Linear Algebra, Theory And Applications, 2012a

366 NORMS FOR FINITE DIMENSIONAL VECTOR SPACES
Next show u (t) =cos(bt) andv (t) =sin(bt) work in the above and that there is at
most one solution to
w ′′ + b 2 w =0w (0) = α, w ′ (0) = β.
Thus z (t) =cos(bt)+i sin (bt) andsoy (t) =e at (cos (bt)+i sin (bt)). To show there
is at most one solution to the above problem, suppose you have two, w 1 ,w 2 . Subtract
them. Let f = w 1 − w 2 . Thus
f ′′ + b 2 f =0
and f is real valued. Multiply both sides by f ′ and conclude
d
dt
(
(f ′ ) 2
2
+ b 2 f 2
2
)
=0
Thus the expression in parenthesis is constant. Explain why this constant must equal
0.
17. Let A ∈L(R n , R n ) . Show the following power series converges in L (R n , R n ).
∞∑ t k A k
k=0
You might want to use Lemma 14.4.2. This is how you can define exp (tA). Next show
using arguments like those of Corollary 14.4.3
d
exp (tA) =A exp (tA)
dt
so that this is a matrix valued solution to the differential equation and initial condition
k!
Ψ ′ (t) =AΨ(t) , Ψ (0) = I.
This Ψ (t) is called a fundamental matrix for the differential equation y ′ = Ay. Show
t → Ψ(t) y 0 gives a solution to the initial value problem
y ′ = Ay, y (0) = y 0 .
18. In Problem 17 Ψ (t) is defined by the given series. Denote by exp (tσ (A)) the numbers
exp (tλ) whereλ ∈ σ (A) . Show exp (tσ (A)) = σ (Ψ (t)) . This is like Lemma 14.4.7.
Letting J be the Jordan canonical form for A, explain why
∞∑ t k A k
Ψ(t) ≡
k!
k=0
= S
∞∑
k=0
t k J k
S −1
k!
and you note that in J k , the diagonal entries are of the form λ k for λ an eigenvalue
of A. AlsoJ = D + N where N is nilpotent and commutes with D. Argue then that
∞∑
k=0
t k J k
k!
is an upper triangular matrix which has on the diagonal the expressions e λt where
λ ∈ σ (A) . Thus conclude
σ (Ψ (t)) ⊆ exp (tσ (A))