# 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.