Announcements

Ï Quiz 4 (last quiz of the term) tomorrow on sec 3.3, 5.1 and5.2.

Ï Three problems on tomorrow's quiz(Cramer's rule/adjugate,�nding eigenvector(s) given one or more eigenvalue(s), �ndingchar polynomial/eigenvalues of a 2×2 or a nice 3×3 matrix.)

Ï You must show all relevant work on the quiz. Calculatoranswers are not acceptable.

Chapter 6 Orthogonality

Objectives

1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn

2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)

3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.

Chapter 6 Orthogonality

Objectives

1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn

2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)

3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.

Chapter 6 Orthogonality

Objectives

1. Extend the idea of simple geometric ideas namely �length�,�distance� and �perpendicularity� from R2 and R3 into Rn

2. Useful in "�tting" experimental data of a system Ax= b. If x1is an acceptable solution, we want the distance between b andAx1 to be minimum (or minimize the error)

3. The solution above is called the "least-squares" solution and iswidely used where experimental data is scattered over a widerange and you want to "�t" a straight line.

Inner product

Let u and v be two vectors in Rn.

u=

u1u2...un

,v=

v1v2...vn

Both u and v are n×1 matrices.

uT = [u1 u2 . . . un

]This is an 1×n matrix. Thus we can de�ne the product uTv as

uTv= [u1 u2 . . . un

]

v1v2...vn

Inner product

Let u and v be two vectors in Rn.

u=

u1u2...un

,v=

v1v2...vn

Both u and v are n×1 matrices.

uT = [u1 u2 . . . un

]

This is an 1×n matrix. Thus we can de�ne the product uTv as

uTv= [u1 u2 . . . un

]

v1v2...vn

Inner product

Let u and v be two vectors in Rn.

u=

u1u2...un

,v=

v1v2...vn

Both u and v are n×1 matrices.

uT = [u1 u2 . . . un

]This is an 1×n matrix. Thus we can de�ne the product uTv as

uTv= [u1 u2 . . . un

]

v1v2...vn

1×n n×1

Match

Size of uTv

The product will be a 1×1 matrix or it is just a number (not avector) and is given by

u1v1+u2v2+ . . .+unvn

Nothing but sum of the respective components multiplied.

1×n n×1

Match

Size of uTv

The product will be a 1×1 matrix or it is just a number (not avector) and is given by

u1v1+u2v2+ . . .+unvn

Nothing but sum of the respective components multiplied.

1×n n×1

Match

Size of uTv

The product will be a 1×1 matrix or it is just a number (not avector) and is given by

u1v1+u2v2+ . . .+unvn

Nothing but sum of the respective components multiplied.

Inner Product

1. The number uTv is called the inner product of u and v.

2. Inner product of 2 vectors is a number.

3. Inner product is also called dot product (in Calculus II)

4. Often written as u �v

Example

Let

w= 4

12

,x= 5

0−3

Find w �x, w �w and w�x

w�w

w �x=wTx= [4 1 2

] 50−3

= (4)(5)+ (1)(0)+ (2)(−3)= 14

w �w=wTw= [4 1 2

] 412

= (4)(4)+ (1)(1)+ (2)(2)= 21

w �xw �w

= 14

21= 2

3.

Example

Let

w= 4

12

,x= 5

0−3

Find w �x, w �w and w�x

w�w

w �x=wTx= [4 1 2

] 50−3

= (4)(5)+ (1)(0)+ (2)(−3)= 14

w �w=wTw= [4 1 2

] 412

= (4)(4)+ (1)(1)+ (2)(2)= 21

w �xw �w

= 14

21= 2

3.

Example

Let

w= 4

12

,x= 5

0−3

Find w �x, w �w and w�x

w�w

w �x=wTx= [4 1 2

] 50−3

= (4)(5)+ (1)(0)+ (2)(−3)= 14

w �w=wTw= [4 1 2

] 412

= (4)(4)+ (1)(1)+ (2)(2)= 21

w �xw �w

= 14

21= 2

3.

Example

Let

w= 4

12

,x= 5

0−3

Find w �x, w �w and w�x

w�w

w �x=wTx= [4 1 2

] 50−3

= (4)(5)+ (1)(0)+ (2)(−3)= 14

w �w=wTw= [4 1 2

] 412

= (4)(4)+ (1)(1)+ (2)(2)= 21

w �xw �w

= 14

21= 2

3.

Properties of Inner Product

1. u �v= v �u

2. (u+v) �w= u �w+v �w

3. (cu) �v= u � (cv)

4. u �u≥ 0, and u �u= 0 if and only if u=0

Length of a Vector

Consider any point in R2, v=[a

b

]. What is the length of the line

segment from (0,0) to v?y

x0

|a|

|b|

(a,b)

pa2+b2

Length of a Vector

Consider any point in R2, v=[a

b

]. What is the length of the line

segment from (0,0) to v?y

x0 |a|

|b|

(a,b)

pa2+b2

Length of a Vector

Consider any point in R2, v=[a

b

]. What is the length of the line

segment from (0,0) to v?y

x0 |a|

|b|

(a,b)

pa2+b2

Length of a Vector

Consider any point in R2, v=[a

b

]. What is the length of the line

segment from (0,0) to v?y

x0 |a|

|b|

(a,b)

pa2+b2

Length of a Vector

Consider any point in R2, v=[a

b

]. What is the length of the line

segment from (0,0) to v?y

x0 |a|

|b|

(a,b)

pa2+b2

Length of a Vector

We can extend this idea to Rn, where v=

v1v2...vn

.De�nition

The length (or the norm) of v is the nonnegative scalar ‖v‖ de�nedby

‖v‖ =pv �v=

√v21+v2

2+ . . .+v2n

Since we have sum of squares of the components, the square root isalways de�ned.

Length of a Vector

If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.

De�nition

A vector of length 1 is called a unit vector.

If we divide a vector v by its length ‖v‖ (or multiply by 1

‖v‖), we getthe unit vector u in the direction of v.

The process of getting u from v is called normalizing v.

Length of a Vector

If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.

De�nition

A vector of length 1 is called a unit vector.

If we divide a vector v by its length ‖v‖ (or multiply by 1

‖v‖), we getthe unit vector u in the direction of v.

The process of getting u from v is called normalizing v.

Length of a Vector

If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.

De�nition

A vector of length 1 is called a unit vector.

If we divide a vector v by its length ‖v‖ (or multiply by 1

‖v‖), we getthe unit vector u in the direction of v.

The process of getting u from v is called normalizing v.

Length of a Vector

If c is a scalar, the length of cv is c times the length of v. If c > 1,the vector is stretched by c units and if c < 1, c shrinks by c units.

De�nition

A vector of length 1 is called a unit vector.

If we divide a vector v by its length ‖v‖ (or multiply by 1

‖v‖), we getthe unit vector u in the direction of v.

The process of getting u from v is called normalizing v.

Example 10, sec 6.1

Find a unit vector in the direction of v= −6

4−3

.

To compute the length of v, �rst �nd

v �v= (−6)2+42+ (−3)2 = 36+16+9= 61

Then,‖v‖ =

p61

The unit vector in the direction of v is

u= 1

‖v‖v=1p61

−64−3

=

−6/p61

4/p61

−3/p61

Example 10, sec 6.1

Find a unit vector in the direction of v= −6

4−3

.

To compute the length of v, �rst �nd

v �v= (−6)2+42+ (−3)2 = 36+16+9= 61

Then,‖v‖ =

p61

The unit vector in the direction of v is

u= 1

‖v‖v=1p61

−64−3

=

−6/p61

4/p61

−3/p61

Example 10, sec 6.1

Find a unit vector in the direction of v= −6

4−3

.

To compute the length of v, �rst �nd

v �v= (−6)2+42+ (−3)2 = 36+16+9= 61

Then,‖v‖ =

p61

The unit vector in the direction of v is

u= 1

‖v‖v=1p61

−64−3

=

−6/p61

4/p61

−3/p61

Distance in Rn

In R (the set of real numbers), the distance between 2 numbers iseasy.

The distance between 4 and 15 is |4−14| = |−10| = 10 or|14−4| = |10| = 10.

Similarly the distance between -5 and 5 is |−5−5| = |−10| = 10 or|5− (−5)| = |10| = 10

Distance has a direct analogue in Rn.

Distance in Rn

In R (the set of real numbers), the distance between 2 numbers iseasy.

The distance between 4 and 15 is |4−14| = |−10| = 10 or|14−4| = |10| = 10.Similarly the distance between -5 and 5 is |−5−5| = |−10| = 10 or|5− (−5)| = |10| = 10

Distance has a direct analogue in Rn.

Distance in Rn

De�nition

For any two vectors u and v in Rn, the distance between u and vwritten as dist(u,v) is the length of the vector u-v.

dist(u,v)= ‖u-v‖

Example 14, sec 6.1

Find the distance between u= 0

−52

and v= −4

−18

.

To compute the distance between u and v, �rst �nd

u−v= 0

−52

− −4

−18

= 4

−4−6

Then,

‖u-v‖ =p16+16+36=

p68

Orthogonal Vectors

0

-v

v

u‖u-v‖

‖u-(-v)‖

If the 2 green lines are perpendicular, u must have the samedistance from v and -v

Orthogonal Vectors

0

-v

v

u

‖u-v‖

‖u-(-v)‖

If the 2 green lines are perpendicular, u must have the samedistance from v and -v

Orthogonal Vectors

0

-v

v

u‖u-v‖

‖u-(-v)‖

If the 2 green lines are perpendicular, u must have the samedistance from v and -v

Orthogonal Vectors

‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares

‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)

= u � (u+v)+v � (u+v)

= u �u+u �v+v �u+v �v

= ‖u‖2+‖v‖2+2u �v

Interchange -v and v and we get

‖u-v‖2 = ‖u‖2+‖v‖2−2u �v

Orthogonal Vectors

‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares

‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)

= u � (u+v)+v � (u+v)

= u �u+u �v+v �u+v �v

= ‖u‖2+‖v‖2+2u �v

Interchange -v and v and we get

‖u-v‖2 = ‖u‖2+‖v‖2−2u �v

Orthogonal Vectors

‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares

‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)

= u � (u+v)+v � (u+v)

= u �u+u �v+v �u+v �v

= ‖u‖2+‖v‖2+2u �v

Interchange -v and v and we get

‖u-v‖2 = ‖u‖2+‖v‖2−2u �v

Orthogonal Vectors

‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares

‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)

= u � (u+v)+v � (u+v)

= u �u+u �v+v �u+v �v

= ‖u‖2+‖v‖2+2u �v

Interchange -v and v and we get

‖u-v‖2 = ‖u‖2+‖v‖2−2u �v

Orthogonal Vectors

‖u-(-v)‖ = ‖u-v‖To avoid square roots, let us work with the squares

‖u-(-v)‖2 = ‖u+v‖2 = (u+v) � (u+v)

= u � (u+v)+v � (u+v)

= u �u+u �v+v �u+v �v

= ‖u‖2+‖v‖2+2u �v

Interchange -v and v and we get

‖u-v‖2 = ‖u‖2+‖v‖2−2u �v

Orthogonal Vectors

Equate the 2 expressions,

‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v

=⇒ 2u �v=−2u �v

=⇒ u �v= 0

If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if

u �v= 0

Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.

Orthogonal Vectors

Equate the 2 expressions,

‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v

=⇒ 2u �v=−2u �v

=⇒ u �v= 0

If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if

u �v= 0

Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.

Orthogonal Vectors

Equate the 2 expressions,

‖u‖2+‖v‖2+2u �v= ‖u‖2+‖v‖2−2u �v

=⇒ 2u �v=−2u �v

=⇒ u �v= 0

If u and v are points in R2, the lines through these points and (0,0)are perpendicular if and only if

u �v= 0

Generalize this idea of perpendicularity to Rn. We use the wordorthogonality in linear algebra for perpendicularity.

Orthogonal Vectors

De�nition

Two vectors u and v in Rn are orthogonal (to each other) if

u �v= 0

The zero vector 0 is orthogonal to every vector in Rn.

Theorem

Two vectors u and v are orthogonal if and only if

‖u+v‖2 = ‖u‖2+‖v‖2

This is called the Pythagorean theorem.

Orthogonal Vectors

De�nition

Two vectors u and v in Rn are orthogonal (to each other) if

u �v= 0

The zero vector 0 is orthogonal to every vector in Rn.

Theorem

Two vectors u and v are orthogonal if and only if

‖u+v‖2 = ‖u‖2+‖v‖2

This is called the Pythagorean theorem.

Example 16, 18 section 6.1

Decide which pair(s) of vectors are orthogonal

16)u= 12

3−5

,v= 2

−33

u �v= (12)(2)+ (3)(−3)+ (−5)(3)= 24−9−15= 0.

Thus u and v are orthogonal.

18)y=

−3740

,z=

1−815−7

y �z= (−3)(1)+(7)(−8)+(4)(15)+(0)(−7)=−3−56+60−0= 1 6= 0.

Thus y and z are not orthogonal.

Example 16, 18 section 6.1

Decide which pair(s) of vectors are orthogonal

16)u= 12

3−5

,v= 2

−33

u �v= (12)(2)+ (3)(−3)+ (−5)(3)= 24−9−15= 0.

Thus u and v are orthogonal.

18)y=

−3740

,z=

1−815−7

y �z= (−3)(1)+(7)(−8)+(4)(15)+(0)(−7)=−3−56+60−0= 1 6= 0.

Thus y and z are not orthogonal.

Orthogonal Complement

Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .

There could be more than one such vector z which is orthogonal toW .

De�nition

A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .

The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".

Orthogonal Complement

Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .

There could be more than one such vector z which is orthogonal toW .

De�nition

A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .

The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".

Orthogonal Complement

Let W be a subspace of Rn. If any vector z is orthogonal to everyvector in W , we say that z is orthogonal to W .

There could be more than one such vector z which is orthogonal toW .

De�nition

A collection of all vectors that are orthogonal to W is called theorthogonal complement of W .

The orthogonal complement of W is denoted by W⊥ and is read as"W perpendicular" or "W perp".

Orthogonal Complement

1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .

2. W⊥ is a subspace of Rn.

3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)

4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)

Orthogonal Complement

1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .

2. W⊥ is a subspace of Rn.

3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)

4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)

Orthogonal Complement

1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .

2. W⊥ is a subspace of Rn.

3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)

4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)

Orthogonal Complement

1. A vector x is in W⊥ if and only if x is orthogonal to everyvector that spans (generates) W .

2. W⊥ is a subspace of Rn.

3. If A is an m×n matrix, the orthogonal complement of Col A isNul AT . (Useful in part (d) of T/F questions, prob 19)

4. If a vector is in both W and W⊥, then that vector must bethe zero vector. (The only vector perpendicular to itself is thezero vector)

Section 6.2 Orthogonal Sets

Consider a set of vectors{u1,u2, . . . ,up

}in Rn. If each pair of

distinct vectors from the set is orthogonal (that is u1 �u2 = 0,u1 �u3 = 0, u2 �u3 = 0 etc etc) then the set is called an orthogonalset.

Example 2 section 6.2

Decide whether the set

1−21

,

012

,

−5−21

is orthogonal.

1−21

�

012

= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0

012

�

−5−21

= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0

1−21

�

−5−21

= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0

Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)

Example 2 section 6.2

Decide whether the set

1−21

,

012

,

−5−21

is orthogonal.

1−21

�

012

= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0

012

�

−5−21

= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0

1−21

�

−5−21

= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0

Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)

Example 2 section 6.2

Decide whether the set

1−21

,

012

,

−5−21

is orthogonal.

1−21

�

012

= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0

012

�

−5−21

= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0

1−21

�

−5−21

= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0

Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)

Example 2 section 6.2

Decide whether the set

1−21

,

012

,

−5−21

is orthogonal.

1−21

�

012

= (1)(0)+ (−2)(1)+ (1)(2)=−2+2= 0

012

�

−5−21

= (0)(−5)+ (1)(−2)+ (2)(1)=−2+2= 0

1−21

�

−5−21

= (1)(−5)+ (−2)(−2)+ (1)(1)=−5+4+1= 0

Since all pairs are orthogonal, we have an orthogonal set. (If onepair fails, and all other pairs are orthogonal, it FAILS to be anorthogonal set)

Orthogonal set and Linear Independence

Theorem

Let S = {u1,u2, . . . ,up

}be an orthogonal set of NONZERO vectors

in Rn. S is linearly independent and is a basis for the subspace

spanned (generated) by S.

Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.

Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that

1. spans W and

2. is linearly independent

Orthogonal set and Linear Independence

Theorem

Let S = {u1,u2, . . . ,up

}be an orthogonal set of NONZERO vectors

in Rn. S is linearly independent and is a basis for the subspace

spanned (generated) by S.

Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.

Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that

1. spans W and

2. is linearly independent

Orthogonal set and Linear Independence

Theorem

Let S = {u1,u2, . . . ,up

}be an orthogonal set of NONZERO vectors

in Rn. S is linearly independent and is a basis for the subspace

spanned (generated) by S.

Make sure that the zero vector is NOT in the set. Otherwise theset is linearly dependent.

Remember the de�nition of basis? For any subspace W of Rn, a setof vectors that

1. spans W and

2. is linearly independent

Orthogonal Basis

An orthogonal basis for a subspace W of Rn is a set

1. spans W and

2. is linearly independent and

3. is orthogonal

Theorem

Let{u1,u2, . . . ,up

}be an orthogonal basis for a subspace W of Rn.

For each y in W , the weights in the linear combination

y= c1u1+c2u2+ . . .+cpup

are given by

c1 = y �u1u1 �u1

,c2 = y �u2u2 �u2

,c3 = y �u3u3 �u3

. . .

Orthogonal Basis

An orthogonal basis for a subspace W of Rn is a set

1. spans W and

2. is linearly independent and

3. is orthogonal

Theorem

Let{u1,u2, . . . ,up

}be an orthogonal basis for a subspace W of Rn.

For each y in W , the weights in the linear combination

y= c1u1+c2u2+ . . .+cpup

are given by

c1 = y �u1u1 �u1

,c2 = y �u2u2 �u2

,c3 = y �u3u3 �u3

. . .

Orthogonal Basis

If we have an orthogonal basis

1. Computing the weights in the linear combination becomesmuch easier.

2. No need for augmented matrix/ row reductions.

Example 8, section 6.2Show that {u1,u2} is an orthogonal basis and express x as a linear

combination of the u's where u1 =[31

],u2 =

[ −26

],x=

[ −63

]

Solution: You must verify whether the set is orthogonal.[31

]�[ −2

6

]= (3)(−2)+ (1)(6)= 0

. So we have an orthogonal set. By the theorem, we also have anorthogonal basis.

To �nd the weights so that we can expressx= c1u1+c2u2, we need

x �u1 =[ −6

3

]�[31

]=−18+3=−15

u1 �u1 =[31

]�[31

]= 9+1= 10

Example 8, section 6.2Show that {u1,u2} is an orthogonal basis and express x as a linear

combination of the u's where u1 =[31

],u2 =

[ −26

],x=

[ −63

]

Solution: You must verify whether the set is orthogonal.[31

]�[ −2

6

]= (3)(−2)+ (1)(6)= 0

. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2, we need

x �u1 =[ −6

3

]�[31

]=−18+3=−15

u1 �u1 =[31

]�[31

]= 9+1= 10

Example 8, section 6.2

c1 = x �u1u1 �u1

= −1510

=−1.5

x �u2 =[ −6

3

]�[ −2

6

]= 12+18= 30

u2 �u2 =[ −2

6

]�[ −2

6

]= 4+36= 40

c2 = x �u2u2 �u2

= 30

40= 0.75

Thusx=−1.5u1+0.75u2.

Example 8, section 6.2

c1 = x �u1u1 �u1

= −1510

=−1.5

x �u2 =[ −6

3

]�[ −2

6

]= 12+18= 30

u2 �u2 =[ −2

6

]�[ −2

6

]= 4+36= 40

c2 = x �u2u2 �u2

= 30

40= 0.75

Thusx=−1.5u1+0.75u2.

Example 8, section 6.2

c1 = x �u1u1 �u1

= −1510

=−1.5

x �u2 =[ −6

3

]�[ −2

6

]= 12+18= 30

u2 �u2 =[ −2

6

]�[ −2

6

]= 4+36= 40

c2 = x �u2u2 �u2

= 30

40= 0.75

Thusx=−1.5u1+0.75u2.

Example 10, section 6.2

Show that {u1,u2,u3} is an orthogonal basis for R3 and express x asa linear combination of the u's where

u1 = 3

−30

,u2 = 2

2−1

,u3 = 1

14

,x= 5

−31

Solution: You must verify whether the set is orthogonal (check allpairs). 3

−30

�

114

= (3)(1)+ (−3)(1)+ (0)(4)= 0

. 114

�

22−1

= (1)(2)+ (1)(2)+ (4)(−1)= 0

Example 10, section 6.2

3−30

�

22−1

= (3)(2)+ (−3)(2)+ (0)(4)= 0

. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2+c3u3, we need

x �u1 = 5

−31

�

3−30

= 15+9= 24

u1 �u1 = 3

−30

�

3−30

= 9+9= 18

c1 = x �u1u1 �u1

= 24

18= 4

3

Example 10, section 6.2

3−30

�

22−1

= (3)(2)+ (−3)(2)+ (0)(4)= 0

. So we have an orthogonal set. By the theorem, we also have anorthogonal basis. To �nd the weights so that we can expressx= c1u1+c2u2+c3u3, we need

x �u1 = 5

−31

�

3−30

= 15+9= 24

u1 �u1 = 3

−30

�

3−30

= 9+9= 18

c1 = x �u1u1 �u1

= 24

18= 4

3

Example 10, section 6.2

x �u2 = 5

−31

�

22−1

= 10−6−1= 3

u2 �u2 = 2

2−1

�

22−1

= 4+4+1= 9

c2 = x �u2u2 �u2

= 3

9= 1

3

x �u3 = 5

−31

�

114

= 5−3+4= 6

u3 �u3 = 1

14

�

114

= 1+1+16= 18

c3 = x �u3u3 �u3

= 6

18= 1

3

Thus

x= 4

3u1+ 1

3u2+ 1

3u3.

Section 6.2 Orthonormal Sets

Consider a set of vectors{u1,u2, . . . ,up

}. If this is an orthogonal

set (pairwise dot product =0) AND if each vector is a unit vector(length 1), the set is called an orthonormal set. A basis formed byorthonormal vectors is called an orthonormal basis (linearlyindependent by the same theorem we saw earlier).

Example 20 section 6.2

Decide whether the set u= −2/3

1/32/3

,v= 1/3

2/30

is an

orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set.

−2/31/32/3

�

1/32/30

=−2

3+ 2

3+0= 0

The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9=

√9

9= 1.

Thus u has unit length.

Example 20 section 6.2

Decide whether the set u= −2/3

1/32/3

,v= 1/3

2/30

is an

orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set. −2/3

1/32/3

�

1/32/30

=−2

3+ 2

3+0= 0

The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9=

√9

9= 1.

Thus u has unit length.

Example 20 section 6.2

Decide whether the set u= −2/3

1/32/3

,v= 1/3

2/30

is an

orthonormal set. If only orthogonal, normalize the vectors toproduce an orthonormal set. −2/3

1/32/3

�

1/32/30

=−2

3+ 2

3+0= 0

The set is orthogonal. Find the length of each vector to checkwhether it is orthonormal.

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9

‖u‖ =pu �u=

√4

9+ 1

9+ 4

9=

√9

9= 1.

Thus u has unit length.

Example 20 section 6.2

‖v‖ =pv �v=

√1

9+ 4

9+0=

p5p9=

p5

3.

Since this is not of unit length we have to divide each component

of v by its length which isp5

3. This gives

1

3/p5

3

2

3/p5

3

0

=

1p52p5

0