14.2: Orthogonal and Orthonormal Bases

There are many other bases that behave in the same way as the standard basis. As such, we will study:

1. \(\textit{Orthogonal bases}\) \(\{v_{1}, \ldots, v_{n} \}\):
\[
v_{i}\cdot v_{j}=0 \textit{ if } i\neq j\, .
\]
In other words, all vectors in the basis are perpendicular.

2. \(\textit{Orthonormal bases}\) \(\{u_{1}, \ldots, u_{n} \}\):
\[
u_{i}\cdot u_{j} = \delta_{ij}.
\]
In addition to being orthogonal, each vector has unit length.

Suppose \(T=\{u_{1}, \ldots, u_{n} \}\) is an orthonormal basis for \(\Re^{n}\). Because \(T\) is a basis, we can write any vector \(v\) uniquely as a linear combination of the vectors in \(T\):
\[
v=c^{1}u_{1}+\cdots c^{n}u_{n}.
\]
Since \(T\) is orthonormal, there is a very easy way to find the coefficients of this linear combination. By taking the dot product of \(v\) with any of the vectors in \(T\), we get:
\begin{eqnarray*}
v\cdot u_{i} &=& c^{1}u_{1}\cdot u_{i} + \cdots + c^{i}u_{i}\cdot u_{i} + \cdots + c^{n}u_{n}\cdot u_{i} \\
& = & c^{1}\cdot 0 + \cdots + c^{i}\cdot 1 + \cdots + c^{n}\cdot 0 \\
& = & c^{i}, \\
\Rightarrow\, c^{i} &=& v\cdot u_{i} \\
\Rightarrow\ v &=& (v\cdot u_{1}) u_{1} + \cdots + (v\cdot u_{n})u_{n}\\
&=& \sum_{i} (v\cdot u_{i})u_{i}.
\end{eqnarray*}

This proves the theorem:

Theorem

For an orthonormal basis \(\{u_{1}, \ldots, u_{n} \}\), any vector \(v\) can be expressed as
\[
v =\sum_{i} (v\cdot u_{i})u_{i}.
\]

 

Contributor

• David Cherney, Tom Denton, and Andrew Waldron (UC Davis)