Modern Control Engineering

Example Problems and Solutions 711

Solving this last equation for f(A), we obtain

wherem=4. Thus, we have shown the equivalence of Equations (9–102) and (9–103). Although

we assumed m=4, the entire argument can be extended to an arbitrary positive integer m.(For

the case when the matrix Ainvolves multiple eigenvalues, refer to Problem A–9–13.)

A–9–13. Consider Sylvester’s interpolation formula in the form given by Equation (9–104):

This formula for the determination of f(A)applies to the case where the minimal polynomial of

Ainvolves only distinct roots.

Suppose that the minimal polynomial of Ainvolves multiple roots. Then the rows in the

determinant that correspond to the multiple roots become identical, and therefore modification

of the determinant in Equation (9–104) becomes necessary.

Modify the form of Sylvester’s interpolation formula given by Equation (9–104) when the

minimal polynomial of Ainvolves multiple roots. In deriving a modified determinant equation,

assume that there are three equal roots in the minimal polynomial of Aand that

there are other roots that are distinct.

Solution.Since the minimal polynomial of Ainvolves three equal roots, the minimal polynomial

f(l)can be written as

An arbitrary function f(A)of an n*nmatrixAcan be written as

where the minimal polynomial f(A)is of degree manda(A)is a polynomial in Aof degree

m-1or less. Hence we have

wherea(l)is a polynomial in lof degree m-1or less, which can thus be written as

a(l)=a 0 +a 1 l+a 2 l^2 +p+am- 1 lm-^1 (9–105)

f(l)=g(l)f(l)+a(l)

f(A)=g(A)f(A)+a(A)

=Al-l 1 B^3 Al-l 4 BAl-l 5 BpAl-lmB

f(l)=lm+a 1 lm-^1 +p+am- 1 l+am

Al 4 , l 5 ,p, lmB

Al 1 =l 2 =l 3 B

7

1 1    1 I

l 1

l 2







lm

A

l^21

l^22







l^2 m

A^2

p

p

p

p

p

lm 1 -^1

lm 2 -^1







lmm-^1

Am-^1

fAl 1 B

fAl 2 B







fAlmB

f(A)

7 = 0

= a

m

k= 1

fAlkB

AA-l 1 IBpAA-lk- 1 IBAA-lk+ 1 IBpAA-lm IB

Alk-l 1 BpAlk-lk- 1 BAlk-lk+ 1 BpAlk-lmB

+fAl 3 B

AA-l 1 IBAA-l 2 IBAA-l 4 IB

Al 3 - l 1 BAl 3 - l 2 BAl 3 - l 4 B

+fAl 4 B

AA-l 1 IBAA-l 2 IBAA-l 3 IB

Al 4 - l 1 BAl 4 - l 2 BAl 4 - l 3 B

f(A)=fAl 1 B

AA-l 2 IBAA-l 3 IBAA-l 4 IB

Al 1 - l 2 BAl 1 - l 3 BAl 1 - l 4 B

+fAl 2 B

AA-l 1 IBAA-l 3 IBAA-l 4 IB

Al 2 - l 1 BAl 2 - l 3 BAl 2 - l 4 B