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TMC1874 Mathematics For Computing

Tutorial Solution
Chapter 5: Inner Product Spaces

1
...

(a)

1
−1

0
0

(b)

(c)

3
2

Solution

(a)
(b)
(c)

12 + (−1)2 =

2

02 + (0)2 = 0
32 + (2)2 =

13

2
...

(a) u =

1
0

,v =

2
1

(b) u =

0
0

,v =

3
−1

Solution

(a)
(b)

u−v =

(1 − 2)2 + (0 − 1)2 =

u−v =

=

(0 − 3)2 + (0 + 1)2

2
10

3
...

(a) u =

1
2

,v =

4
−5

(b) u =

1
2

,v =

8
−5

Solution
The distance between u and v:
(a)
(b)

u−v =

(1 − 4)2 + (2 + 5)2 =

58

u−v =

=

98

(1 − 8)2 + (2 + 5)2

The cosine of the angle θ between u and v is given by
cos θ =

W
...
Tiong

·

u v
u v

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TMC1874 Mathematics For Computing

(a) cos θ =
(b) cos θ =

(1)(4) + (2)(−5)

12 + 22 +

=

42 + (−5)2

(1)(8) + (2)(−5)

12 + 22 +

=

82 + (−5)2

−6

5 41
−2

5 89


2
4
...


2

c + 5 = 9,

5
...
Compute (u, v)
...
Let the inner product space of continuous functions on [0,1] defined as
1

( f , g) =

f (t)g(t) dt
...
K
...

0

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TMC1874 Mathematics For Computing

1

(b) ( f , g) =

0

(t)(e t ) dt = −

1
0

te t dt = (te t − e t )|1 = (e − e) − (0 − 1) = 1
...
Let V be the inner product space defined as
1

(p(t), q(t)) =

p(t)q(t) dt
...

(a) p(t) = 1, q(t) = 2t + 1
2

3

(b) p(t) = t , q(t) = 2t −

(c) p(t) = sin t, q(t) = cos t

4
3t

Solution
The cosine of the angle is given by
cos θ =

(u, v)

...


0
1

(p(t), p(t)) =

dt = 1
...

3

3
13

(b)
1

(p(t), q(t)) =

∴ cos θ =

W
...
Tiong

4
(t2 )(2t3 − t) dt = 0
...

p(t) q(t)

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TMC1874 Mathematics For Computing

(c)
1

(p(t), q(t)) =

(sin t)(cos t) dt =

0
1

(p(t), p(t)) =

0
1

(q(t), q(t)) =

∴ cos θ =

0

sin2 dt =
cos2 dt =

1
0
1

1
1
1
sin2 t = sin2 1
...

2 4

=

1 1
+ sin 2
...


8
...

0

Compute the distance between the given vectors
(b) t2 , t3

(a) sin t, cos t

Solution

(a)
1

d(sin t, cos t) = sin t − cos t

=

0
1

=

0

1/2

(sin t − cos t)2 dt

1/2,

(sin2 t − 2 sin t cos t + cos2 t) dt
1/2

1

=
=
=

0

,

(1 − 2 sin t cos t) dt

(t − sin2 t)

,

1 1/2
,
0

1 − sin2 1
...

105

9
...

W
...
Tiong

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TMC1874 Mathematics For Computing







(a) 






1

1
  − 2
2  
0 ,
0
 
1  
1
 

2

2

Solution



Let v1 = 








0 


, 1 




0 







1


−

0
2 
2 



0  , v2 = 
0  , v3 =  1 
...

2
2
2
1

Hence the given vectors are orthogonal
...


2
1

2

2
)(−

1
2

2

)(

) + (0)(0) + (

2

1
2

)(

1
2

1/2

)

= 1,

Therefore, the given vectors are orthonormal
...
Then

3

3

(v1 , v2 ) = (

W
...
Tiong

2

3

)(−

1
3

)+(

1
3

)(

2
3

) = 0
...
Next,
2

v1

=

(v1 , v1 ) = (

v2

=

(v2 , v2 ) = (−

3

)(

1
3

2
3
)(−

1

)+(
1
3

3

1

)(

)+(

3

2
3

)(

1/2

)

=

2
3

5
,
3

1/2

)

=

5

...


 
 

1
0
0 

(c)  −1  ,  1  ,  0 


0
1
1

Solution







1
0
0
Let v1 =  −1  , v2 =  1  , v3 =  0 
...


So the vectors are not orthogonal and thus they cannot be orthonormal
...




2
−1
10
...
For what values of a
3
4
are u and v orthogonal?


Solution
If u and v orthogonal, then (u, v) = 0
...


For the following questions, the Euclidean spaces Rn and
mathb f R n have the standard inner products on them
...
K
...

0

Page 6 of 13

TMC1874 Mathematics For Computing

2
1

,

2
1

11
...
Then

...



...

Solution

w1
w2

=

v1
=
v1

=

v2
=
v2

1
(2)(2) + (1)(1)

2
1

1
(−2)(−2) + (4)(4)

Hence the orthonormal basis is

2
5
1
5

,

=
−2
4

1

2
1

5

1

=

−1
5
2
5

5

=
−1
2

2
5
1
5

=

−1
5
2
5


...


12
...
Use the Gram-Schmidt process to obtain an orthonormal basis for W
...
K
...
Then

1]
...

−1

5 11]}
...


13
...
Find an orthonormal basis
for W
...
Then
(i) Let v1 = u1 = t
...
K
...


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TMC1874 Mathematics For Computing

Hence the orthogonal basis is t, 1 − 3 t
...


Therefore, the orthonormal basis is { 3t, 2 − 3t}
14
...



2
1
3
1
Solution 




2
1
Note that  2  = 2  1 
...
Thus we need to omit
2
1


2
 2  and let
2

 
 

0
1 
 1
S = { u1 , u2 , u3 } =  1  ,  0  ,  2 
...
Then we have




 

1
0
1
0
a1  1  + a2  0  + a3  2  =  0 
...

− − − −→
− R 3 + R 1 →R 1
0 0 1 0

1
 1
1


0
0
1

1
2
3

0
0
1

1
1
−2



0
1
R 2 ↔R 3
0 −− −  0
− −→
0
0

0
1
0

1
−2
1


0
0 
0

Since the rref of the matrix is I 3
...
Next, apply
Gram-Schmidt process to find the orthogonal basis
...
K
...

1



 
  1 




−3
0
1
1
0
(u2 , u1 )
 0  − (1)(0)+(1)(0)+(1)(1)  1  =  0  − 1  1  =  − 1 
...
Next, find the orthonormal
Hence, the orthogonal basis is {
2
1
2
0
basis


1


1
 13 
1 
v1
1  =  3 ,
=
w1 =


v1
3 1
1
3




− 1
−1
6

v2
1 
1 
−1  =  − 6  ,
w2 =
=


v2
6
2
2
6

 1  
− 22
−2
1  1  
v3
2 
...



0 

2
2
2
2

15
...
K
...

0

Page 10 of 13

TMC1874 Mathematics For Computing

Solution
First, we need to find a basis for the solution space of the homogeneous system
...


So every solution has the form



−3r
−3
x =  4r  = r  4 
...
To find orthonormal basis,
1


w1

=

u
=
u



−
−3

1 
4 =

26
1


3
26
4
26
4
26





...




3
16
...
Find a basis for W ⊥
...
Then (u, w) = 3a + b − 2c = 0 =⇒ a = − 1 b + 2 c
...

b
u=
c
0
1


W
...
Tiong

Page 11 of 13

TMC1874 Mathematics For Computing

Therefore, the set
 1   2 
 −3

3
S =  1 , 0 


0
1

spans W ⊥
...


17
...


Find a basis for W ⊥
...
Since the vector u must be orthogonal to each of the vectors w1 , w2 , w3 , w4 , w5 , we have a system of homogeneous linear
equations
(u, w1 ) = 2a 1 − a 2 + a 3 + 3a 4 = 0,
(u, w2 ) =

a 1 + 2a 2 + a 4 − 2a 5 = 0,

(u, w3 ) = 4a 1 + 3a 2 + a 3 + 5a 4 − 4a 5 = 0,
(u, w4 ) = 3a 1 + a 2 + 2a 3 − 1a 4 + a 5 = 0,

(u, w5 ) = 2a 1 − a 2 + 2a 3 − 2a 4 + 3a 5 = 0
...



2
1
4
3
2

1
0
1
2
2

3
1
5
−1
−2

0
−2
−4
1
3

0
0
0
0
0





−1
2
3
1
−1

1
0
0
0
0

2
−5
−5
−5
−5

0
1
1
2
2

1
1
1
−4
−4

−2
4
4
7
7

0
0
0
0
0

















W
...
Tiong







 R 1 ↔R 2 
− −→
−− − 





1
2
4
3
2


 − R 2 + R 3 →R 3 
 − 1 R 2 →R 2 


5
− − − −→
−− − − − 
 − R 4 + R 5 →R 5 
 − R 2 + R 4 →R 4 

2
−1
3
1
−1
1
0
0
0
0

0
1
1
2
2
2
1
0
0
0

1
3
5
−1
−2

−2
0
−4
1
3

0
0
0
0
0

0

1

−1
5
0
1
0

−1
5
0
−5
0

−2
−4
5
0
3
0



 −2 R 1 + R 2 → R 2

 −4 R 1 + R 3 → R 3
−−−−−
 −− − − −→
 −3 R 1 + R 4 → R 4
 −2 R 1 + R 5 → R 5

0
0
0
0
0





 R 3 ↔R 4
− −→
−− −



Page 12 of 13

TMC1874 Mathematics For Computing



1
0
0
0
0

2
1
0
0
0

0

1

−1
5
1
0
0

−1
5
−5
0
0



1
0
0
0
0

0
1
0
0
0

0
0
1
0
0















− 17
5
−6
5
−5
0
0

−2
−4
5
3
0
0
−8
5
−1
5
3
0
0

0
0
0
0
0
0
0
0
0
0







 1 R 3 + R 2 →R 2 
 5

− − − −→
−− − − − 





1
0
0
0
0

2
1
0
0
0

0
0
1
0
0

1
−6
5
−5
0
0

−2
−1
5
3
0
0

0
0
0
0
0







 −2 R 2 + R 1 → R 1
−−−−−
 −− − − −→










So the solution is
a5

=

a4

=

a3

s,
r,

= 5r − 3s,
6
1
=
r + s,
5
5
17
8
=
r + s
...

5
5
5
5
Since they are not multiples of each other, they are linearly independent and thus form
a basis for W ⊥
...
K

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