338 CHAPTER 13. NORMS

Definition 13.4.1 Let {Ak}∞k=1 be a sequence in L (X,Y ) where X,Y are finite dimen-sional normed linear spaces. Then limn→∞Ak = A if for every ε > 0 there exists N suchthat if n > N, then

||A−An|| < ε.

Here the norm refers to any of the norms defined on L (X,Y ) . By Corollary 13.0.8 andTheorem 8.2.3 it doesn’t matter which one is used. Define the symbol for an infinite sum inthe usual way. Thus

∞∑k=1

Ak ≡ limn→∞

n∑k=1

Ak

Lemma 13.4.2 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 (13.10)

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 13.10 are a Cauchy sequence. Therefore byCorollary 13.0.7 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 13.4.3 Let X be a finite dimensional inner product space and let A ∈ L (X,X) .Define

Φ (t) ≡∞∑k=0

tkAk

k!