308 Nonlinear Programming II: Unconstrained Optimization Techniques
whereλiis the ith eigenvalue anduiis the corresponding eigenvector. In the present
case, the eigenvalues,λi, are given by
∣
∣
∣
∣12 −λi − 6
− 6 4 −λi∣
∣
∣
∣=λ2
i−^61 λi+^21 =^0 (E^4 )which yieldλ 1 = 8 +√
52 = 15 .2111 andλ 2 = 8 −√
52 = 0 .7889. The eigenvector
uicorresponding toλican be found by solving Eq. (E 3 ):
[
12 −λ 1 − 6
− 6 4 −λ 1]{
u 11
u 21}
=
{
0
0
}
or ( 12 −λ 1 )u 11 − 6 u 21 = 0or u 21 = − 0. 5332 u 11that is,u 1 ={
u 11
u 21}
=
{
1. 0
− 0. 5332
}
and
[
12 −λ 2 − 6
− 6 4 −λ 2]{
u 12
u 22}
=
{
0
0
}
or ( 12 −λ 2 )u 12 − 6 u 22 = 0or u 22 = 1. 8685 u 12that is,u 2 ={
u 12
u 22}
=
{
1. 0
1. 8685
}
Thus the transformation that reduces [A] to a diagonal form is given byX=[R]Y=[u 1 u 2 ]Y=[
1 1
− 0 .5352 1. 8685
]{
y 1
y 2}
(E 5 )
thatis,x 1 =y 1 +y 2
x 2 = − 0. 5352 y 1 + 1. 8685 y 2This yields the new quadratic term as^12 YT[ A ̃]Y,where[A ̃]=[R]T[ [A]R]=
[
19 .5682 0. 0
0. 0 3. 5432
]
and hence the quadratic function becomesf (y 1 , y 2 )=BT[R]Y+^12 YT[A ̃]Y= 0. 0704 y 1 − 4. 7370 y 2 +^12 ( 91. 8682 )y 12 + 21 ( 4323. 5 )y^22 (E 6 )