182 CHAPTER 6. SPECTRAL THEORY

40. In the above problem, consider the question whether all solutions to

x′ = Ax (6.25)

are obtained in the form Φ (t) c for some choice of c ∈ Fn. In other words, is thegeneral solution to this equation Φ (t) c for c ∈ Fn? Prove the following theorem usinglinear algebra.

Theorem 6.10.1 Suppose Φ (t) is an n × n matrix which satisfies Φ′ (t) = AΦ (t) .

Then the general solution to 6.25 is Φ (t) c if and only if Φ (t)−1

exists for some t.

Furthermore, if Φ′ (t) = AΦ (t) , then either Φ (t)−1

exists for all t or Φ (t)−1

neverexists for any t.

(det (Φ (t)) is called the Wronskian and this theorem is sometimes called the Wronskianalternative.)

Hint: Suppose first the general solution is of the form Φ (t) c where c is an arbitrary

constant vector in Fn. You need to verify Φ (t)−1

exists for some t. In fact, show

Φ (t)−1

exists for every t. Suppose then that Φ (t0)−1

does not exist. Explain whythere exists c ∈ Fn such that there is no solution x to the equation c = Φ(t0)x. Bythe existence part of Problem 38 there exists a solution to

x′ = Ax, x (t0) = c

but this cannot be in the form Φ (t) c. Thus for every t, Φ (t)−1

exists. Next suppose

for some t0,Φ (t0)−1

exists. Let z′ = Az and choose c such that

z (t0) = Φ (t0) c

Then both z (t) ,Φ (t) c solve

x′ = Ax, x (t0) = z (t0)

Apply uniqueness to conclude z = Φ(t) c. Finally, consider that Φ (t) c for c ∈ Fn

either is the general solution or it is not the general solution. If it is, then Φ (t)−1

exists for all t. If it is not, then Φ (t)−1

cannot exist for any t from what was justshown.

41. Let Φ′ (t) = AΦ (t) . Then Φ (t) is called a fundamental matrix if Φ (t)−1

exists for allt. Show there exists a unique solution to the equation

x′ = Ax+ f , x (t0) = x0 (6.26)

and it is given by the formula

x (t) = Φ (t) Φ (t0)−1

x0 +Φ(t)

∫ t

t0

Φ (s)−1

f (s) ds

Now these few problems have done virtually everything of significance in an entire un-dergraduate differential equations course, illustrating the superiority of linear algebra.The above formula is called the variation of constants formula.

Hint: Uniquenss is easy. If x1,x2 are two solutions then let u (t) = x1 (t)−x2 (t) andargue u′ = Au, u (t0) = 0. Then use Problem 38. To verify there exists a solution, you