Cayley hamilton

MASSACHUSETTS INSTITUTE OF TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING

2.151 Advanced System Dynamics and Control

Computing the Matrix Exponential The Cayley-Hamilton Method 1
The matrix exponential eAt forms the basis for the homogeneous (unforced) and the forced response of LTI systems. We consider here a method of determining eAt based on the the Cayley-Hamiton theorem. Consider asquare matrix A with dimension n and with a characteristic polynomial ∆(s) = |sI − A| = sn + cn−1 sn−1 + . . . + c0 , and define a corresponding matrix polynomial, formed by substituting A for s above ∆(A) = An + cn−1 An−1 + . . . + c0 I where I is the identity matrix. The Cayley-Hamilton theorem states that every matrix satisfies its own characteristic equation, that is ∆(A) ≡ [0] where [0] is thenull matrix. (Note that the normal characteristic equation ∆(s) = 0 is satisfied only at the eigenvalues (λ1 , . . . , λn )).

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The Use of the Cayley-Hamilton Theorem to Reduce the Order of a Polynomial in A

Consider a square matrix A and a polynomial in s, for example P (s). Let ∆(s) be the characteristic polynomial of A. Then write P (s)in the form P (s) = Q(s)∆(s) + R(s) where Q(s) isfound by long division, and the remainder polynomial R(s) is of degree (n − 1) or less. At the eigenvalues s = λi , i = 1, . . . , n by definition ∆(s) = 0, so that P (λi ) = R(λi ). Now consider the corresponding matrix polynomial P (A): P (A) = Q(A)∆(A) + R(A) But Cayley-Hamilton states that ∆(A) ≡ [0], therefore P (A) = R(A). where the coefficients of R(A) may determined from Eq. (1), or by longdivision. (2) (1)

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D. Rowell 10/16/04

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Example
Reduce the order of P (A) = A4 + 3A3 + 2A2 + A + I for the matrix A= Solution: ∆(s) = |sI − A| = s2 − 5s + 5 P (s) s4 + 3s3 + 2s2 + s + 1 = ∆(s) s2 − 5s + 5 146s − 184 = s2 + 8s + 37 + s2 − 5s + 5 2 P (s) = (s + 8s + 37)∆(s) + 146s − 184 or R(s) = 146s − 184. Then for the given A, P (A) = R(A), or P (A) = A4 + 3A3 + 2A2 + A + I = 146A −184. 3 1 1 2

Summary: A matrix polynomial, of a matrix A of degree n, can always be expressed as a polynomial of degree (n − 1) or less.

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The Use of Cayley-Hamilton to Determine Analytic Functions of a Matrix

Assume that a scalar function f (s) is analytic in a region of the complex plane. Then in that region f (s) may be expressed as a polynomial
∞

f (s) =
k=0

βk sk .

LetA be a square matrix of dimension n, with characteristic polynomial ∆(s) and eigenvalues λi . Then as above f (s) may be written f (s) = ∆(s)Q(s) + R(s) where R(s) is of degree (n − 1) or less. In particular, for s = λi f (λi ) = R(λi )
n−1

=
k=0

αk λk i

(3)

Since the λi , i = 1 . . . n are known, Eq. (3) defines a set of simultaneous linear equations that will generate the coefficientsα0 , . . . , αn−1 . 2

The matrix function f (A is defined to have the same series expansion as f (s), that is
∞

f (A) =
k=0

βk sk
∞

= ∆(A)
k=0

βk sk

+

R(A)

= R(A) by Cayley-Hamilton, since ∆(A) ≡ [0]. then
n−1

f (A) =
k=0

αk Ak

(4)

where the αi ’s may be found from Eq. (3). Thus the defined analytic function of a matrix A of dimension n may be expressed asa polynomial of degree (n − 1) or less.

Example
Find sin(A) where A= Solution: For A ∆(s) = |sI − A| = (s + 2)(s + 3) giving λ1 = −3 and λ1 = −2. Since n = 2, R(s) must be of degree 1 or less. Let R(s) = α0 + α1 s. and from Eq. (3) sin(λ1 ) = α0 + α1 λ1 sin(λ2 ) = α0 + α1 λ2 . Substituting for λ1 and λ2 , and solving for α0 and α2 gives α0 = 3 sin(−2) − 2 sin(−3) α1 = sin(−2) − sin(−3).Substituting in Eq. (4) gives sin(A) = (3 sin(−2) − 2 sin(−3))I + (sin(−2) − sin(−3))A or sin −3 1 0 −2 = sin(−3) sin(−2) − sin(−3) 0 sin(−2) −3 1 0 −2 .

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Computation of the Matrix Exponential eAt
n−1 k=0

The matrix exponential is simply one case of an analytic function as described above. eAt = αk Ak (5)

where the αi ’s are determined from the set of equations given by the...