V$ be an inner product space.

Two vectors $u,v\in V$ are orthogonal if $\langle u, v \rangle=0$.

$u\bot v$ denotes this orthogonality.

Properties

$\vec 0$ is orthogonal to every vector

$$ \forall v \in V, \vec 0 \bot v $$

Proof:

$$ Recall the properties of a vector space by definition. By additive identity, $\vec 0 +\vec 0 = \vec 0$

Therefore,

$$ \langle \vec 0, u \rangle = \langle \vec 0+\vec 0, u\rangle $$

Recall the properties of the inner product by definition, by linearity in the 1st slot,

$$ \langle \vec 0+\vec 0, u\rangle = \langle\vec 0, u \rangle+\langle \vec 0, u\rangle $$

Subtracting $\langle \vec 0, u\rangle$ from both sides gives

$$ \langle \vec 0, u \rangle=0. $$