Two vectors are considered orthogonal if \(\langle u,v \rangle = 0\), that is, their inner product is \(0\).

See also orthogonality test.

orthogonality and \(0\)

• \(0\) is orthogonal to every vector in \(v\) because \(\langle 0,v \rangle=0\) for every \(v\) because of the properties of inner product
• \(0\) is the only vector orthogonal to itself as, by inner product definiteness, \(\langle v,v \rangle=0\) implies \(v=0\).

A list of vectors is orthonormal if each vector is orthogonal to every other vector, and they all have norm 1.

In other words:

\begin{equation} \langle e_{j}, e_{k} \rangle = \begin{cases} 1, j = k\\ 0, j \neq k \end{cases} \end{equation}

The vectors should inner-product with itself to \(1\), and be orthogonal to all others.

Additional Information

orthonormal basis

See also orthonormal basis

Norm of an Orthogonal Linear Combination

\begin{equation} \| a_1e_1 + \dots + a_{m}e_{m} \|^{2} = |a_1|^{2} + \dots + |a_{m}|^{2} \end{equation}

An Orthonormal basis is defined as a basis of a finite-dimensional vector space that’s orthonormal.

Additional Information

orthonormal list of the right length is a basis

An orthonormal list is linearly independent, and linearly independent list of length dim V are a basis of V. \(\blacksquare\)

Writing a vector as a linear combination of orthonormal basis

According to Axler, this result is why there’s so much hoopla about orthonormal basis.

Result and Motivation

For any basis of \(V\), and a vector \(v \in V\), we by basis spanning have:

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action outcomes are uncertain