452 CHAPTER 18. LINEAR TRANSFORMATIONS

partial sums of the i jth entry of ∑∞k=0

Aktkk

k! form a Cauchy sequence. Hence passing to thelimit, it follows from calculus that∣∣∣∣∣∣

(∞

∑k=0

Aktk

k!

)i j

∣∣∣∣∣∣≤ exp(p(|a|+ |b|)∥A∥∞)

Since i j is arbitrary, this establishes the inequality. ■Next consider the derivative of Φ(t). Why is Φ(t) a solution to the above 18.14? By

the mean value theorem,

Φ(t +h)−Φ(t)h

=1h

∞

∑k=0

(t +h)k− tk

k!Ak =

1h

∞

∑k=0

k (t +θ kh)k−1 hk!

Ak

= A∞

∑k=1

(t +θ kh)k−1

(k−1)!Ak−1 = A

∞

∑k=0

(t +θ kh)k

k!Ak,θ k ∈ (0,1)

Does this sum converge to ∑∞k=0

tk

k! Ak ≡ Φ(t)? If so, it will have been shown that Φ′ (t) =AΦ(t). By the mean value theorem again,∥∥∥∥∥ ∞

∑k=0

(t +θ kh)k

k!Ak−

∞

∑k=0

tk

k!Ak

∥∥∥∥∥∞

=

∥∥∥∥∥h∞

∑k=0

k (t +ηkh)k−1θ k

k!Ak

∥∥∥∥∥∞

, ηk ∈ (0,1)

≤

∥∥∥∥∥h∞

∑k=0

k (t +ηkh)k−1

k!Ak

∥∥∥∥∥∞

Now for |h| ≤ 1, the expression tk ≡ t +ηkh ∈ [t−1, t +1] and so by Theorem 18.6.4,∥∥∥∥∥ ∞

∑k=0

(t +θ kh)k

k!Ak−

∞

∑k=0

tk

k!Ak

∥∥∥∥∥∞

=

∥∥∥∥∥h∞

∑k=0

k (t +ηkh)k−1

k!Ak

∥∥∥∥∥∞

≤C |h| (18.20)

for some C. Then∥∥∥∥Φ(t +h)−Φ(t)h

−AΦ(t)∥∥∥∥

∞

=

∥∥∥∥∥A∞

∑k=0

(t +θ kh)k

k!Ak−A

∞

∑k=0

tk

k!Ak

∥∥∥∥∥∞

This converges to 0 as h→ 0 by Lemma 18.6.3 and 18.20. Hence the i jth entry of thedifference quotient

Φ(t +h)−Φ(t)h

converges to the i jth entry of the matrix A∑∞k=0

tk

k! Ak = AΦ(t) . In other words,

Φ′ (t) = AΦ(t)

Now also it follows right away from the formula for the infinite sum that Φ(0) = I. Thisproves the following theorem.