86
a) The eigenvalue method.
Ifλ= 2, thenA−2I=
−2 2 5 −5
. By taking the cross vector we see that we can select the eigenvectorv1= (1,1).
Ifλ=−5, thenA+ 5I= 5 2
5 2
. By taking the cross vector we see that we can select the eigenvectorv2= (−2,5).
The complete solution of the homogeneous system is x=c1e2t
1 1
+c2e−5t −2
5
=
e2t −2e−5t e2t 5e−5t
c1
c2
. It follows that a fundamental matrix is given by
e2t −2e−5t e2t 5e−5t
.
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88
b) Theexponential matrix found by means of Caley-Hamilton’s theorem.
SinceP(λ) =
λ+3 2
2
− 7
2 2
it follows by Cayley-Hamilton’s theorem that
A+3 2I
2
− 7
2 2
I=0, thus B2=
A+3 2I
2
= 7
2 2
I,
where we have putB=A+3 2I.
Now,AandIcommute, so we get by the exponential series that exp(At) = exp
−3 2I+
A+3
2I
t
= exp
−3 2t
exp(Bt)
= exp
−3 2t
∞ n=0
1
(2n)!B2n+ ∞ n=0
1
(2n+ 1)!B2n+1
= exp
−3 2t
∞ n=0
1 (2n)!
7 2
2n I+
∞ n=0
1 (2n+ 1)!
7 2
2n B
= exp
−3
2t cosh 7
2t
I+2 7sinh
7 2t
B
=
⎛
⎜⎝ 1 2e2t+1
2e−5t 0
0 1
2e2t+1 2e−5t
⎞
⎟⎠+1
7(e2t−e−5t)
⎛
⎜⎝ 3
2 2
5 −3 2
⎞
⎟⎠
=
⎛
⎜⎜
⎝ 5 7e2t+2
7e−5t 2 7e2t−2
7e−5t 5
7e2t−5
7e−5t 2 7e2t+5
7e−5t
⎞
⎟⎟
⎠,
which is the exponential matrix corresponding to the problem.
2) Sinceet are not of the same type ase2tor e−5t, a reasonable guess is a solution of the form x=
a b
where dx dt =
a b
et.
Then by insertion into the matrix equation, followed by a multiplication by e−t, a
b
=
0 2 5 −3
a b
− 6
0
, dvs.
−1 2 5 −4
a b
= 6
0
. By Cramer’s formulæ or just some fumbling we find
a= 4 and b= 5, and the complete solution is
x(t) = 4
5
et+c1
1 1
e2t+c2
−2 5
e−5t, t∈R, wherec1 andc2 are arbitrary constants.
Alternativelyone may apply the unfortunate solution formula x0(t) =Φ(t)
Φ(t)−1u(t)dt in order to obtain a particular solution.
a) It follows from Φ(t) =
e2t −2e−5t e2t 5e−5t
where detΦ(t) = 7e−3t that
Φ(t)−1=e3t 7
5e−5t 2e−5t
−e2t e2t
=1 7
5e−2t 2e−2t
−e5t e5t
.
b) The integrand is Φ(t)−1u(t) = 1
7
5e−2t 2e−2t
−e5t e5t
−6et 0
= 1 7
−30e−t 6e6t
. c) Then by integrating each coordinate separately,
Φ(t)−1u(t)dt= 1 7
30e−t e6t
. d) A particular solution is
Φ(t)
Φ(t)−1u(t)dt= 1 7
e2t −2e−5t e2t 5e−5t
30e−t e6t
= 1 7
28et 35et
= 4
5
et.
e) The complete solution is x(t) =
4 5
et+
e2t −2e−5t e2t 5e−5t
c1
c2
, t∈R, wherec1 andc2∈Rare arbitrary constants.
90 Example 4.3 Let
A=
a b c d
be any(2×2)matrix, and let
p(λ) = det(A−λI) =λ2−(a+d)λ+(ad−bc) = (λ−λ1)(λ−λ2) be the corresponding characteristic polynomial. Prove that
p(A) := (A−λ1I)(A−λ2I) =0.
When we identify the coefficients of the characteristic polynomial we get λ1+λ2=a+d and λ1λ2=ad−bc.
A computation of p(A) gives
p(A) = (A−λ1I)(A−λ2I) =A2−(λ1+λ2)A+λ1λ2I
= A2−(a+d)A+ (ad−bc)I=A{A−(a+d)I}+ (ad−bc)I
=
a b c d
a b c d
−
a+d 0 0 a+d
+ detA·I
=
a b c d
−d b c −a
+ detA·I=A{−detA·A−1}+ detA·I=0, where we have applied the result of Example 4.1.
Example 4.4 LetA be any(2×2) matrix.
1) Prove that
A2= (λ+μ)A−λμI,
whereλandμare the two roots of the characteristic polynomial for A.
Hint: Apply the result of Example 4.3.
2) Conclude from (1) that there exist functionsϕ(t)andψ(t), such that exp(At) =ϕ(t)I+ψ(t)A,
whereϕ(0) = 1andψ(0) = 0.
1) It follows from Example 4.3 that
p(A) = (A−λI)(A−μI) =A2−(λ+μ)A+λμI=0. Then by a rearrangement,
A2= (λ+μ)A−λμI.
2) The exponential matrix is defined by the matrix series exp(At) =I+
∞ n=1
1
n!Antn=I+At+ ∞ n=2
1 n!Antn, where the radius of convergence is ∞.
According to (1) every An, n≥2, can be written as a linear combination of Iand A, by using the recursion formula
An=A2An−2= (λ+μ)An−1−λμAn−2, n≥2.
The coefficients only increase as a polynomial, while the denominatorn! is of factorial size. There- fore, this process cannot change the radius of convergence, so we conclude that exp(At) can also be written as a linear combination of IandA, where the coefficients ϕ(t) and ψ(t) are given by power series int of radius of convergence∞,
exp(At) =ϕ(t)I+ψ(t)A. If we here putt= 0, then
exp(A0) =I=ϕ(0)I+ψ(0)A,
and it follows that ϕ(0) = 1 and ψ(0) = 0, because we may assume that I and A are linearly independent. (There is nothing to prove if this is not the case).
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92
Example 4.5 LetA be a(2×2)matrix. We proved in Example 4.4 that A2= (λ+μ)A−λμI,
and
exp(At) =ϕ(t)I+ψ(t)A, ϕ(0) = 1 andψ(0) = 0.
Prove thatϕandψ are the unique solutions of the system of equations
⎧⎨
⎩
ϕ(t) =−λμψ(t),
ϕ(t) = (λ+μ)ϕ(t)−λμϕ(t) = 0,
ϕ(0) = 1, ψ(0) = 0. Then find exp(At)explicitly.
Hint: Split the investigation into the to cases (1) λ=μ, and (2)λ=μ. We shall only check the differential equation for exp(At),
d
dtexp(At) = d dt
I+
∞ n=1
1 n!Antn
= ∞ n=1
1
(n−1)!Antn−1=Aexp(At). Since
d
dtexp(At) =ϕ(t)I+ψ(t)A and
Aexp(At) =ϕ(t)A+ψ(t)A2= (λ+μ)ψ(t)A−ψ(t)λμI+ϕ(t)A, it follows by identification of the coefficients that
ϕ(t) =−λμψ(t),
ψ(t) = (λ+μ)ψ(t)+ϕ(t),
ϕ(0) =−λμψ(0) = 0, ψ(0) = 0 +ϕ(0) = 1. When the latter equation is differentiated, we get
ψ(t) = (λ+μ)ψ(t)+ϕ(t) = (λ+μ)ψ(t)−λμψ(t), hence by a rearrangement,
d2ψ
dt2 −(λ+μ)dψ
dt +λμψ(t) = 0, ψ(0) = 0, ψ(0) = 1, and of course
ϕ(t) =−λμψ(t), ϕ(0) = 1.
The characteristic polynomial for the differential equation ofψis R2−(λ+μ)R+λμ= (R−λ)(R−μ),
with the rootsλandμ.
1) Ifλ=μ, then
ψ(t) =c1eλt+c2eμt where
ψ(0) =c1+c2= 0 and ψ(0) =λc1+μc2= 1, hence c1= 1
λ−μ andc2=− 1
λ−μ, thus ψ(t) = 1
λ−μ(eλt−eμt). We have found that
ϕ(t) =−λμψ(t) = −μ
λ−μ·λeλt+ λ
λ−μ·μeμt, ϕ(0) = 1, so we get by an integration,
ϕ(t) = 1 + t
0
− μ
λ−μλeλτ+ λ λ−μμeμτ
dτ
= 1 + λ
λ−μeμt− μ λ−μeλt−
λ
λ−μ− μ
λ−μ
= λ
λ−μeμt− μ λ−μeλt. Summing up we have
exp(At) =ϕ(t)I+ψ(t)A= λ
λ−μeμt− μ λ−μeλt
I+ 1
λ−μ(eλt−eμt)A. 2) Ifμ=λ, then
ψ(t) =c1eλt+c2teλt, whereψ(0) =c1= 0, and
ψ(t) =c2eλt+c2λteλt, whereψ(0) =c2= 1. Then
ψ(t) =teλt.
Now,ϕ(t) =−λ2ψ(t) andψ(0) = 1, so ϕ(t) = 1−λ2
t
0 τ eλτdτ = 1−λ"
τ eλτ#t
0+"
eλτ#t
0= 1−λteλt+eλt−1 = (1−λt)eλt. Summing up we get
exp(At) =ϕ(t)I+ψ(t)A= (1−λt)eλtI+teλtA.
94
Example 4.6 Let A be a (2×2) matrix, the characteristic polynomial of which has the root λ of multiplicity 2. Computeexp(At) by using that
exp(At) = exp(λtI+(A−λI)t) =eλtexp((A−λI)t).
When λis a double root, then the characteristic polynomial is (R−λ)2, and it follows from either Cayley-Hamilton’s theorem or Example 4.3 that
(A−λI)2=0. Then
exp((A−λI)t) =I+ (A−λI)t+ ∞ n=2
1
n!(A−λI)ntn=I+ (A−λI)t+0= (1−λt)I+tA, and hence
exp(At) =eλtexp((A−λI)t) = (1−λt)eλtI+teλtA, cf. the result of Example 4.5 (2).
Example 4.7 Let A be a (2×2) matrix, the characteristic polynomial of which has the root λ of multiplicity 2. Prove that
Φ(t) =eλtI+teλt(A−λI) is a solution of
dΦ
dt =A Φ(t), Φ(0) =I, and conclude that
exp(At) =eλtI+teλt(A−λI). Since
dΦ
dt =λeλtI+ (1 +λt)eλt(A−λI) =−λ2teλtI+ (1 +λt)eλtA and
A Φ(t) = eλtA+teλt(A−λI)(A−λI+λI) =eλtA+λteλt(A−λI)
= −λ2teλtI+ (1 +λt)eλtA, it follows that
dΦ
dt =A Φ(t). Since
Φ(0) =I+ 0·e0(A−λI) =I,
the initial conditions are also fulfilled. Due to the uniqueness the only solution is exp(At), so exp(At) =eλtI+teλt(A−λI).
Example 4.8 LetA be a(3×3)matrix, the characteristic polynomial of which p(x) = (x−λ)3 has the rootλ of multiplicity 3.
1) Prove thatp(A) := (A−λI)3=0.
2) Let
Φ(t) =eλtI+teλt(A−λI) +1
2t2eλt(A−λI)2. Prove thatΦ(t) = exp(At).
1) This follows immediately form Cayley-Hamilton’s theorem.
96 2) Clearly,Φ(0) =I. Then we just check,
dΦ
dt =λeλtI+(1+λt)eλt(A−λI)+
t+λ
2t2
(A−λI)2 and
A Φ(t) = eλt(A−λI+λI) +teλt(A−λI)(A−λA+λI) +1
2t2eλt(A−λI)2(A−λI+λI)
= λeλtI+eλt(A−λI) +λteλt(A−λI) +teλt(A−λI)2+λ
2t2eλt(A−λI)2+1
2t2eλt(A−λI)3
= λeλtI+(1+λt)eλt)(A−λI)+
t+λ
2t2
(A−λI)2+0, henceΦ(t) also satisfies the matrix differential equation
dΦ
dt =A Φ(t), Φ(0) =I.
Now, the exponential matrix is the unique solution of this matrix differential equation, so we conclude that
exp(At) =Φ(t) =eλtI+teλt(A−λI)+1
2t2eλt(A−λI)2.
Example 4.9 LetAbe a(3×3)matrix, the characteristic polynomial of whichp(x) = (x−λ)2(x−μ) has a double root λand a simple real rootμ(=λ).
1) Prove that
p(A) := (A−λI)2(A−μI) =0. 2) Prove that
Φ(t) := −eλt
(λ−μ)2(A−λI)(A−μI) + eλt
λ−μ(A−μI) + teλt
λ−μ(A−μI)(A−μI) + eμt
(λ−μ)2(A−λI)2 is a solution of
dΦ
dt =A Φ(t), Φ(0) =I, and then conclude thatΦ(t) = exp(At).
1) This follows immediately from Cayley-Hamilton’s theorem.
2) We get by differentiation, dΦ
dt =
−λ
(λ−μ)2+ 1 λ−μ
eλt(A−λI)(A−μI) + λ
λ−μeλt(A−μI) + λ
λ−μteλt(A−λI)(A−μI) + μ
(λ−μ)2eμt(A−λI)2, which should be compared with
A Φ(t) = − eλt
(λ−μ)2(A−λI)(A−μ)(A−λI+λI) + eλt
λ−μ(A−μI)(A−λI+λI) +teλt
λ−μ(A−λI)(A−μI)(A−λI+λI) + eμt
(λ−μ)2(A−λI)2(A−μI+μI), where we have used that all matrices commute. Now,
(A−λI)2(A−μI) =0, so this expression is reduced to
A Φ(t) = −λ
(λ−μ)2eλt(A−λI)(A−μI) + 1
λ−μeλt(A−λI)(A−μI) + λ
λ−μeλt(A−μI)− λ
λ−μteλt(A−μI)(A−λI) + μ
(λ−μ)2eμt(A−λI)2.
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98
It follows by the comparison thatΦ(t) satisfies the matrix differential equation dΦ
dt =A Φ(t). Since
Φ(0) = −1
(λ−μ)2(A−λI)(A−μI) + 1
λ−μ(A−μI) + 1
(λ−μ)2(A−λI)2
= 1
(λ−μ)2(A−λI)(μ−λ)I+ 1
λ−μ(A−μI)
= − 1
λ−μ(A−λI) + 1
λ−μ(A−μI) = λ−μ λ−μI=I,
it follows thatΦ(t) fulfils the same initial conditions as exp(At). Since this solution is unique, we conclude that
Φ(t) = exp(At).
Example 4.10 Let A be an (n×n)matrix, the characteristic polynomial p(x) has λ as ann-tuple root.
1) Prove thatp(A) := (A−λI)n=0.
Hint: Apply a coordinate transformationS, such thatA=SΛS−1, whereΛis an upper triangular matrix with only λs in the diagonal.
2) Prove that
exp(At) =eλtexp((A−λI)t) =eλtI+
n−1
j=1
1
j!tjeλt(A−λI)j.
1) By using the hint we get
(A−λI)n = (SΛS−1−λnSIS−1)n=Sn(Λ−λI)nS−n.
Now,Λ−λIis an upper triangular matrix with only zeros in the diagonal. Hence, (Λ−λI)n=0, and the claim is proved.
2) SinceAandIcommute, and
(A−λI)j = (A−λI)n(A−λI)j−n=0 forj≥n, it follows that
exp(At) = exp(λIt+ (A−λI)t) =eλtexp((A−λI)t)
= eλt ∞ j=0
1
j!tj(A−λI)j =eλt
n−1
j=0
1
j!tj(A−λI)j.
Example 4.11 Let A be an (n×n) matrix, the characteristic polynomial of which p(x) has the n mutually different eigenvalues λ1,. . .,λn, everyone of multiplicity 1.
1) Prove thatp(A) :=$n
j=1(A−λjI) =0.
2) Letq(x)be the polynomial of degree gradn−1defined by
q(x) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
x−λk
λj−λk
⎫⎪
⎬
⎪⎭.
Prove thatq(λj) = 1forj= 1, . . . , n, and conclude thatq(x) = 1 identically.
3) Conclude from (2) that
q(A) :=
n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭=I.
4) Prove that
Φ(t) :=
n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭eλjt
is a solution of dΦ
dt =A Φ(t), Φ(0) =I, and conclude that
exp(At) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭eλjt.
1) There exists a coordinate transformationS, such that A=SΛS−1,
whereΛis a diagonal matrix with the diagonal elementsλj and zeros outside the diagonal. Then
p(A) =Sn
⎧⎨
⎩
&n j=1
(Λ−λjI)
⎫⎬
⎭S−n=0,
becauseΛ−λjIonly contains elements= 0 in the diagonal, and because thej-th diagonal element is also 0. Sincej goes through{1, . . . , n}, it follows that the product matrix is0.
100 2) When we putx=λj0, then
q(λj0) = n j=1
⎧⎪
⎨
⎪⎩
&
k=jk=1
}nλj0−λk
λj−λk
⎫⎪
⎬
⎪⎭=
&n k=jk=10
λj0−λk
λj0−λk = 1.
Clearly, the polynomialq(x)−1 has at most degreen−1, and since it hasnzeros,λ1, . . . , λn, we must have q(x)−1 = 0 identically, soq(x) = 1 identically.
3) It follows immediately fromx→Aand 1→Ithat
q(A) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭=I.
4) Then by (3),
Φ(0) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭=q(A) =I.
Furthermore, dΦ
dt = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
λj
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭eλjt
and
A Φ(t) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λjI)
⎫⎪
⎬
⎪⎭(A−λjI+λjI)eλjt
= 0+ n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
λj
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭eλjt= dΦ dt .
SinceΦ(t) and exp(At) fulfil the same differential equation and same initial conditions, and since the solution is unique, we conclude that
exp(At) =Φ(t) = n j=1
⎧⎪
⎨
⎪⎩
&n k=jk=1
1
λj−λk(A−λkI)
⎫⎪
⎬
⎪⎭eλjt.
Example 4.12 Computeexp(At)for (1) A=
1 1 0 1
, (2) A=
0 1
−2 3
, (3) A=
0 1
−1 2
.
1) The characteristic polynomial is (λ−1)2, where λ = 1 is a double root. Then it follows from Example 4.7 that
exp(At) =etI+tet(A−I) =
et 0 0 et
+
0 tet 0 0
=
et tet 0 et
. 2) The characteristic polynomial is
−λ 1
−2 3−λ
=λ(λ−3)+2 =λ2−3λ+2 = (λ−1)(λ−2).
whereλ= 1 and λ= 2 are simple roots. We get by using the formula of Example 4.11 that exp(At) = 1
1−2(A−2I)et+ 1
2−1(A−I)e2t=
2 −1 2 −1
et+
−1 1
−2 2
e2t
=
⎛
⎝ 2et−e2t e2t−et 2et−2e2t 2e2t−et
⎞
⎠.
102 3) The characteristic polynomial is
−λ 1
−1 2−λ
=λ(λ−2)+1 =λ2−2λ+1 = (λ−1)2,
whereλ= 1 is a double root. Then by Example 4.7, exp(At) =etI+tet(A−I) =et
1 0 0 1
+et
−t t
−t t
=
(1−t)et tet
−tet (1+t)et
.
Example 4.13 Computeexp(At)for
(1) A=
⎛
⎝1 1 0 0 1 0 0 0 1
⎞
⎠, (2) A=
⎛
⎝1 1 0 0 1 0 0 0 2
⎞
⎠.
1) Clearly,λ= 1 is a root of multiplicity 3, hence by Example 4.8, exp(At) = etI+tet(A−I) +1
2t2et(A−I)2
= et
⎛
⎝1 0 0 0 1 0 0 0 1
⎞
⎠+tet
⎛
⎝0 t 0 0 0 0 0 0 0
⎞
⎠+1 2t2et
⎛
⎝0 0 0 0 0 0 0 0 0
⎞
⎠
=
⎛
⎝et tet 0 0 et 0 0 0 et
⎞
⎠.
2) Clearly,λ= 1 is a double root, andμ= 2 is a simple root. It then follows from Example 4.9 that exp(At) = − et
(1−2)2(A−I)(A−2I) + et
1−2(A−2I) + tet
1−2(A−I)(A−2I) + e2t
(1−2)2(A−I)2
= −(t+1)et
⎛
⎝0 1 0 0 0 0 0 0 1
⎞
⎠
⎛
⎝ −1 1 0 0 −1 0
0 0 0
⎞
⎠−et
⎛
⎝ −1 1 0 0 −1 0
0 0 0
⎞
⎠
+e2t
⎛
⎝0 1 0 0 0 0 0 0 1
⎞
⎠
⎛
⎝ 0 1 0 0 0 0 0 0 1
⎞
⎠
=
⎛
⎝et tet 0 0 et 0 0 0 e2t
⎞
⎠.
Example 4.14 Compute exp(At)for
A=
⎛
⎝ 0 1 0
−4 7 −2
−5 7 −1
⎞
⎠.
First we find the characteristic polynomial
−λ 1 0
−4 7−λ −2
−5 7 −1−λ
= −λ(λ−7)(λ+1)+10−4(λ+1)−14λ
= −λ3+6λ2+7λ+10−4λ−4−14λ=−λ3+6λ2−11λ+6
= −(λ−1)(λ−2)(λ−3). The eigenvalues areλ= 1, 2, 3. They are all simple.
The equations of the eigenvectors are
⎧⎪
⎪⎪
⎪⎨
⎪⎪
⎪⎪
⎩
−λx1+x2 = 0,
−4x1+ (7−λ)x2−2x3 = 0,
−5x1+ 7x2−(λ+ 1)x3 = 0,
dvs.
⎧⎪
⎨
⎪⎩
x2=λx1, x3= 7λ−5
λ+ 1 x1. Ifλ= 1, then we can choose the eigenvector (1,1,1).
Ifλ= 2, then we can choose the eigenvector (1,2,3).
Ifλ= 3, then we can choose the eigenvector (1,3,4).
The complete solution of the corresponding system of differential equations is
x=c1et
⎛
⎝1 1 1
⎞
⎠+c2e2t
⎛
⎝1 2 3
⎞
⎠+c3e3t
⎛
⎝1 3 4
⎞
⎠=
⎛
⎝et e2t e3t et 2e2t 3e3t et 3e2t 4e3t
⎞
⎠
⎛
⎝c1
c2
c3
⎞
⎠.
Hence a fundamental matrix is Φ(t) =
⎛
⎝et e2t e3t et 2e2t 3e3t et 3e2t 4e3t
⎞
⎠, medΦ(0) =
⎛
⎝1 1 1 1 2 3 1 3 4
⎞
⎠.
Here,
detΦ(0) =
1 1 1 1 2 3 1 3 4
=
1 0 0 1 1 1 1 2 1
=−1. Then by an inversion of the matrix,
Φ(0)−1=
⎛
⎝ 1 1 −1 1 −3 2
−1 2 −1
⎞
⎠.
104 Finally, we get
exp(At) = Φ(t)Φ(0)−1=
⎛
⎝et e2t e3t et e2t 3e3t et 3e2t 4e3t
⎞
⎠
⎛
⎝ 1 1 −1
1 −3 2
−1 2 −1
⎞
⎠
=
⎛
⎝ et+e2t−e3t et−3e2t+2e3t −et+2e2t−e3t et+2e2t−3e3t et−6e2t+6e3t −et+4e2t−3e3t et+3e2t−4e3t et−9e2t+8e3t −et+6e2t−4e3t
⎞
⎠.
Example 4.15 Compute exp(At)for
A=
⎛
⎝2 −3 2 2 −5 4 3 −9 7
⎞
⎠.
The characteristic polynomial is
1−λ −3 2
2 −5−λ 4
3 −9 7−λ
=
2−λ −3 2
2λ−2 1−λ 0 1 −4 +λ 3−λ
= (λ−1)
2−λ −3 2
2 −1 0
1 −4 +λ 3−λ
= (λ−1){(λ−2)(3−λ) + 4λ−16 + 2−6λ+ 18}
= (λ−1){(λ−2)(3−λ)−2λ+ 4}
= (λ−1)(λ−2)(1−λ) =−(λ−1)2(λ−2). Clearly,λ= 1 is a double root, andμ= 2 is a simple root. Then by Example 4.9,
exp(At) = − eλt
(λ−μ)2(A−λI)(A−μI) + eλt
λ−μ(A−μI) +teλt
λ−μ(A−λI)(A−μI) + eμt
(λ−μ)2(A−λI)2
=−(1+t)et(A−I)(A−2I)−et(A−2I) +e2t(A−I)2
=−etA(A−2I)−tet(A−I)(A−2I)+e2t(A−I)2
=−et{(A−I)2−I}−tet{(A−I)2−(A−I)}+e2t(A−I)2. A small computation gives
(A−I)2=
⎛
⎝1 −3 2 2 −6 4 3 −9 6
⎞
⎠
⎛
⎝1 −3 2 2 −6 4 3 −9 6
⎞
⎠=
⎛
⎝1 −3 2 2 −6 4 3 −9 6
⎞
⎠=A−I.