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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 a square matrix A with dimension n and with a characteristic polynomial
∆(s) = |sI − A| = sn^ + cn− 1 sn−^1 +... + c 0 ,
and define a corresponding matrix polynomial, formed by substituting A for s above
∆(A) = An^ + cn− 1 An−^1 +... + c 0 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 the null matrix. (Note that the normal characteristic equation ∆(s) = 0 is satisfied only at the eigenvalues (λ 1 ,... , λn)).
1 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) is found 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). (1)
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). (2)
where the coefficients of R(A) may determined from Eq. (1), or by long division.
(^1) D. Rowell 10/16/
Example
Reduce the order of P (A) = A^4 + 3A^3 + 2A^2 + A + I for the matrix
A =
[ 3 1 1 2
]
Solution:
∆(s) = |sI − A| = s^2 − 5 s + 5 P (s) ∆(s)
s^4 + 3s^3 + 2s^2 + s + 1 s^2 − 5 s + 5 = s^2 + 8s + 37 +
146 s − 184 s^2 − 5 s + 5 P (s) = (s^2 + 8s + 37)∆(s) + 146 s − 184
or R(s) = 146s − 184. Then for the given A, P (A) = R(A), or
P (A) = A^4 + 3A^3 + 2A^2 + A + I = 146 A − 184.
Summary: A matrix polynomial, of a matrix A of degree n, can always be expressed as a polynomial of degree (n − 1) or less.
2 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=
βksk.
Let A 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=
αkλki (3)
Since the λi, i = 1... n are known, Eq. (3) defines a set of simultaneous linear equations that will generate the coefficients α 0 ,... , αn− 1.
3 Computation of the Matrix Exponential eAt
The matrix exponential is simply one case of an analytic function as described above.
eAt^ =
n∑− 1
k=
αkAk^ (5)
where the αi’s are determined from the set of equations given by the eigenvalues of A.
eλit^ =
n∑− 1
k=
αkλki (6)
Example
Find eAt^ for A =
[ 0 1 − 2 − 3
] .
Solution: The characteristic equation is s^2 + 3s + 2 = 0, and the eigenvalues are λ 1 = −1, λ 2 = −2. From Eq. (5) eAt^ = α 0 I + α 1 A From Eq. (6), for λ 1 = −1 and λ 2 = − 2 e−t^ = α 0 − α 1 e−^2 t^ = α 0 − 2 α 1 , or α 0 = (2e−t^ − e−t) and α 1 = (e−t^ − e−^2 t). Then eAt^ = (2e−t^ − e−t)I + (e−t^ − e−^2 t)A
=
[ 2 e−t^ − e−^2 t^ e−t^ − e−^2 t − 2 e−t^ + 2e−^2 t^ −e−t^ + 2e−^2 t
]
Example
Find eAt^ for A =
[ 0 1 − 1 0
] .
Solution: The characteristic equation is s^2 + 1 = 0, and the eigenvalues are λ 1 = +j, λ 2 = −j. From Eq. (5) eAt^ = α 0 I + α 1 A From Eq. (6), for λ 1 = +j and λ 2 = −j ejt^ = cos(t) + j sin(t) = α 0 + α 1 j e−jt^ = cos(t) − j sin(t) = α 0 − α 1 j, or α 0 = cos(t) and α 1 = sin(t). Then eAt^ = cos(t)I + sin(t)A
=
[ cos(t) sin(t) − sin(t) cos(t)
]
Note: If one or more of the eigenvalues is repeated (λi = λj , i 6 = j, then Eqs. (6) will yield two or more identical equations, and therefore will not be a set of n independent equations. For an eigenvalue of multiplicity m, the first (m − 1) derivatives of ∆(s) all vanish at the eigenvalues, therefore
f (λi) =
(n ∑−1)
k=
αkλki = R(λi)
df dλ
∣∣ ∣∣ λ=λi
dR dλ
∣∣ ∣∣ λ=λi .. .
dm−^1 f dλm−^1
∣∣ ∣∣ ∣ λ=λi
dm−^1 R dλm−^1
∣∣ ∣∣ ∣ λ=λi
form a set of m linearly independent equations, which when combined with the others will yield the required set os n equations to solve for the α’s.
