MA2223: SOLUTIONS TO ASSIGNMENT 4
1. Prove directly that the following three norms on R2 are equivalent.
(a) kxk1 = |x1 | + |x2 |
p
(b) kxk2 = (x1 )2 + (x2 )2
(c) kxk∞ = maxi=1,2 |xi |
where x = (x1 , x2 ).
Solution: First we will show that the 1-norm and the maximum norm are
equivalent. We claim that for all points x = (x1 , x2 ) in R2 we have
kxk∞ ≤ kxk1 ≤ 2kxk∞
To show the first inequality note that
kxk∞ = max{|x1 |, |x2 |} ≤ |x1 | + |x2 | = kxk1
For the second inequality we have
kxk1 = |x1 | + |x2 | ≤ kxk∞ + kxk∞ = 2kxk∞
Next we will show that the maximum norm and the Euclidean norm are
equivalent. We claim that for all points x = (x1 , x2 ) in R2 we have
kxk∞ ≤ kxk2 ≤
√
2kxk∞
To show the first inequality note that
2
x21
2
x22
|x1 | =
|x2 | =
≤
x21
≤
x21
q
+
x22
=⇒ |x1 | ≤
+
x22
q
=⇒ |x2 | ≤ x21 + x22
x21 + x22
Combining these two observations gives
kxk∞ = max{|x1 |, |x2 |} ≤
1
q
x21 + x22 = kxk2
2
MA2223: SOLUTIONS TO ASSIGNMENT 4
For the second inequality note that
(x1 )2 = |x1 |2 ≤ kxk2∞
(x2 )2 = |x2 |2 ≤ kxk2∞
Combining these two observations gives
(x1 )2 + (x2 )2 ≤ kxk2∞ + kxk2∞ = 2kxk2∞
We have
kxk2 =
p
√
(x1 )2 + (x2 )2 ≤ 2kxk∞
Note that equivalence of norms is transitive so now we automatically
have the Euclidean norm is equivalent to the 1-norm. To see this directly
our claim is that
kxk2 ≤ kxk1 ≤ 2kxk2
To show the first inequality note that
(x1 )2 + (x2 )2 = |x1 |2 + |x2 |2 ≤ |x1 |2 + 2|x1 ||x2 | + |x2 |2 = (|x1 | + |x2 |)2
=⇒ kxk2 =
p
(x1 )2 + (x2 )2 ≤ |x1 | + |x2 | = kxk1
For the second inequality note that
|x1 |2 = (x1 )2 ≤ (x1 )2 + (x2 )2 =⇒ |x1 | ≤
p
|x2 |2 = (x2 )2 ≤ (x1 )2 + (x2 )2 =⇒ |x2 | ≤
p
(x1 )2 + (x2 )2 = kxk2
(x1 )2 + (x2 )2 = kxk2
Combining these two observations gives
kxk1 = |x1 | + |x2 | ≤ kxk2 + kxk2 = 2kxk2
MA2223: SOLUTIONS TO ASSIGNMENT 4
3
2. Compute the Frobenius norm, 1-norm, maximum norm and spectral norm
of the following matrices.

A=
2
3
1 −4



0 2 0


B= 0 0 3

0 0 0


1
1 
 1 1


 1 −1 −1 1 


C=

 1 1 −1 −1 




1 −1 1 −1





Solution:
(a)

At A = 
2
1
3 −4


2
3
1 −4


=
13
−10
−10
17
The Frobenius norm is
kAkF =
p
trace(At A) =
√
30
The 1-norm is
kAk1 = max{3, 7} = 7
The maximum norm is
kAk∞ = max{5, 5} = 5
The characteristic polynomial of At A is
(13 − λ)(17 − λ) − 100 = 0
=⇒ λ2 − 30λ + 121 = 0
√
√
30 ± 900 − 484
=⇒ λ =
= 15 ± 2 26
2


4
MA2223: SOLUTIONS TO ASSIGNMENT 4
The spectral norm is
kAk2 =
p
=
q
max eigenvalue of At A
√
15 + 2 26
' 5.02
(b)

0 0 0


BtB =  2 0 0

0 3 0

0 2 0


 0 0 3

0 0 0


0 0 0
 
 
= 0 4 0
 
0 0 9
The Frobenius norm is
kBkF =
p
trace(B t B) =
√
13
The 1-norm is
kBk1 = max{2, 3, 0} = 3
The maximum norm is
kBk∞ = max{0, 2, 3} = 3
The spectral norm is
kBk2 =
p
= 3
max eigenvalue of B t B





MA2223: SOLUTIONS TO ASSIGNMENT 4
5
(c)

1
 1 1

 1 −1 1

C tC = 
 1 −1 −1


1 1 −1

 4 0 0 0

 0 4 0 0

= 
 0 0 4 0


0 0 0 4

1 


−1 



1 


−1

1
1
1
1 

1 −1 −1 1 


1 1 −1 −1 


1 −1 1 −1








The Frobenius norm is
p
trace(C t C) = 4
kCkF =
The 1-norm is
kCk1 = max{4, 4, 4, 4} = 4
The maximum norm is
kCk∞ = max{4, 4, 4, 4} = 4
The spectral norm is
kCk2 =
p
= 2

max eigenvalue of C t C