Principles of Mathematics in Operations Research

230 Solutions

Problems of Chapter 5

5.1

Proof. Let Q~lAQ = A and Q~l = QT,

yTAy Xiyf + • • • + XnVl

x = Qy => R(x) =

yTy vi + '-' + vl

2/1 = 1,2/2 = • • • = yn = 0 => Ai < i?(x) since

Ai(2/f + ' •' + Vl) < Aiy? + •' • + A„2/£ «= Ai = min {AJf=1.

Similarly, A„(A) — max||x||=i xTAx. D

5.2

i. xTAx > 0, Vx ^ 0;

xr.4x = [xix 2 x 3 ] ——

100

"2 10'

12 1

01 1

Xi

^2

.X3.

— [2x1 + xix2 + xxx 2 + 2x2 + ^2X3 + x 2 x 3 + X3]

100

-^ [(xi + x 2 f + (x 2 + x 3 )^2 + xi] > 0, Vx ^ 6\

ii. All the eigen values of A satisfy A, > 0;

det(,I-A) = —

100s - 2 -1 0

-1 100s-2 -1

0 -1 100s - 1

= 0<£>

s^3 -0.05s^2 +0.0006s-0.000001 = (s-0.002)(s-0.01552)(s-0.03248) = 0

=> Ai = 0.002 > 0, A 2 = 0.01552 > 0, A 3 = 0.03248 > 0!

iii. All the submatrices Ak have nonnegative determinants;
Since each entry of A is nonnegative, all 1 x 1 minors are OK.

= 1>0

= 1 >0

= 2>0

2 1

1 2

2 1

01

12

01

= 3>0,

= 2>0,

= 1>0,

20

1 1

20

01

1 1

01

= 2>0,

= 2>0,

= 1 >0,

10

21

10

1 1

2 1

1 1

All 2 x 2 minors are OK.