
# 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
$v =\sum_{i} (v\cdot u_{i})u_{i}.$