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.
More References and links
- Vector Spaces - Questions with Solutions
- Linear Algebra and its Applications - 5 th Edition - David C. Lay , Steven R. Lay , Judi J. McDonald
- Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
- Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
- Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
- Linear Algebra with Applications - 2012 - Gareth Williams
