18.6. THE MATRIX EXPONENTIAL, DIFFERENTIAL EQUATIONS ∗ 455

18.6.1 Computing A Fundamental MatrixIn case that A is p× p and nondefective, you can easily compute Φ(t). Recall that in thiscase, there exists a matrix S such that

S−1AS = D =

λ 1

. . .

λ p

where D is a diagonal matrix. Then A = SDS−1. Now Φ(t) =

∞

∑k=0

SDkS−1

k!tk = S

∞

∑k=0

Dk

k!tkS−1 = S

eλ 1t

. . .

eλ pt

S−1

Thus you can easily compute Φ(t) in this case. For the case where a λ k is complex, seethe above observation for the meaning of eλ t . Thus one can explicitly find the fundamentalmatrix in this case.

In fact you can find it whenever you can compute the eigenvalues exactly. Suppose thatthe matrix A is defective. A chain based on the eigenvector v1 is an ordered list of nonzerovectors

(v1,v2, · · · ,vm)

where (A−λ I)vk+1 = vk and (A−λ I)v1 = 0. Given such a chain, m > 1, consider

x(t)≡m

∑k=1

tm−k

(m− k)!vkeλ t

Then

x′ (t) = λ

m

∑k=1

tm−k

(m− k)!vkeλ t +

m−1

∑k=1

(m− k) tm−(k+1)

(m− k)!vkeλ t

= λ

m

∑k=1

tm−k

(m− k)!vkeλ t +

m−1

∑k=1

tm−(k+1)

(m− (k+1))!vkeλ t

= λ

m

∑k=1

tm−k

(m− k)!vkeλ t +

m

∑k=2

tm−k

(m− k)!vk−1eλ t

Recalling that Avk−vk−1 = λvk for k > 1, and Av1−λv1 = 0,

=m

∑k=1

tm−k

(m− k)!Avkeλ t −

m

∑k=2

tm−k

(m− k)!vk−1eλ t +

m

∑k=2

tm−k

(m− k)!vk−1eλ t

= Am

∑k=1

tm−k

(m− k)!vkeλ t = Ax(t)

Thus each such chain results in a solution to the system of equations. Also, each suchchain of length m results in m solutions to the differential equation. Just consider the chains

v1,(v1,v2) , · · · ,(v1,v2, · · · ,vm) ,