9.12.1. Theorem 9.10.
The Cayley- Hamilton theorem shows that for an n  n matrix A, all integral powers
of A can be written as a linear combination of Ak , where k  0,1, , n  1 .
Consequently, we can write
n 1
etA   qk  t  Ak
(9.27)
k 0
where the qk’s are scalar functions.
Starting from this, E. J. Putzer developed 2
useful methods for expressing e tA as a polynomial in A. The simpler one is as
follows.
Theorem 9.10.
Let A be an n  n matrix with eigenvalues 1 ,
, n . Define a sequence of
polynomials in A by
P0  A  I
k
Pk  A    A  m I 
and
for k  1,2,
,n
m 1
Then we have
n 1
etA   rk 1  t  Pk  A
(9.29)
k 0
where the scalar coefficients rk are determined recursively from the following
triangular system of linear 1st order differential equations:
r1 t   1r1  t 
r1  0  1
rk1  t   k 1rk 1  t   rk  t 
rk 1  0  0
for k  1,2,
,n 1
(9.30)
Proof
Let
r  t 
k
be the set of scalar functions determined by (9.30). Define
n 1
F  t    rk 1  t  Pk  A
(9.31)
k 0
where the Pk’s are given by (9.28).
By the uniqueness theorem, we can prove
F  t   etA ( and hence the theorem ) by showing that F satisfies the same initial
problem as e tA , namely
F   t   AF  t 
with
F  0  I
To begin, we set t  0 in (9.31), and by means of (9.30) and (9.28), we get
F  0  r1  0 P0  A  I
Hence, the initial condition is satisfied. Next, taking the derivative of (9.31) gives
n 1
F   t    rk1  t  Pk  A
k 0
n 1
 1r1  t  P1  A   k 1rk 1  t   rk  t   Pk  A
[ (9.30) used ]
k 1
n 1
  k 1rk 1  t   rk  t   Pk  A
[ r0  t   0 ]
k 0
n 1
n 1
k 1
k 0
  rk  t  Pk  A   k 1rk 1  t  Pk  A
n 2
n 1
k 0
k 0
  rk 1  t  Pk 1  A   k 1rk 1  t  Pk  A
Subtracting from this
n 1
n F  t    n rk 1  t  Pk  A
k 0
gives
n 2
n 2
k 0
k 0
F   t   n F  t    rk 1  t  Pk 1  A    k 1  n  rk 1  t  Pk  A
n 2
  rk 1  t   Pk 1  A   k 1  n  Pk  A 
(9.32)
k 0
Now, from (9.28), we see that
Pk 1  A   A  k 1I  Pk  A
so that
Pk 1  A   k 1  n  Pk  A   A  n I  Pk  A
Therefore, (9.32) becomes
n 2
F   t   n F  t    A  n I   rk 1  t Pk  A
k 0
  A  n I   F  t   rn  t  Pn 1  A 
[ (9.31) used ]
  A  n I  F  t   rn  t  Pn  A
[ (9.28) used ]
  A  n I  F  t 

F   t   AF  t 
QED.
[ Cayley- Hamilton theorem used ]