Principles of Mathematics in Operations Research

234 Solutions

Problems of Chapter 6

6.1 The norm of a matrix A is denned as ||.4|| = -y/largest eigen value of AT A.

If Q is orthogonal then QT = Q~l •£> QTQ = I and the unique eigen value of

QTQ is 1. Hence

\\Q\\ = \\QT\\ = i.

Furthermore,

c=\\Q\\\\Q-^1 \\ = \\Q\\^2 = l.

Hence for orthogonal matrices,

c=||Q|| = l.

IIS|| = ||QT|| = i,

C=||Q||||Q-

(^1) ||=a||Q||i||Q|| = ||Q|| (^2) = l.
For orthogonal matrices, ||Q|| = c(Q) = 1. Orthogonal matrices and their
multipliers (aQ) are only perfect condition matrices. It is left as an exercise
to prove the only part.
Let Q = aQ. Then QT
Thus,
and
6.2 A = QQRQ, where
Qo =
i?0 =
'-0.4083 -0.3762 -
0.9129 -0.1882 -
0 0.9111 -
0 0-
0 0
0 0
-1.2247 83.7098
0 -87.8778
0 0
0 0
0 0
0 0
-0.5443 0.5452 -0.3020 0.0843'
-0.2434 0.2438 -0.1351 0.0377
-0.2696 0.2701 -0.1496 0.0418
-0.7562 -0.5672 0.3142 -0.0877
0 -0.4986 -0.8349 0.2331
0 0 0.2689 0.9632
-73.0929 0 0 0
87.3454 3.8183 0 0
-5.5417 -3.0895 -0.1695 0
0 -0.4497 -0.1898 0.0050
0 0 -0.0372 0.0095
0 0 0 0.0016
Ax = R 0 Qo =
Ai = QiRi, where
-76.9159 80.2207
80.2207 94.3687
0 -5.0493
0 0
0 0
0 0
0 0 0 0
-5.0493 0 0 0
3.8305 0.3400 0 0
0.3400 0.3497 0.0185 0
0 0.0185 0.0336 0.0004
0 0 0.0004 0.0016