Estimation of Homography Dynamics on the Special Linear Group - Visual Servoing via Advanced Numerical Methods - page 133

Plugging (8.12) into (8.11), one obtains:

D )= 1 D ( I −

3 trace( D ( I

− D )) I .

Knowing that det( D )=1, the λ i 's satisfy the following equations:

λ

1 (1

− λ

1 )=

λ

2 (1

− λ

2 )

(8.13)

λ

2 (1

− λ

2 )=

λ

3 (1

− λ

3 )

(8.14)

λ 3 = 1 / (λ 1 λ 2 ) (8.15)

which can also be written as follows:

λ 1 − λ 2 =( λ 1 − λ 2 )( λ 1 + λ 2 ) (8.16)

λ 1 − λ 3 =( λ 1 − λ 3 )( λ 1 + λ 3 ) (8.17)

λ 3 = 1 / ( λ 1 λ 2 ) . (8.18)

First of all, let us remark that if λ 1 = λ 2 = λ 3 then λ 1 = λ 2 = λ 3 = 1. This solution

is associated with the equilibrium point E s =( I , 0).

If λ 1 = λ 2 < λ 3 then:

1 = λ 2 + λ 3 (8.19)

2
2 )

λ 3 = 1 / ( λ (8.20)

3
2

2
2 + 1 = 0.

where λ 2 ≈− 0 . 7549 is the unique real solution of the equation λ − λ

This solution is associated with the equilibrium set E u .
If λ 1 < λ 2 = λ 3 then:

1 = λ 1 + λ 2 (8.21)

2
2

λ 1 = 1 / λ (8.22)

2 2 + 1 = 0. But this is impossible

so that λ 2 is also solution of the equation λ − λ

since we supposed λ 1 < λ 2 and the solution of the equation is such that − 1 < λ 2 < 0

2

and 0 < λ 1 = 1 / λ < 1.

If λ 1 = λ 2 = λ 3 , then:

1 = λ 1 + λ 2 (8.23)

1 = λ 1 + λ 3 (8.24)

λ 3 = 1 / ( λ 1 λ 2 ) (8.25)

which means that λ 3 . This is in contradiction with our initial hypothesis.

In conclusion, H has two equal negative eigenvalues

λ 2 =

λ 1 = λ 2 = λ ( λ ≈− 0 . 7549

3 2 + 1 = 0) and the third one is

is the unique real solution of the equation λ − λ

2 . Writing the diagonal matrix D as follows:

λ 3 = 1 / λ

λ − 3 1) e 3 e 3 ) D = λ ( I +( −