Download MATH 6370 Homework Chapter 4: Cauchy-Schwarz Inequality and Matrix Norms and more Assignments Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!
Department of Mathematics Fall 2007
MATH 6370, Section 10583 MW 5:30-7:00PM
A. Caboussat
Answers Homework Chapter 4
HOMEWORK FROM THE TEXT
Problem 1 (Chapter 4 - Problem 1) 4 1
(a) Check with the definitions.
(b)
||x||
2 ∞ = max |xi|
2 6
n ∑
i=
|xi|
2 = ||x||
2 2
||x||
2 2
n ∑
i=
|xi|
2 6
n ∑
i=
|xi|
= ||x||
2 1 , by induction on n.
Equality can hold, for ei for instance.
(c)
n ∑
i=
|xi|
2 6 max i
|xi|
2 · n.
n ∑
i=
|xi| · 1
n ∑
i=
|xi|
2
n ∑
i=
, (Cauchy-Schwartz).
Equality can hold, for x = (1,... , 1)
T for instance.
(d) We want to compute
|||A|||
1 = max x∈R n ,||x|| 16 =
||Ax|| 1
||x|| 1
Thus
|||A|||
1
n ∑
i=
n ∑
k=
aik xk
n ∑
k=
n ∑
i=
|aik| |xk|
n ∑
k=
|xk|
n ∑
i=
|aik|
n ∑
k=
|xk| max j
n ∑
i=
|aij |
= ||x|| 1 max j
n ∑
i=
|aij |.
Then
||A||
1 6 max j
n ∑
i=
|aij |
and equality is obtained for el, with l the index such that maxj
∑n
i= |aij | =
∑n
i= |ail|.
(e) By setting y = A − 1 x 6 = 0, when x 6 = 0:
∣A−^1
∣ (^) = max x∈Rn,||x||6=
A
− 1 x
||x||
= max y∈Rn,||y||6=
||y||
||Ay||
So
|||A−^1 |||
maxy∈Rn^ ,||y||6=
||y|| ||Ay||
= min y∈Rn,||y||6=
||Ay||
||y||
ADDITIONAL HOMEWORK
Problem 1 5
Since the inverse of a (e.g. lower) triangular matrix is also (lower) triangular, A
− 1 is (lower) triangular. Moreover
A
H is (upper) triangular, so A
− 1 = A
H is diagonal. Same for A.
Problem 2 6
By induction, we have H
T 0 H
− 1 0 =^ I^0 = 1. Assume^ HkH
T k = ckIk , where ck is a constant.
Obviously Hk+1 has entries ±1. Then we check that
Hk+1H
T k+
Hk Hk
Hk −Hk
H
T k
H
T k H
T k
−H
T k
2 HkH
T k
0 2 HkH
T k
2 ckIk 0
0 2 ckIk
= 2ckIk+1.
Therefore
H
T k+1 =^
2 ck
H
− 1 k+
Problem 5 7
5.a) Direct proof.
5.b) The Gauss elimination (without pivoting) consists in the matrix multiplication A
n = GA
0 = GA, where
G = GnGn− 1 · · · G 1 =
−l 21 1
−l 31 −l 32 1
. .
. 0
−lm 1 −lmn 0 0 1
G 11 0
G 12 I
Therefore after n steps
A
n
R 11 ⋆
0 B
and
bij =
m ∑
k=
gikakj
and the product GA after n steps gives:
A
n = GA =
R 11 ⋆
G 21 A 11 + A 21 A 22 + G 21 A 12
We conclude by noting that G 21 A 11 + A 21 = 0, thus G 21 = −A 21 A
− 1 11 and^ B^ =^ A^22 −^ A^21 A
− 1 11 A^12.
