A pair of vectors in an inner product space are orthogonal if their inner product is zero. This could be useful, for example, to check for linear independence.

Definition

Orthogonal Vectors & Sets ^formula1

A pair of vectors, in an inner product space are orthogonal iff:

\langle \vec{u}, \vec{v} \rangle = 0 \quad\quad \vec{u} \neq \vec{v}$$ Note that in a *complex* inner product space, two vectors can be orthogonal and *not* be at 90 degree angles, since the [definition for angles only checks real components](Geometry%20from%20Inner%20Products.md#Angle) \ A [set](Set%20-%20Maths.md) of vectors, $\{ \vec{v}_{1}, \vec{v}_{2} , \dots \vec{v}_{n} \}$ is **orthogonal** iff:

\langle \vec{v}{1} \vec{v}{j}\rangle = 0 \quad\quad i \neq j, i,j \in [0, n]

We also have a subcategory of orthogonal vectors, known as orthonormal vectors.

Orthonormal Vectors & Sets

Orthonormal Vectors & Sets ^formula2

If all of these vectors have a norm of 1, then the set is orthonormal:

\langle \vec{v}_{i}, \vec{v}_{j} \rangle = \begin{cases}

0, & i \neq j \ 1, & i = j \end{cases}

Properties

Linear Independence of Orthogonal Sets ^theorem

Every orthogonal set of non-zero vectors in an inner product space is linearly independent.

Applications

Expressing a vector in a vector space

If the basis of an inner product space is an orthogonal set, then we can easily represent any vector in the space using the inner product:

Vector expressed through an orthonormal basis ^theorem

If an inner product space, , has an ordered, orthogonal basis, , then:

I.e. a vector can be expressed using the inner product.

Pythagorean Theorem

Using orthogonal vectors, we can extend the Pythagorean theorem to work for all real inner product spaces:

#todo