Principles of Mathematics in Operations Research

6.1 Solution of Ax = b 85

Proposition 6.1.13 The norm of A is the square root of the largest eigen

value of ATA. The vector that is amplified the most is the corresponding eigen

vector of ATA.

Jtj 4L /lily JU ''TTIH.X**^

An U\.

Example 6.1.14 Let us further continue the previous example:

1 K

0 1

and A l =^1 -K

0 1

ATA =

1 K

K K? + 1

S — 1 K

— K S — K^2 — 1 = 0 => s

(^2) - (K (^2) + 2)s + 1 = 0
A^2 = (K^2 + 2)^2 -4(1)1 = K^2 (K^2 +4)
^max —
-(-K^2 -2)+y/K^2 (K^2 +4)
2(1)
K^2 =$• \A\ = \/Amax « K.
Similarly, \A^1 || = \/Amax[(j4^1 )TA :] « «. 77ms, i/ie relative amplification
is controlled by \A\ ||-A-1|| « «^2 -
Remark 6.1.15 //A is symmetric, then ATA = A^2 and \A\ = max|A;|.
Let us consider now the changes in the coefficient matrix.
Proposition 6.1.16 If we perturb A, then
'IAJI. \AA\<
\x + Ax
<c^1 where c •• \A\ A-
Proof.
Ax = b
(A + AA){x + Ax) 4 =»
AAX + AA(x + Ax) = 0&Ax = -A-l{AA){x + Ax).
|Ac||<IU-^1 ||||ZlA||||:C + Ac||^
Example 6.1.17
A =
I A,
\x + Ax
< A -H
1 10 100
Wffi 1
1 -i- -i-
x 10 100.
, b =
'-JATA- AI-
111
in
10
in

. 100.
31329

X = Xi

\*A\

U\

D

13

= 100.5099, ||x|| = VS,