The matrix of a type of M ¼
1 1
0 1
has been fully investigated in Sect. 12.6
and characterized as a matrix that cannot be diagonalized. In such a case, we directly
calculate exp tM. Here, remember that product of triangle matrices of same kind (i.e.,
an upper triangle matrix or a lower triangle matrix; see Sect. 12.1) is a triangle matrix
of the same kind.
We have
M
2
¼
1 1
0 1
1 1
0 1
¼
1 2
0 1
, M
3
¼
1 2
0 1
1 1
0 1
¼
1 3
0 1
, Á Á Á:
Repeating this, we get
M
n
¼
1 n
0 1
:
ð15:134Þ
Thus, we have
exp tM ¼ ℜ 0, t
ð Þ ¼ E þ tM þ
1
2!
t
2 M
2
þ Á Á Á þ
1
ν!
t
ν M
ν
þ Á Á Á
¼
1 0
0 1
!
þ t
1 1
0 1
!
þ
1
2!
t
2
1 2
0 1
!
þ Á Á Á þ
1
ν!
t
ν
1 ν
0 1
!
þ Á Á Á
¼
e
t
t 1 þ t þ
1
2!
t
2
þ Á Á Á þ
1
ν À 1
ð
Þ!
t
νÀ1
þ Á Á Á
!
0
e
t
0
B
@
1
C
A ¼
e
t
te
t
0 e
t
!
:
ð15:135Þ
With the resolvent matrix we get
R s, t
ð Þ ¼ exp ÀsM
ð
Þexp tM ¼
e
tÀs
t À s
ð
Þe
tÀs
0
e
tÀs
:
ð15:136Þ
Using (15.107), we get
x t
ð Þ ¼ σ τ
ð ÞR 0, t
ð Þ þ
Z t
0
ds c d
ð ÞR s, t
ð Þ
½
Š :
ð15:137Þ
Thus, we obtain the solution described by
606
15 Exponential Functions of Matrices