Solutions to Review Problems for Chapter 6 (6.1-6.6) - College of Arts ...

Solutions to Review Problems for Chapter 6 (6.1-6.6)

The Final Exam is on Wednesday, December 13, 2006, 1:00 ? 3:00 PM at Disque 103

1

-1

1. Let u = 5 , v =

0

.

Describe

the

set

of

all

vectors

in

R3

that

are

orthogonal

to

both

0

1

u and v.

x1

Let

x=

x2

be

a

vector

orthogonal

to

both

u

and

v.

Then

x?u=0

and

x ? v = 0.

Thus,

x3

x1 + 5x2 -x1 + x3

= =

0 0

1

150 -1 0 1

150 051

1 0 -1

01

1 5

x1 = x3

x2

=

-

1 5

x3

x3 = x3 free.

Thus,

x

=

x3

-

1 5

,

where

x3

is any real number.

1

1

The vectors in R3

orthogonal

to both u and v

are the

scalar multiples of

-

1 5

.

1

2. Let S be the two-dimensional subspace of R3 spanned by

1

-1

u = 5 , v = 0 .

0

1

Find a basis for S. Give a geometric description of S and S.

1

This is just question (1).

We

have that S =Span

-

1 5

.

1

1

A basis for S

is

-

1 5

.

1

S is the plane in R3 spanned by the vectors u and v, and S is the line through the origin and

1

the vector

-

1 5

.

1

3. Let y =

2 3

, u=

4 -7

. Let

L =Span{u}.

(a) Find the orthogonal projection of y onto L. (b) Write y as a sum of a vector in L and a vector orthogonal to L. (c) What is the distance from y to L?

1

(a) y^ = projLy =

y?u u?u

u

=

-13 65

4 -7

=

-

1 5

4 -7

.

(b) y = y^ + z, y^ =

-

4 5

7

, z=

2 3

-

-

4 5

7

=

14 5

8

.

5

5

5

(c)

y^ - y = z =

(

14 5

)2

+

(

8 5

)2

=

260 5

.

0

2

-1

4. Let x = 6 , u = -1 , v = 2 , and W = Span{u, v}. Note that u ? v = 0.

4

1

4

Find a vector a in W and a vector b that is orthogonal to W , such that x = a + b.

a = projW x =

x?u u?u

u+

x?v v?v

v,

b = x - a.

-2 2

28 -1

-1 2

4 -1

-2

a = -1 + 2 = -1 + 2 = 3 ,

6

1

21

4

3

1

3

4

5

2

b = x-a = 3 . -1

5. Given a vector x =

2 4

, and a line L =

x1 x2

: 4x1 + 3x2 = 0 in R2:

(a) Find the vector in L that is closest to x; that is, find the orthogonal projection of x onto the line L.

(b) Write x as a vector sum x = p + z, where p is in L and z is in L.

We have that L = Span{u} where u =

-3 4

.

(a) x^ = projLx =

x?u u?u

u

=

10 25

-3 4

=

2 5

-3 4

=

-

6 5

8

.

5

(b) x = x^ + z, x^ =

-

6 5

8

, z=

2 4

-

-

6 5

8

=

16 5

12

.

5

5

5

1

6.

Let

A

=

1

1

1

3

5

2

,

b

=

2

.

1

-2

2

3

(a) Write the normal equations for the least-squares solution of Ax = b. (b) Find the least-squares solution of Ax = b, using the normal equations. (c) Find the closest point to b in the column space of A.

2

(a) The normal equations are ATAx = ATb, where

1 3

AT A =

1111

1

2

=

4

8,

3 2 1 2 1 1 8 18

12

and 5

AT b =

1111 3212

2 -2

=

8 23

.

3

(b) Row reduction:

488 8 18 23

488 027

12 2 0 1 7/2

1 0

0 1

-5 7/2

.

The least-squares solution is x^ =

-5 7/2

.

(c) The least-squares solution x^ has the property that Ax^ is the closest point in Col A to b. So

1 3

11/2

Ax^

=

1 1

2

1

-5 7/2

=

2 -3/2

.

12

2

7. Describe all least-squares solutions of Ax = b:

1 1 0

1

A

=

1

1

0

,

b=

3

.

1

0

1

8

101

2

The normal equations are ATAx = ATb, where

1 AT A = 1

0

1 1 0

1 0 1

1 0 1

1 1 1 1

1 1 0 0

0 0 1 1

=

4 2 2

2 2 0

2 0 ,

2

and

1

1111

14

AT b = 1

1

0

0

3

=

4

.

8

0 0 1 1 10

2

4 2 2 14

1

Row reduction: 2 2 0 4 ? ? ? 0

2 0 2 10

0

015 1 -1 -3 .

000

The least-squares solutions are vectors of the form

5

-1

x^

=

-3

+

x3

1 ,

0

1

where x3 is any real number.

3

 1

8. Let A = 2 2

3 -1 2 -7 . Find an orthonormal basis for the column space of A.

13

1

3

-1

A

=

[x1

x2

x3],

where

x1

=

2

,

x2

=

2

,

x3

=

-7

.

2

1

3

Let's use the Gram-Schmidt process to orthogonalize {x1, x2, x3} : 1

v1

=

x1

=

2

2

3

1

2

v2

=

x2 -

x2?v1 v1?v1

v1

=

2

-

9 9

2

=

0

1

2

-1

-1

1

2

2

v3

=

x3 -

x3?v1 v1?v1

v1 -

x3?v2 v2?v2

v2

=

-7

-

-9 9

2

-

-5 5

0 = -5 .

3

2

-1

4

An orthonormal basis for the column space of A is

v1 , v2 , v3 v1 v2 v3

1

3

2

5

2

45

=

2 3

,

0

,

- 5

45

.

2

3

- 1

5

4

45

1

9. Let A =

1 0

-1

1 0

2

0 1

-1 1

,

b=

5 6

.

1 -1

6

(a) Find the orthogonal projection of b onto Col A. (b) Use (a) to find the least-squares solution of Ax = b.

(a) Let A = [u1 u2 u3]. Then

u1 ? u2 = 0, u1 ? u3 = 0 and u2 ? u3 = 0,

so {u1, u2, u3} is an orthogonal basis for Col A. Thus,

b^ = projCol Ab =

b?u1 u1?u1

u1 +

b?u2 u2?u2

u2 +

b?u3 u3?u3

u3

1

1

0

=

1

3

1 0

+

14

3

0 1

+

-5

3

-1 1

-1

1

-1

5

=

2 3

.

6

4

(b) From (a), we have

b^

=

1 3 u1

+

14 3 u2

-

5 3 u3.

The least-squares solution x^ is the solution to Ax^ = b^, that is, the entries of x^ are the weights when the projection b^ is written as a combination of the columns of A. Thus,

1

3

x^

=

14 3

.

-

5 3

10. Find the equation y = c0 +c1x of the least-squares line for the data: (1, 3), (2, 2), (4, 1), and (5, 1).

We need to find c0 and c1 that best approximates

1 1

3

1 2

c0

=

2

.

1

4

c1

1

15

1

A

b

The normal equations are: AT Ax^ = AT b, where

1 1

AT A =

1111 1245

1 1

2 4

=

4 12 12 46

,

15

and 3

AT b =

1111

2

=

7

.

1 2 4 5 1 16

1

Row reduction:

4 12 7 12 46 16

4 12 7 0 10 -5

13

7 4

0

1

-

1 2

10

13

4.

0

1

-

1 2

The

least-squares

line

that

best

approximates

the

data

points

is

y

=

13 4

-

1 2

x.

5

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