Orthonormal Basis

  

Orthonormal Basis Definition

A set of vectors is orthonormal if each vector is a unit vector ( length or norm is equal to $1$) and all vectors in the set are orthogonal to each other. Therefore a basis is orthonormal if the set of vectors in the basis is orthonormal. The vectors in a set of orthogonal vectors are linearly independent.

Examples With Solutions

Example 1
Show that the vectors $\textbf i = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ , $\textbf j = \begin{bmatrix} 0\\ 1\\ 0 \end{bmatrix}$ and $\textbf k = \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$ form an ortonormal basis.

Solution to Example 1
Show that the inner product of each pair of vectors is equal to 0.
$\textbf i \cdot \textbf j = (1)(0)+(0)(1)+(0)(0) = 0$
$\textbf i \cdot \textbf k = (1)(0)+(0)(0)+(0)(1) = 0$
$\textbf j \cdot \textbf k = (0)(0)+(1)(0)+(0)(1) = 0$
Therefore the three vectors are orthogonal and also linearly independent.
Show that the length or norm of each vector is equal to 1.
$||i|| = \sqrt {1^2+0^2+0^2} = \sqrt 1 = 1$
$||j|| = \sqrt {0^2+ 1^2+0^2} = \sqrt 1 = 1$
$||j|| = \sqrt {0^2+ 0^2+1^2} = \sqrt 1 = 1$
The three vectors are orthogonal to each other and the length of each is equal to 1; they therefore form an orthonormal basis.

Example 2
a) Show that the vectors $\; \textbf u = K \begin{bmatrix} 1\\ 0 \\ 2 \end{bmatrix}$ , $\; \textbf v = L \begin{bmatrix} 4 \\ 1 \\ -2 \end{bmatrix}$ and $\; \textbf w = M \begin{bmatrix} -2 \\ 10 \\ 1 \end{bmatrix}$, where $K$, $L$ and $M$ are real constants, are orthogonal.
b) Find positive values for the constants $K$, $L$ and $M$ so that the vectors $\textbf u$ , $\textbf v$ and $\textbf w$ form an orthonormal basis.
Solution to Example 2
a)
Calculate the inner product of each pair of vectors.
$\textbf u \cdot \textbf v = KL(1)(4)+KL(0)(1)+KL(2)(-2) = 4 KL - 4 KL = 0$
$\textbf u \cdot \textbf w = KM(1)(-2)+KM(0)(10)+KM(2)(1) = -2 KM + 2 KM = 0$
$\textbf v \cdot \textbf w = LM(4)(-2)+LM(1)(10)+LM(-2)(1) = -8LM + 10 LM - 2LM = 0$
Therefore vectors $\textbf u$, $\textbf v$ and $\textbf w$ are orthogonal and linearly independent.
b)
We now need to find the length (or norm) of each vectors and set it equal to $1$.
$||\textbf u|| = \sqrt {k^2 + 0^2 + (2k)^2} = 1$
which gives the equation
$\sqrt {5 k^2} = 1$
Square both sides of the equation above and solve for $K$ positive.
$5 k^2 = 1$ gives $K = \dfrac{1}{\sqrt 5}$

$||\textbf v|| = \sqrt {(4L)^2 + L^2 + (-2L)^2} = 1$
which gives the equation
$\sqrt {21 L^2} = 1$
Square both sides of the equation above and solve for $L$ positive.
$21 L^2 = 1$ gives $L = \dfrac{1}{\sqrt {21}}$

$||\textbf w|| = \sqrt {(-2M)^2 + (10M)^2 + M^2} = 1$
which gives the equation
$\sqrt {105 M^2} = 1$
Square both sides of the equation above and solve for $M$ positive.
$105 M^2 = 1$ gives $M = \dfrac{1}{\sqrt {105}}$

Example 3
Proove that the vectors $\textbf u = \begin{bmatrix} \sin(\theta) \cos (\phi) \\ \sin(\theta) \sin (\phi) \\ \cos (\theta) \end{bmatrix}$ , $\textbf v = \begin{bmatrix} \cos(\theta) \cos (\phi) \\ \cos(\theta) \sin (\phi) \\ -\sin (\theta) \end{bmatrix}$ and $\textbf w = \begin{bmatrix} -\sin (\phi) \\ \cos (\phi) \\ 0 \end{bmatrix}$ form an orthonormal basis.
Solution to Example 3
In example 6 of
orthogonal vectors , it has been prooved that vectors $\textbf u$, $\textbf v$ and $\textbf w$ given above are orthogonal and therefore linearly independent.
We now need to show that the norm of each vector is equal to $1$.
$||\textbf u || = \sqrt { (\sin(\theta) \cos (\phi))^2 +(\sin(\theta) \sin (\phi))^2 +(\cos (\theta))^2 }$
Expand
$||\textbf u || = \sqrt { \sin^2(\theta) \cos^2 (\phi) + \sin^2(\theta) \sin^2 (\phi) +\cos^2 (\theta)) }$
Factor $\sin^2(\theta)$ out from the first and second term of the radicand
$||\textbf u || = \sqrt { \sin^2(\theta) [ \cos^2 (\phi) + \sin^2 (\phi) ] +\cos^2 (\theta)) }$
Use trigonometric identity $\cos^2 (\phi) + \sin^2 (\phi) = 1$ to simplify the above to
$||\textbf u || = \sqrt { \sin^2(\theta) +\cos^2 (\theta)) }$
Use trigonometric identity $\cos^2 (\theta) + \sin^2 (\theta) = 1$ to simplify the above to
$||\textbf u || = \sqrt 1 = 1$

$||\textbf v || = \sqrt { (\cos(\theta) \cos (\phi))^2 +(\cos(\theta) \sin (\phi))^2 +(-\sin (\theta))^2 }$
Expand
$||\textbf v || = \sqrt { \cos^2(\theta) \cos^2 (\phi) + \cos^2(\theta) \sin^2 (\phi) +\sin^2 (\theta)) }$
Factor $\cos^2(\theta)$ out from the first and second term of the radicand
$||\textbf v || = \sqrt { \cos^2(\theta) [ \cos^2 (\phi) + \sin^2 (\phi) ] +\sin^2 (\theta)) }$
Use trigonometric identity $\cos^2 (\phi) + \sin^2 (\phi) = 1$ to simplify the above to
$||\textbf v || = \sqrt { \cos^2(\theta) +\sin^2 (\theta)) }$
Use trigonometric identity $\cos^2 (\theta) + \sin^2 (\theta) = 1$ to simplify the above to
$||\textbf v || = \sqrt 1 = 1$

$||\textbf w || = \sqrt { (-\sin (\phi))^2 +(\cos (\phi))^2 + 0^2 }$
Simplify
$||\textbf w || = \sqrt { \sin^2 (\phi) + \cos^2 (\phi) }$
Use trigonometric identity $\cos^2 (\phi) + \sin^2 (\phi) = 1$ to simplify the above to
$||\textbf w || = \sqrt 1 = 1$
The vectors $\textbf u$ , $\textbf v$ and $\textbf w$ form an orthonormal basis.