394 CHAPTER 14. ANALYSIS OF LINEAR TRANSFORMATIONS

Lemma 14.3.2 Suppose {Ak}∞

k=1 is a sequence in L (X ,Y ) where X ,Y are finite dimen-sional normed linear spaces. Then if ∑

∞k=1 ||Ak||< ∞, It follows that

∞

∑k=1

Ak (14.6)

exists (converges). In words, absolute convergence implies convergence. Also,∥∥∥∥∥ ∞

∑k=1

Ak

∥∥∥∥∥≤ ∞

∑k=1∥Ak∥

Proof: For p≤ m≤ n, ∥∥∥∥∥ n

∑k=1

Ak−m

∑k=1

Ak

∥∥∥∥∥≤ ∞

∑k=p∥Ak∥

and so for p large enough, this term on the right in the above inequality is less than ε. Sinceε is arbitrary, this shows the partial sums of 14.6 are a Cauchy sequence. Therefore byCorollary 10.6.4 it follows that these partial sums converge. As to the last claim,∥∥∥∥∥ n

∑k=1

Ak

∥∥∥∥∥≤ n

∑k=1∥Ak∥ ≤

∞

∑k=1∥Ak∥

Therefore, passing to the limit, ∥∥∥∥∥ ∞

∑k=1

Ak

∥∥∥∥∥≤ ∞

∑k=1∥Ak∥ . ■

Why is this last step justified? (Recall the triangle inequality |∥A∥−∥B∥| ≤ ∥A−B∥. )Now here is a useful result for differential equations.

Theorem 14.3.3 Let X be a finite dimensional inner product space and let A ∈L (X ,X) .Define

Φ(t)≡∞

∑k=0

tkAk

k!

Then the series converges for each t ∈ R. Also

Φ′ (t)≡ lim

h→0

Φ(t +h)−Φ(t)h

=∞

∑k=1

tk−1Ak

(k−1)!= A

∞

∑k=0

tkAk

k!= AΦ(t)

Also AΦ(t) = Φ(t)A and for all t,Φ(t)Φ(−t) = I so Φ(t)−1 = Φ(−t), Φ(0) = I. (It isunderstood that A0 = I in the above formula.)

Proof: First consider the claim about convergence.

∞

∑k=0

∥∥∥∥ tkAk

k!

∥∥∥∥≤ ∞

∑k=0

|t|k ∥A∥k

k!= e|t|∥A∥ < ∞