Section 9–5 / Some Useful Results in Vector-Matrix Analysis 669
where Since
(lI-A)adj(lI-A)=Cadj(lI-A)D(lI-A)=|lI-A|I
we obtain
From this equation, we see that Aand (i=1, 2,p,n)commute. Hence, the product
of(lI-A)and becomes zero if either of these is zero. If Ais substitut-
ed for lin this last equation, then clearly lI-Abecomes zero. Hence, we obtain
An+a 1 An–1+p+an–1A+anI= 0
This proves the Cayley–Hamilton theorem, or Equation (9–44).
Minimal Polynomial. Referring to the Cayley–Hamilton theorem, every n*n
matrixAsatisfies its own characteristic equation. The characteristic equation is not,
however, necessarily the scalar equation of least degree that Asatisfies. The least-degree
polynomial having Aas a root is called the minimal polynomial. That is, the minimal
polynomial of an n*nmatrixAis defined as the polynomial of least degree,
f(l)=lm+a 1 lm–1+p+am–1l+am,mn
such that or
f(A)=Am+a 1 Am–1+p+am–1A+amI= 0
The minimal polynomial plays an important role in the computation of polynomials in
ann*nmatrix.
Let us suppose that a polynomial in l, is the greatest common divisor of all the
elements of We can show that if the coefficient of the highest-degree term
inlof is chosen as 1, then the minimal polynomial is given by
(9–45)
[See Problem A–9–8for the derivation of Equation (9–45).]
It is noted that the minimal polynomial of an n*nmatrixAcan be determined
by the following procedure:
1.Form and write the elements of as factored polynomials
inl.
2.Determine as the greatest common divisor of all the elements of
Choose the coefficient of the highest-degree term in lof to be 1. If there is no
common divisor,
3.The minimal polynomial is then given as divided by
Matrix Exponential In solving control engineering problems, it often becomes
necessary to compute If matrix Ais given with all elements in numerical values,
MATLAB provides a simple way to compute eAT, where Tis a constant.
eAt.
eAt.
f(l) ∑l I-A∑ d(l).
d(l)=1.
d(l)
d(l) adj(l I-A).
adj(l I-A) adj(l I-A)
f(l)
f(l)=
∑l I-A∑
d(l)
d(l) f(l)
adj(l I-A).
d(l),
f(A)= 0 ,
f(l)
adj(l I-A)
Bi
=AB 1 ln-^1 +B 2 ln-^2 +p+Bn- 1 l+BnB(l I-A)
=(l I-A)AB 1 ln-^1 +B 2 ln-^2 +p+Bn- 1 l+BnB
∑l I-A∑ I=Iln+a 1 Iln-^1 +p+an- 1 Il+an I
B 1 =I.
