pp. 247-248
1. For each of the following matrices, determine a basis for each of the subspaces
R(AT), N(A), R(A), and N(AT)
3 4
(a) A  

6 8 
Change A to row-echelon form
3 4 R2  2R1 3 4 1 4 / 3
6 8  0 0  0 0 



 

 1 
Basis of R(AT) is the basis of row space of A which is 
.
 4 / 3
N(A) is the null space of A which is the solution of
3 4  x1  0
Ax  
  x   0 .
6
8

 2   
3 4 0 1 4 / 3 0



6 8 0 0 0 0
 4 / 3
4
x1   x2  x2 

3
 1 
 4 / 3
Hence, the basis of N(A) is 

 1 
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
3 6 R2  4 R1 / 33 6
1 2
 

4 8




0 0
0 0
1 
Basis of R(A) is the basis of row space of AT which is   .
2
T
N(A ) is the null space of A which is the solution of
3 6  x1  0
AT x  
     .
4 8  x2  0
3 6 0 1 2 0



4 8 0 0 0 0
  2
x1  2 x2  x2  
1
 2
Hence, the basis of N(AT) is   .
1 
1 3 1 
(b) A  

 2 4 0
To find the basis of R(AT) we have A to its reduced row-echelon form.
1
1 3 1
1 3
1 3 1
1 0  2
2 4 0  0  2  2  0 1 1  0 1 1 








1 
Bases of R(A ) is the basis of row space of A which are  0  .and
  2
N(A) is the null space of A which is the solution of
 x1 
1 3 1     0 
Ax  
  x2    
 2 4 0   x  0 
 3
T
0 
1 
 
1 
1 3 1 0
1 0  2 0



 2 4 0 0
0 1 1 0
 2 x3 
2
x1  2 x3 , x2   x3   x3   x3  1
 x3 
 1 
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
1 2
1 2 
1 2
1 0
3 4  0  2  0 1  0 1








1 0
0  2
0 0
0 0
1 
0 
Bases of R(A) is the basis of row space of AT which are   .and  
0 
1 
Since, the rank of A is 2 which is equal to number of column of 2, the dimension of
N(AT) is equal to number of column minus number of rank = 2-2 =0. This means that
N(AT) consists of 0 only. Hence, N(AT) does not have basis.
 4  2
1 3 

(c) A  
2 1 


3 4 
To find the basis of R(AT) we have A to its reduced row-echelon form.
3 
 4  2
1 3 
1
1
1 3 
 4  2
0  14




  0
2 1 
2 1 
0  5 
0







3 4 
3 4 
0  5 
0
3
1

0
1

0
0


0
0
0
1
0

0
1 
Bases of R(AT) is the basis of row space of A which are   .and
0 
0 
1 
 
Since, the rank of AT is 2 which is equal to number of column of 2, the dimension of
N(A) is equal to number of column minus number of rank = 2-2 =0. This means that
N(A) consists of 0 only. Hence, N(A) does not have basis.
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
2
3  1 1 / 4 1 / 2 3 / 4  1 0 5 / 14 5 / 14
 4 1 2 3  4 1
 2 3 1 4  0 7 / 2 2 11 / 2  0 1 4 / 7 11 / 7  0 1 4 / 7 11 / 7 

 
 
 

 1 
 0 
T
 .and
Bases of R(A) is the basis of row space of A which are 
5 / 14 


5 / 14 
N(AT) is the null space of AT which is the solution of
 x1 
 
 4 1 2 3   x 2  0 
AT x  
 

  2 3 1 4   x 3  0 
 
 x4 
 4 1 2 3 0 1 0 5 / 14 5 / 14 0


.
 2 3 1 4 0 0 1 4 / 7 11 / 7 0
Here, we have
5
5
x1   x3 
x4
14
14
 0 
 1 


4/7


11 / 7
4
11
x3  x 4 ,
7
7
which lead to
5
 5

 5
 5
  14 x3  14 x 4 
  14
  14
 4

 4
 11 
11
x    x3  x 4   x3     x 4    .
7
 7

 7
 7
x3


 1 
 0 




 1 
x4
 0 




x2  
 5   5 
    
  14  14  
4
11 

Hence, N ( AT )  span   ,    
 7  7
 1   0 
 
  1  
0

 


1
0
(d) A  
0

1
0 0 0
1 1 1
0 1 1

1 2 2
To find the basis of R(AT) we have A to its reduced row-echelon form.
1
0

0

1
0 0 0 1
1 1 1 0

0 1 1  0
 
1 2 2  0
0 0 0 1
1 1 1 0

0 1 1  0
 
1 2 2  0
0 0 0 1
1 1 1 0

0 1 1  0
 
0 1 1  0
0 0 0 1
1 1 1 0

0 1 1  0
 
0 0 0  0
1 
0 
T
Bases of R(A ) is the basis of row space of A which are   ,
0 
 
0 
N(A) is the null space of A which is the solution of
1 0 0 0  x1  0
0 1 1 1   x   0 
 2    
Ax  
0 0 1 1   x 3   0 

   
1 1 2 2  x 4  0
which can be written in the form of augmented matrix as
0 
1 
  .and
0 
 
0 
0 
0 
 
1 
 
1 
0 0 0
1 0 0
0 1 1

0 0 0
1

0
0

1
0 0 0 0 1
 
1 1 1 0  0

0 1 1 0  0
 
1 2 2 0 0
0 0 0 0

1 0 0 0
0 1 1 0

0 0 0 0
Here, we have
x1  0
x2  0 ,
and
x3   x 4
which lead to
 0 
0
 0 
 
x 0
x
4
 x 4 
 1


 
1
 x4 
 0 
 
 0 
Hence, N ( A)  span   
 1
 
 1 
 
To find basis of R(A), and N(AT), reduce AT to reduce row-echelon form
1
0

0

0
0 0 1 1
1 0 1 0

1 1 2  0
 
1 1 2  0
0 0 1 1
1 0 1 0

0 1 1 0
 
0 1 1 0
0 0 1
1 0 1
0 1 1

0 0 0
1 
0 
Bases of R(A) is the basis of row space of AT which are   ,
0 
 
1 
N(AT) is the null space of AT which is the solution of
1
0
AT x  
0

0
0 0 1  x1  0
1 0 1  x 2  0

1 1 2   x 3  0 
   
1 1 2   x 4  0 
0 
1 
  .and
0 
 
1 
0 
0 
 
1 
 
1 
1

0
0

0
0 0 1 0 1
 
1 0 1 0  0

1 1 2 0  0
 
1 1 2 0 0
0 0 1 0

1 0 1 0
.
0 1 1 0

0 0 0 0
Here, we have
x1   x4
x2   x4 ,
and
x3   x 4
which lead to
which lead to
 x 4 
 1
 x 
 1
4

x
 x4  
 x 4 
 1


 
1
 x4 
  1 
 
 1 
T
Hence, N ( A )  span   
 1
 
 1 
 
2. Let S be the subspace spanned by x  1  1 1T
Let A= 1  1 1 be a matrix of size 1 by 3. We observe that R(AT), row space of A,
 

is spanned by x. By Theorem 5.2.1, the N ( A)  R AT  S  .
Null space of A is determined by solving
 y1 
1  1 1 y 2   0
 y 3 
From the above equation, we have
y1  y 2  y 3
or
 y 2  y3 
1
 1




y   y 2   y 2 1  y 3  0 
 y 3 
0
 1 
 
T 
Hence, the basis of N ( A)  R A
1 
 1


 S  is 1  and  0  .
0 
 1 
4. Let S be the subspace in  4 spanned by x1  1 0  2 1T ,
x 2  0 1 3  2T . Find a basis of S 
1 0  2 1 
T
Let A= 
 be a matrix of size 2 by 4. We observe that R(A ), row
0
1
3

2


 

space of A, is spanned by x. By Theorem 5.2.1, the N ( A)  R AT  S  .
Null space of A is determined by solving
 y1 
 
1 0  2 1   y 2  0
 0 1 3  2   y   0  ,

 3
 
 
 y4 
which can be written in the augmented form as
1 0  2 1 0


0 1 3  2 0
From the above equation we can conclude that
y1  2 y3  y 4
and
y 2  3 y3  2 y 4
This leads to
 2 y3  y 4 
2
 1
 3 y  2 y 
 3
2
3
4



y
 y3
 y4  


1
0
y3


 
 
y4
0
1


 1
2
2


 3
T 


Hence, the basis of N ( A)  R A
and   .
 S is
0
1
 
 
1
0
 
pp. 258-259
1. Find the least square solution to each of the following systems.
(a)
1 1 
 3


A  2  3 and b  1
0 0 
2
We have that
 3
1
2
0


1    5 
AT b  

   
1  3 0 2 0
 
and
1 1 
1 2 0 
 5  5
A A
2  3  

.

1  3 0 0 0   5 10 


The least square solution is determined by solving
T
AT Ax  AT b ,
which can be written in the augmented matrix form as
 5  5 5 5  5 5 5 0 10 1 0 2




.
 5 10 0 0 5 5 0 5 5  0 1 1
Hence, the solution is x1  2 and x2  1
(b)
 1 1 
10 


A 2
1  and b   5 
 1  2
20
We have that
10 
 1 2 1     20 
A b
 5   

 1 1  2 20  25
 
T
and
 1 1 
 1 2 1  
   6  1 .
A A
2
1



 
 1 1  2  1  2   1 6 


The least square solution is determined by solving
T
AT Ax  AT b ,
which can be written in the augmented matrix form as
 6  1 20 
 1 6  25
 1 6  25 
 1 6  25 





 1 6  25
 6  1 20 
 0 35  130
 0 1  26 / 7
.
 1 0  19 / 7 
1 0 19 / 7 



 0 1  26 / 7
0 1  26 / 7
Hence, the solution is x1  19 / 7 and x2  26 / 7
(c)
1 1
 1 1
A
 0 1

0
1
1
1
1

1
 4
0 
b 
1 
 
 2
We have that
 4
1  1 0 1   6
0
AT b  1 1  1 0    3
1 
1 1
1 1   7
 2
and
1 1
1  1 0 1 
1 1
AT A  1 1  1 0 
 0 1
1 1 1 1 
0
1
1
3 0 1 
1 
 0 3 1 .
1
 1 1 4
1 
The least square solution is determined by solving
AT Ax  AT b ,
which can be written in the augmented matrix form as
3 0 1 6 
1 1 4 7
1 1
1
4 7 







1 3   0
 0 3 1 3    0 3 1 3   0 3
1 1 4 7
3 0 1 6
0  3  11  15
0
1 1 4 7 
1 1 0 11 / 5
1





 0 3 1 3   0 3 0 9 / 5   0
0 0 1 6 / 5
0 0 1 6 / 5 
0
.
Hence, the solution is x1  18 / 5 , x2  3 / 5 and x3  6 / 5
7 

3 
0  10  12
1
3
4
1
1 0 0 8 / 5
1 0 11 / 5



1 0 3 / 5   0 1 0 3 / 5 
0 0 1 6 / 5
0 1 6 / 5 
2.
For (1.a), we have
1 1 
 3
2


pˆ  Axˆ  2  3    1
1
0 0    0
and
3 3 0
r xˆ   b  pˆ  1  1  0 .
2 0 2
Check that r xˆ   N ( AT ) by determining AT r x̂  . If AT r xˆ   0 then
r xˆ   N ( AT ) .
0 
1 2 0   0
A r x̂   
 0    
1  3 0 2 0
 
T
For (1.b), we have
 1 1 
 45 / 7
 19 / 7  



pˆ  Axˆ   2
1 
   12 / 7 

26
/
7
  71 / 7 
 1  2 


and
10   45 / 7 115 / 7
r xˆ   b  pˆ   5    12 / 7    23 / 7  .
20  71 / 7   69 / 7 
Check that r xˆ   N ( AT ) by determining AT r x̂  . If AT r xˆ   0 then
115 / 7
 1 2 1  
  0
A r x̂   
23
/
7


  
 1 1  2  69 / 7  0


T
For (1.c), we have
 1 1 1
17 / 5
 1 1 1 8 / 5  1 / 5 
 3 / 5   

pˆ  Axˆ  
  3/ 5 
 0  1 1 

 6 / 5 

1
0
1


14 / 5
and
4 17 / 5  3 / 5 
0   1 / 5    1 / 5 

.
ˆ
ˆ
r x   b  p     
1   3 / 5   2 / 5 
  
 

2 14 / 5  4 / 5
Check that r xˆ   N ( AT ) by determining AT r x̂  . If AT r xˆ   0 then
 3/ 5 
1  1 0 1 
0 
 1 / 5   


T

A r x̂   1 1  1 0
 0
 2/5   
1 1 1 1 
 0
 4 / 5
5.
(a) Find the best least square fit by a linear function to the data
Liner function implies that y  ax  b where a and b are to be determined by the data.
Hence, we write
0  a ( 1)  b
1  a ( 0)  b
,
3  a (1)  b
9  a ( 2)  b
or
 1
0

1

2
1
0 

1 a  1

 Ax  b .
1 b  3

 
1
9
Obviously, the above equation is inconsistent. We use the least square approach to
solve the problem.
0 
 
 1 0 1 2 1 21
T
A b
 

 1 1 1 1 3 13
 
9
and
 1 1


 1 0 1 2  0 1 6 2
T
A A



 1 1 1 1  1 1 2 4


 2 1
The least square solution is determined by solving
AT Ax  AT b ,
which can be written in the augmented matrix form as
6 2 21
1 1 / 3 7 / 2 1 1 / 3 7 / 2
1 1 / 3 7 / 2 1 0 29 / 10






2
4
13
2
4
13
0
10
/
3
6
0
1
9
/
5



 


 0 1 9 / 5 
Hence, the solution is a  29 / 10 and b  9 / 5
(b)
10
Data
Least Square Approximation
8
y
6
4
2
0
-2
-1
-0.5
0
0.5
x
1
1.5
2
7. Given a collection of points (x1,y1)… (xn,yn) and Let
x  x1
x 2  x n T and y   y1
y 2  y n T
and let y  c0  c1 x be the linear function that gives the best least square fit to the
points.
We can write in the form of equations as
y1  c1 x1  c0
y 2  c1 x2  c0

y n  c1 xn  c0
or in the form of matrix as
 y1   x1 1
 y   x 1 c
 2    2   1   b  Ax
      c0 
  

 y n   xn 1
Obviously, the above equation is inconsistent. We use the least square approach to
solve the problem.
x
AT b   1
1
 y1   n

xi y i 




 T 
x2  xn   y 2  i 1

  x y 

  ny 
1  1      n
    yi  
 y n   i 1 
and
 x1 1  n

 x x
 x1 x2  xn   x2 1 i 1 i i
T
A A
 n

 1 1  1     

   xi
 xn 1  i 1
Since x  0 we have
n

 xi   T

i 1   x x nx

n 

xT x 0 
AT A  

n 
 0
The least square solution is determined by solving
AT Ax  AT b ,
which can be written in the augmented matrix form as

 nx

n 
x T x 0 xT y 
1 0 xT y xT x



n ny 
y
 0
0 1

Hence, the solution is c1  xT y xT x and c0  y .
pp. 268
7.
(a)
1
1
0
0
e x , e  x   e x e  x dx   1dx  1
(b)
1
x, sin x   x sin xdx  
0
11
 xd cos x  
0
(c)
1
1
0
0
x 2 , x 3   x 2 x 3 dx   x 5 dx 
1
6
8.
(a)
1 
1
1.1   1.1dx  1
0
x 
1
1
0
0
x.x   x.xdx   x 2 dx 
1
1
0
0
1, x   1.xdx   xdx 
cos  
1
2
1, x
1 x

3
2
 3
  /6
Hence,   cos 1 

2


(b) p is the projection of 1 onto x
Hence, we have
1, x
1/ 2
3
p
x
x x
x, x
1/ 3
2
1
3
x cos x

1
0

11
1
 cos xdx  
0
(c)
1 p  1
2
2
1
3
2 1 3   3  1
 3 
x   1  x  dx    1  x  d 1  x  
2
2 
3 0 2   2  2
0
2
1 3
3
91 2
3
 
p  x    x  dx 
x dx 

2
40
2
0 2 
So, we have
1
2
 1  1 p
2
 p
2

1 3
  1.
4 4
9
cos mx, sin nx 
1 
 cos mx sin nxdx 
 
1 
(  sin (n  m) x   sin (n  m) x dx
2  
1 
1 


sin
(
n

m
)
x
dx


 sin (n  m) x dx
2  
2  

1
1


cos(n  m) x   
cos(n  m) x  
(n  m)2
(n  m)2
1
1
cos k1  cos k1  
cos k 2  cos k 2 

(n  m)2
(n  m)2
0

Hence, cos mx and sin nx are orthogonal
cos mx 
sin nx 
1 
1 
 
 
 cos mx cos mxdx 
2
 cos mxdx 
1 
1 
 
 
 sin nx sin nxdx 
2
 sin nxdx 
1 
 1  cos 2mxdx  1
2  
1 
 1  cos 2nx dx  1
2  
Since cos mx and sin nx are orthogonal, the Pythagorean law holds. As a result, we
have
cos mx  sin nx
2
 cos mx
2
  sin nx
Hence distance between two vectors is
2
 cos mx
2.
2
 sin nx
2
 11  2
pp. 286-289
2
(a)
 1/ 3 2 


u1   1 / 3 2  ,
 4 / 3 2 


 1/ 2 
2 / 3


u 2  2 / 3 , u 3   1 / 2 
 0 
1 / 3 


1 1 16
 
1
18 18 18
1 2
1 2
4 1 224
 u1T u 2 



0
3 23 3 23 3 23
9 2
1 1
1 1
4
11
 u1T u 3 


0
0
6
3 2 2 3 2 2 3 2
4 4 1
uT2 u 2 
  1
9 9 9
2 1 2 1 1
22
 uT2 u 3 

 0
0
3 2 3 2 3
3 2
1 1
uT3 u 3 
  0 1
2 2
u1  u1T u1 
u1 , u 2
u1 , u 3
u2 
u 2 , u3
u3 
(b) x  1 1 1T
c1  x, u1  xT u1 
1
1
1
1
4
1
11 4
3 2
3 2
3 2
2
2
1
5
c 2  x, u 2  xT u 2  1  1  1 
3
3
3
3
1
1
c3  x, u 3  xT u 3  1
1
 1.0  0 ./
2
2
Hence, we have
 1/ 3 2 
 1/ 2 
1
2 / 3
1   2  1 / 3 2   5 2 / 3  0 1 / 2 
 3




3 



1
1 / 3 
 4 / 3 2 
 0 
and
x  c12  c22  c32 
2 25
27

0 
 3
9 9
9
3 2

2
3
9.
(a)
 1  cos 2 x  1  cos 2 x 
sin 4 x  sin 2 x sin 2 x  


2
2



1
1
1
1  1  cos 4 x 
 1  2 cos 2 x  cos 2 2 x  (1)  ( ) cos 2 x  

4
4
2
4
2

1
1
1 cos 4 x
 (1)  ( ) cos 2 x  
4
2
8
8
3 2  1  1
1
.
 


cos
2
x



  cos 4 x

8
 8  2  2


(b.i)

  3 2  1
1 


 
4
 sin x cos xdx    8 .    2  cos 2 x   8  cos 4 x cos xdx

 
 2 

 
1



 1   3 2  1  1 

1
    
.


cos
2
x

cos
4
x
cos
xdx








8
2
8
2


















 3 2  1

1
 1 1
 
cos
xdx


cos
2
x
cos
xdx






 2   
 8     2




 1 1

  8    cos 4 x cos xdx




 3 2  1
 1
1

  
,
cos
x


cos
2
x
,
cos
x



  cos 4 x, cos x
 8  2
 2
8



 3 2   1   1  
0    0   0   0
  
 8   2   8  
(b.ii)

  3 2  1
 1
 1 
4


sin
x
cos
2
xdx

.


cos
2
x



  cos 4 x cos 2 xdx


 

 8 

   8  2  2 

 3 2  1
 1
1
  
, cos 2 x     cos 2 x, cos 2 x    cos 4 x, cos 2 x


 2
8
 8  2







 3 2   1   1  

  
0    1   0  

2

 8   2   8  

(b.iii)

  3 2  1
 1
 1 
4
sin
x
cos
3
xdx


  8 .    2  cos 2 x   8  cos 4 x cos 3xdx

 
 2 

  

 3 2  1
 1
1
  
,
cos
3
x


cos
2
x
,
cos
3
x



  cos 4 x, cos 3x
 2
8
2



8







 3 2   1   1  
0    0   0   0
  
 8   2   8  
(b.iv)

  3 2  1
 1
 1 
4
sin
x
cos
4
xdx


  8 .    2  cos 2 x   8  cos 4 x cos 4 xdx

 
 2 

  

 3 2  1
 1
1
  
, cos 4 x     cos 2 x, cos 4 x    cos 4 x, cos 4 x


 2
8
 8  2

 3 2   1   1  
 
  
0    0   1 


 8   2   8  
 8
26.
(a)
1
1,2 x  1   2 x  1dx 
0
1
1
2 x  12  1 1  1  0
4
4
0
Hence, 1 and 2x-1 are orthogonal .
(b)
1 
1
1,1   1  1dx  1
0
Hence 1  1 , we have u1 ( x)  1





1
1
0
0
1
1
2 x  13
6
0
2 x  1,2 x  1   2 x  12 x  1dx   2 x  12 dx 
2x  1 
1

3
Hence 2 x  1 
, we have u 2 x  
1
3
2x  1
 3 2 x  1 .
2x  1
(c)
c1 
1
x , u1 ( x)   x dx 
0
c2 
2 3/ 2 1 2
x

0
3
3
1
1
1
0
0
0
x , u 2 ( x)   x 3 2 x  1dx  2 3  x 3 / 2 dx  3  x1 / 2 dx
2 3
We can approximate
2
2 2
 3 
3
5
3 15
x as
2 2

3 3 2 x  1
3 15
4 4

 x
15 5
x  c1u1 ( x)  c 2 u 2 ( x) 
1.4
x 1/2
c 1u1(x)+c 2u2(x)
1.2
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
0.7
0.8
0.9
1