orthonormal basis

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Chapter 5: Inner Product Spaces 26

SECTION C Orthonormal Bases

By the end of this section you will be able to

• understand what is meant by an orthonormal basis

• find an orthonormal basis by using the Gram-Schmidt Process

• develop properties of an orthonormal basis

C1 Introduction to an Orthonormal Basis

What do you think the term orthonormal basis means?

Orthonormal basis is a set of vectors which forms a basis for a vector space and each of

these basis vectors are normalized and they are orthogonal to each other.

Definition (5.13).

Let V be a finite dimensional vector space with an inner product. A set of basis vectors

{ }1 2 3, , , , n B = u u u u" for V is called an orthonormal basis if they are

(i) Orthogonal, that is , 0i j =u u for i j≠

(ii) Normalized, that is 1 j =u for 1, 2, 3, , j n= "

What do you notice about this definition?

Same as the definition of an orthonormal set (5.11) given in the last section but this

time the set of vectors are the basis vectors of the vector space.

Examples of orthonormal basis are shown below for : 2 3and\ \ z

x

y

Fig 5

The set of vectors { }1 2,e e form an orthonormal basis for and the set {2\ }1 2 3

, ,e e e

forms an orthonormal basis for .3\

In general the set { }1 2 3, , , , n B = e e e e" forms an orthonormal basis for with

respect to the dot product. Remember

n

\0

1

0

k

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

e

#

#

The set { }1 2 3, , , , n B = e e e e" is not the only orthonormal basis for , we can

create an infinite number of them.

n\

Why do we need an orthonormal basis?

Generally it is easier to work with an orthonormal basis rather than any other basis.

2\ 3\

x

y

e1

e2

e3e2 =

⎝ ⎜⎛ 0

1 ⎠⎟

⎞

e = ⎟1

⎝ ⎜⎛ 1

0 ⎠ ⎞

1 in the kth position and zeros

everywhere else.




Can you remember what the term basis means?

It was defined in the last chapter (4.14) as the set of vectors which span (generate) the

vector space and the vectors are linearly independent.

To check that a given set of vectors forms a basis for a vector space V can be a tedious

task. Why?

Because we need to check that these vectors span V and they are linearly independent.From the last chapter we know it is enough to show that for an n dimensional vector

space then n linear independent vectors form a basis for this n dimension vector space.

Next we show that an orthogonal set is linearly independent.

C2 Properties of an Orthonormal Basis In this subsection we prove that if we have an orthogonal set of vectors then they are

linearly independent. We use this to prove that in an n dimensional vector space any set

of n orthogonal vectors forms a basis for that vector space.

Proposition (5.14). If { }1 2 3 , ,, , nvv v v " is an orthogonal set of non-zero vectors in

an inner product space then they are linearly independent. How do we prove this proposition?

We consider a linear combination and show that all the scalars are zero.

Proof .

Consider the linear combination of the vectors { }1 2 3, ,, ,

nvv v v " and equate them

to the zero vector, O:

1 1 2 2 3 3 n nk k k k + ++ + =vv v v " O

What do we need to prove?

Required to prove that all the scalars are zero:1 2 3

0, 0, 0, and 0nk k k k = = = ="

How can we show all the scalars are 0?

Consider an inner product of an arbitrary vector jv in the set { }1 2 3 , ,, , nvv v v "

with

1 1 2 2 3 3 n nk k k k + ++ + =vv v v " O

What is 1 1 2 2 3 3 ,n n jk k k k + ++ + vv v v v" equal to?

This inner product is zero because by Proposition (5.2) , 0=O v . Therefore

1 1 2 2 3 3, ,n n j jk k k k + + 0+ + =vv v v v O v" = (*)

We are given that { }1 2 3 , ,, , nvv v v " is orthogonal, that means

, 0 if i j i j= ≠v v

Expanding (*) and using this , 0 if i j i j= ≠v v we have

N

N N

( ) ( ) ( )

1 1 1 1

1 1

Taking Out Scalars

1 1

0 00 0

2

1 2

, , , ,

, , ,

, , ,

0 0 0 0 0 0 0

n n j j j j j n n j

j j j j n n

j j j j n n j

j j n

k k k k k

k k k

k k k

k k k k

= == =

++ = + + + +

= + + + +

= + + + +

= + + + + +

vv v v v v v v v

v v v v v v

v v v v v v

v

" " "

" "

" "

" "

j




The last line follows from definition of the norm2

, =u u u . The only non-zero

contribution is the inner product of the vectors j

v and j

v . By (*) we know all this is

equal to zero, which means we have2

0 j jk =v

Since j

v is one of the non-zero vectors amongst the set { }1 2 3, ,, ,

nvv v v "

therefore [2

0 Not Equal to 0 j ≠v ] . Thus to satisfy the above we must have .0 j

k =

Since j

v was an arbitrary vector therefore all the k ’s must be zero which proves that an

orthonormal set of vectors is linearly independent.

■

What does this proposition mean?

Means that orthogonality implies linear independence. For example the vectors

{ }1 2 3, ,e e e are orthogonal therefore linearly independent.

We can go further as the next proposition states.Corollary (5.15). In an n dimensional inner product space V any set of n orthogonal

vectors forms a basis for V .

Proof .

By the above Proposition (5.14) we have that the set of n orthogonal vectors are

linearly independent. By the Proposition of the last chapter we conclude that the n

linear independent vectors form a basis for V .

■

What does this corollary mean?

If we have n dimensional vector space with an inner product then any n orthogonal

vectors form a basis for that vector space.However vectors which are linearly independent may not be orthogonal. For example

the vectors and are linearly independent but not orthogonal as you

can see in the next diagram.

3

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠u

2

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠v

x0.5 1 1.5 2 2.5 3

y

0.5

1

1.5

2

u =

⎝ ⎜⎛ 3

1 ⎠⎟ ⎞

v =

⎝

⎜⎛ 2

1 ⎠

⎟ ⎞

These vectors are linearly

independent but not orthogonal

(not perpendicular).

Fig 6




Example 12

Show that are orthogonal in with respect to the dot1 1

and1 1

− −⎛ ⎞ ⎛ = =⎜ ⎟ ⎜

−⎝ ⎠ ⎝ ⎠u v

⎞⎟

2\

product and write down an orthonormal basis for by normalizing these vectors.2\

Solution. How do we show the given vectors u and v are orthogonal?

Evaluate the inner (dot) product of these vectors:

( )( ) ( )( )1 1

, 1 11 1

− −⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ = − × − + × − =⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠u v u v 1 1 0

Hence vectors u and v are orthogonal. How do we normalize these vectors?

By using Proposition (5.11) which isw

w. What is u equal to?

( )2 2 2

1 1, 1

1 1

− −⎛ ⎞ ⎛ ⎞= = ⋅ = ⋅ = − +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠u u u u u 1 2=

Taking the square root gives 2=u . Similarly we have

( ) ( )2 21 1

, 11 1

− −⎛ ⎞ ⎛ ⎞= = ⋅ = ⋅ = − + −⎜ ⎟ ⎜ ⎟

− −⎝ ⎠ ⎝ ⎠v v v v v

21 2=

Hence 2=v . Thus the normalized vectors are

11

12

−⎛ ⎞= ⎜ ⎟

⎝ ⎠

u

uand

11

12

−⎛ ⎞= ⎜ ⎟

−⎝ ⎠

v

v

The orthonormal basis for is2\1 11 1

,

1 12 2

⎧ − −⎛ ⎞ ⎛ ⎞⎫⎨ ⎬⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠⎩ ⎭

.

We can plot this orthonormal basis:

x-1 -0.5 0.5 1

y

-1

-0.5

0.5

1

u

⎝ ⎜⎜⎜⎛ -1

2

1

2 ⎠⎟⎟⎟

⎞

v

⎝ ⎜⎜⎜⎛ -1

2

-1

2 ⎠⎟⎟⎟

⎞

Fig 7

C3 Gram Schmidt Process

Given any arbitrary basis { }1 2 3, , , ,

nv v v v" for a finite dimensional inner product

space we can find an orthogonal basis { }1 2 3, , , , nw w w w" by the Gram SchmidtProcess described next:




Gram Schmidt Process (5.16).

Let1 1=w v

2 1

2 2 2

1

,= −

v ww v w

w1

3 1 3 2

3 3 1 22 2

1 2

4 1 4 2 4 3

4 4 1 22 2

1 2 3

, ,

, , ,

= − −

= − − −

v w v ww v w w

w w

v w v w v ww v w w w

w w w

# # # # #

32

1 2 3 1

1 2 32 2 2 2

1 2 3 1

, , , ,n n n n n

n n n

n

−−

−

= − − − − −v w v w v w v w

w v w w w ww w w w

" 1

These vectors w’s are orthogonal therefore a basis. What else do we need to do to form

an orthonormal basis?

Normalize { }1 2 3, , , , nw w w w" by using Proposition (5.11) which is j

j

w

w.

Example 13

Transform the basis in to an orthonormal basis for 1 2

2 1and

1 1

⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠v v = 2\ 2\

with respect to the dot product. Also plot this orthonormal basis and the given basis on

two different graphs.

Solution.

Using the Gram Schmidt Process (5.16) above means that we need to find vectors 1w

and which are orthogonal.2

w

Clearly we have . By (5.16) we have1 1

2

1

⎛ ⎞= = ⎜ ⎟

⎝ ⎠w v

2 1

2 2 2

1

,= −

v ww v w

w1 (*)

We need to evaluate each component of (*). What is inner product 2 1,v w equal to?

( ) ( )2 1 2 1

1 2

, 1 21 1

⎛ ⎞ ⎛ ⎞

= ⋅ = ⋅ = × + × =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠v w v w 1 1 3

What is2

1w equal to?

2

1 1 1 1 1

2 2

,

2 22 1 5

1 1

= = ⋅

⎛ ⎞ ⎛ ⎞= ⋅ = + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

w w w w w

Substituting all these, ,2

1

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠v 2 1

, 3=v w ,2

15=w and , into (*) gives1

2

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠w




2 1

2 2 12

1

,

1 23

1 15

1 6 / 5 1/ 5 11

1 3 / 5 2 / 5 25

= −

⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

− − −⎛ ⎞ ⎛ ⎞ ⎛ = = =⎜ ⎟ ⎜ ⎟ ⎜

−⎝ ⎠ ⎝ ⎠ ⎝

v ww v w

w

⎞⎟ ⎠

Our vectors and1

2

1

⎛ ⎞= ⎜ ⎟

⎝ ⎠w 2

11

25

−⎛ = ⎜

⎝ ⎠w

⎞⎟ are an orthogonal basis. But how do we make

these into an orthonormal basis?

By normalizing these vectors which is achieved via dividing the vectors by their norms

(lengths). Call these new vectors and they are given by1

andu2

u

1 21 2

1 2

and= =w w

u u

w w

From above we have 1 5=w . We need to normalize the other vector :2

w

( )

2

2 2 2

2 2

1 11 1,

2 25 5

1 51 2

25 25 5

− −⎛ ⎞ ⎛ ⎞= = ⋅⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎡ ⎤ 1= − + = =

⎣ ⎦

w w w

Thus2

1

5=w . Our normalized vectors are

11 1

1 1

21 1

15

⎛ ⎞

= = = ⎜ ⎟⎝ ⎠

w

u ww w

2 2

2

1 11 1 1 1 1By Law of Indices 5 5

1 2 25 5 5

5

− −⎛ ⎞ ⎛ ⎞ ⎡ ⎤= = = =⎜ ⎟ ⎜ ⎟ ⎢ ⎥

⎣ ⎦⎝ ⎠ ⎝ ⎠u w

w

Hence our orthonormal basis is the set { }1 2,u u where these are given above,

1 2

2 11 1and

1 25 5

−⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠u u

Plotting these orthonormal basis and our given basis in :2\

y

x0.5 1 1.5 2

y

0.2

0.4

0.6

0.8

1

⎝ ⎜⎛ 2

1 ⎠⎟

⎞⎝ ⎜⎛ 1

1 ⎠⎟

⎞

x-1 -0.5 0.5 1

0.2

0.4

0.6

0.8

1

⎝ ⎜⎜⎜⎜⎜⎛ -1

5

2

5 ⎠⎟⎟⎟⎟⎟

⎞

⎝ ⎜⎜⎜⎜⎜⎛ 2

5

1

⎠⎟⎟⎟⎟⎟ ⎞

5

This is our given basisThis is an orthonormal basis for

2\

Fig 8




We can generally ignore the fraction (scalars) because vectors are orthogonalindependent of scalars as was established in question 13 of Exercise 5(b) which says

that the vectors { }1 2 3, , , , nv v v v" are orthogonal if and only if

{ }1 1 2 2 3 3, , , , n nk k k k v v v v" are orthogonal.

For example in the above case we have 21125

−⎛ ⎞= ⎜⎝ ⎠

w ⎟ . Ignore the scalar (fraction) 15

and

normalize the vector ;1

2

−⎛ ⎞⎜ ⎟⎝ ⎠

( )2

2 21 1 1 1

1 2 5 and2 2 2 2

− − − −⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⋅ = − + = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠5

Normalizing this vector gives 2

11

25

−⎛ = ⎜

⎝ ⎠u

⎞⎟ which is the same unit vector as in

Example 13. It makes the arithmetic a lot easier if we ignore fractions or any other scalars.

2u

The orthonormal basis is not unique so if you do not remove the scalar (fraction) you

will have a different orthonormal basis. Also if we apply the Gram Schmidt Process in

reverse order, that is { }2 1,v v then the result will be a different orthonormal basis. In

the next set of exercise (Exercise 5c) question 2 we have switched vectors and of

Example 13. See if you obtain the same or different set of vectors for the orthonormal

basis.

1v 2v

Example 14

Transform the basis in to an orthonormal basis1 2 3

1 1

1 , 1 and 0

1 0

⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

v v v

2

0

3\

for with respect to the dot product.3\Solution.Using the Gram Schmidt Process (5.16) above means that we need to find vectors

1w , and which are orthogonal. The first vector is straightforward because we

2w

3w

only need to write down the given vector . How do we find the vector 1 1

1

1

1

⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟⎝ ⎠

w v

2w ?

By (5.16) we have

2 1

2 2 2

1

,= −

v ww v w

w1 ( )†

What is2

1w equal to?

By the definition of the norm we have

2 2 2 2

1 1 1

1 1

, 1 1 1 1 1 3

1 1

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= = ⋅ = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

w w w =




What is1

w equal to?

Taking the square root of the above gives 1 3=w .

What is2 1,v w equal to?

( )2 2

2 1 2 1

1 1, 1 1 1 1 0

0 1

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= ⋅ = ⋅ = + + × =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v w v w 1 2

Substituting these,2

13=w and 2 1

, 2=v w , into ( )† gives2 2

2

3= −w v w

1.

Substituting vectors into this yields:

2

1 1 1/ 32 1

1 1 1/ 33 3

0 1 2 / 3

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

= − = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

1

1

2− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

w

Clear the fraction to write the vector 2

1

' 1

2

⎛ ⎞⎜

= ⎜⎜ ⎟

⎟⎟

−⎝ ⎠

w . (We have nominated this new vector

2'w ).

What else do we need to find?

The vector . How?3w

By (5.16) we have

3 1 3 2

3 3 12 2

1 2

, , ''

'= − −

v w v ww v w w

w w2 ( )††

Note that we have written rather than because we are ignoring the fraction. We2'w 2w

need to evaluate each of these components:

( ) ( ) ( )3 1 3 1

2 1

, 0 1 2 1 0 1 0 1

0 1

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= ⋅ = ⋅ = × + × + × =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v w v w 2

( ) ( ) ( )( )3 2 3 2

2 1

, ' ' 0 1 2 1 0 1 0 2 2

0 2

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= ⋅ = ⋅ = × + × + × − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

v w v w

What else do we need to work out?

The norm squared2

2'w which is

( )2 22 2

2 2 2

1 1

' ' ' 1 1 1 1 2

2 2

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= ⋅ = ⋅ = + + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

w w w 6

Taking the square root gives 2' 6=w .

Substituting these,3 1, 2=v w ,

2

1 3=w ,3 2, ' 2=v w and

2

2' =w 6 into ( ) ††

gives




3 1 3 2

3 3 1 22 2

1 2

3 1 2

3 1 2

, , ''

'

2 2'

3 6

2 1 2 1' Because

3 3 6 3

= − −

= − −

⎡ ⎤= − − =⎢ ⎥⎣ ⎦

v w v ww v w w

w w

v w w

v w w

Substituting the vectors into this yields:

3 3 1 2

3 1 2

2 1'

3 3

2 1 1 2 1 12 1

0 1 1 Substituting 0 , 1 , ' 13 3

0 1 2 0 1 2

2 2 / 3 1/ 3 2 2 / 30 2 / 3 1/ 3

0 2 / 3 2 / 3

= − −

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − = = =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

w v w w

v w w

1/ 3 10 2 / 3 1 / 3 1

0 2 / 3 2 / 3 0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟− − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟− +⎝ ⎠ ⎝ ⎠

Our orthogonal vectors are ,1

1

1

1

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

w 2

1

' 1

2

⎛ ⎞⎜

= ⎜⎜ ⎟

⎟⎟

−⎝ ⎠

w and3

1

1

0

⎛ ⎞⎜ ⎟

= −⎜⎜ ⎟⎝ ⎠

w ⎟ . To convert these

into an orthonormal basis what do we need to do?

Normalize these vectors. How?

By using Proposition (5.11) which is j

j

w

w(the vector divided by its norm):

We have 2 2 2

1 1 1 1 3= + + =w therefore1

11

13

1

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟⎝ ⎠

u .

Similarly we have 2' =w 6 and the normalized is2'w

2

11

16

2

⎛ ⎞⎜ ⎟

= ⎜ ⎟⎜ ⎟−⎝ ⎠

u .

Lastly we have

( )22 2

3

1 1

1 1 1 1 0

0 0

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= − ⋅ − = + − + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

w 2

Thus 3

11

12

0

⎛ ⎞⎜

= −⎜⎜ ⎟⎝ ⎠

u⎟⎟ . Our orthonormal basis is

1

11

1

3 1

⎛ ⎞⎜ ⎟

= ⎜ ⎟

⎜ ⎟⎝ ⎠

u , 2

11

1

6 2

⎛ ⎞⎜

= ⎜

⎜ ⎟

⎟⎟

−⎝ ⎠

u and 3

11

1

2 0

⎛ ⎞⎜ ⎟

= −⎜ ⎟

⎜ ⎟⎝ ⎠

u




Fig 9 Showing the orthonormal basis of Example 14.

We can also apply the Gram Schmidt Process for subspaces as the next example shows.Example 15

Transform the vectors 1 2 3

1 2

2 1, and

3 1

4 0

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

3

0

1

3

−⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

v v v which spans a subspace of

4\ to an orthonormal basis for this subspace with respect to the dot product.Solution.

The first vector is given . Next we find a vector which is orthogonal to1 1

1

23

4

⎛ ⎞⎜ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟⎝ ⎠

w v

this by using (5.16) which is2 1

2 2 2

1

,= −

v ww v w

w1 . We need to find each of these

components. What is the inner product 2 1,v w equal to?

( ) ( ) ( ) ( )2 1

2 1

1 2, 2 1 1 2 1 3

1 30 4

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= ⋅ = × + × + × + × =

⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v w 0 4 7

We also need to find2

1w :

2 2 2 2 2

1 1 1

1 1

2 21 2 3 4 30

3 3

4 4

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= ⋅ = ⋅ = + + + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

w w w

Substituting these, 2 1, 7=v w and

2

1 30=w , into the above formula gives

3

11

12

0

⎛ ⎞⎜ ⎟= −⎜ ⎟⎜ ⎟⎝ ⎠

u

2

11

16

2

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟−⎝ ⎠

u

1

11

13

1

⎛ ⎞⎜ ⎟= ⎜ ⎟⎜ ⎟

u




2 1

2 2 1 22

1

, 7

30= − = −

v ww v w v w

w1

Substituting the vectors into this yields:

2 2 1 2 1

2 1 2 1

1 2 17 7Substituting and

1 3 130 30

0 4 0 4

2 7 / 30 53 / 30 53

1 14 / 30 16 / 30 161

1 21/ 30 9 / 30 30

0 28 / 30 28 / 30

2

3

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − = − = =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

−⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

−⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟

− −⎝ ⎠ ⎝ ⎠

w v w v w

9

28

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟

−⎝ ⎠

Again ignore the scalar (fraction) to make the computation easier so that .2

53

16'

9

28

⎛ ⎞

⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟

−⎝ ⎠

w

What else do we need to find?

The vector by applying (5.16) which is3

w

3 1 3 2

3 3 12 2

1 2

, , ''

'= − −

v w v ww v w w

w w2 (*)

Evaluating each of the components gives:

( ) ( ) ( ) ( )3 1 3 1

3 1

0 2, 3 1 0 2 1 3

1 3

3 4

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ 3 4 12= ⋅ = ⋅ = × + × + − × + × =⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

v w v w

( ) ( ) ( ) ( )( )3 2 3 2

3 53

0 16, ' ' 3 53 0 16 1 9 3 28 66

1 9

3 28

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= ⋅ = ⋅ = × + × + − × + × − =⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠

v w v w

( )2 22 2 2

2 2 2

53 53

16 16' ' ' 53 16 9 28 3930

9 9

28 28

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟= ⋅ = ⋅ = + + + − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

− −⎝ ⎠ ⎝ ⎠

w w w

Substituting all these,3 1, 12=v w ,

2

130=w ,

3 2, ' 66=v w , and

2

2' 393=w 0

into (*) yields:

3 1 3 2

3 3 1 22 2

1 2

3 1 2 3 1 2

, , ''

'

12 66 2 11 12 2 66 11

' ' Because ,30 3930 5 655 30 5 3930 655

= − −

⎡ ⎤

= − − = − − = =⎢ ⎥⎣ ⎦

v w v ww v w w

w w

v w w v w w




Substituting the vectors 1 2 3

1 53

2 16, ' and

3 9

4 28

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟

3

0

1

3

−⎜ ⎟ ⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

w w v into this

3 3 1 2

3 1 53

0 2 162 11 2 11'

1 3 95 655 5 655

3 4 28

3 2 / 5 583 / 655

0 4 / 5 176 / 655 Carrying Out

1 6 / 5 99 / 655 Scalar Multiplication

3 8 / 5 308 / 655

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − − = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟= − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟−⎜ ⎟ ⎜ ⎟ ⎜ ⎟

−⎝ ⎠ ⎝ ⎠ ⎝ ⎠

w v w w

3 2 / 5 583 / 655 224 /131 224

0 4 / 5 176 / 655 140 /131 1401

1 6 / 5 99 / 655 308 /131 308131

3 8 / 5 308 / 655 245 /131 245

⎡ ⎤⎢ ⎥⎣ ⎦

− −⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

− − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

− +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Ignore the fraction and let . We need to normalize each of the 3 vectors3

224

140'

308

245

⎛ ⎞⎜ ⎟

−⎜=⎜ −⎜ ⎟

⎝ ⎠

w ⎟⎟

1

1

2,

3

4

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟⎝ ⎠

w2

53

16'

9

28

⎛ ⎞⎜ ⎟⎜=⎜⎜ ⎟

−⎝ ⎠

w ⎟⎟

⎟⎟

and . How? 3

224

140'

308

245

⎛ ⎞⎜ ⎟

−⎜=⎜ −⎜ ⎟⎝ ⎠

w

By dividing each vector by its norm. We have 1 30=w :

1 1

1

1

21 1

330

4

⎛ ⎞⎜ ⎟⎜ ⎟= =⎜ ⎟

⎜ ⎟⎝ ⎠

u ww

Similarly we have 2' 393=w 0 :

2 2

2

53

161 1'

9' 3930

28

⎛ ⎞⎜ ⎟⎜ ⎟= =⎜ ⎟⎜ ⎟

−⎝ ⎠

u ww




The norm of the last vector is

( ) ( ) ( )2 2 22

3

224 224

140 140' 224 140 308 245 224665

308 308

245 245

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

− −⎜ ⎟ ⎜ ⎟= ⋅ = + − + − + =⎜ ⎟ ⎜ ⎟− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

w

Therefore 3 3

3

224

1401 1'

308' 224665

245

⎛ ⎞⎜ ⎟

−⎜= =⎜ −⎜ ⎟⎝ ⎠

u ww

⎟⎟

. Thus the orthonormal basis for the given

subspace is 1

1

21

330

4

⎛ ⎞⎜ ⎟⎜ ⎟=⎜ ⎟⎜ ⎟

⎝ ⎠

u , 2

53

161

93930

28

⎛ ⎞⎜ ⎟⎜=⎜⎜ ⎟

−⎝ ⎠

u ⎟⎟

and 3

224

1401

308224665

245

⎛ ⎞⎜ ⎟

−⎜ ⎟=⎜ ⎟−⎜ ⎟

⎝ ⎠

u .

SUMMARY

Definition (5.13).Let V be a finite dimensional vector space with an inner product. A set of basis vectors

{ }1 2 3, , , ,

n B = u u u u"

for V is called an orthonormal basis if they are

(i) Orthogonal, that is , 0i j =u u for i j≠

(ii) Normalized, that is 1 j =u for 1, 2, 3, , j n= "

Proposition (5.14). If { }1 2 3, ,, , nvv v v " is an orthogonal set of non-zero vectors in

an inner product space then they are linearly independent.

Corollary (5.15). In an n dimensional inner product space V any set of n orthogonalvectors forms a basis for V .

Gram Schmidt Process (5.16).

Let 1 1=w v

2 1

2 2 12

1

,= −

v ww v w

w

3 1 3 2

3 3 1 22 2

1 2

4 1 4 2 4 3

4 4 1 2 32 2 2

1 2 3

, ,

, , ,

= − −

= − − −

v w v w

w v w ww w

v w v w v ww v w w w

w w w

# # # # #

1 2 3 1

1 2 3 12 2 2 2

1 2 3 1

, , , ,n n n n n

n n n

n

−−

−

= − − − − −v w v w v w v w

w v w w w ww w w w

"

orthonormal basis

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