13.4. SERIES AND SEQUENCES OF LINEAR OPERATORS 337

Proof: First assume ||·|| is the operator norm with respect to the usual Euclidean metricon Cn. Then letting J denote the Jordan form of A,S−1AS = J, it follows from Lemma13.3.2

lim supn→∞

||An||1/n = lim supn→∞

∣∣∣∣SJnS−1∣∣∣∣1/n ≤ lim sup

n→∞

(∥S∥

∥∥S−1∥∥ ∥Jn∥

)1/n≤ lim sup

n→∞

(||S||

∣∣∣∣S−1∣∣∣∣ ||Jn||

)1/n= ρ

Letting λ be the largest eigenvalue of A, |λ| = ρ, and Ax = λx where ∥x∥ = 1,

∥An∥ ≥ ∥Anx∥ = ρn

and solim inf

n→∞∥An∥1/n ≥ ρ ≥ lim sup

n→∞∥An∥1/n

If follows that lim infn→∞ ||An||1/n = lim supn→∞ ||An||1/n = limn→∞ ||An||1/n = ρ.Now by equivalence of norms, if |||·||| is any other norm for the set of complex p × p

matrices, there exist constants δ,∆ such that

δ ||An|| ≤ |||An||| ≤ ∆ ||An||

Thenδ1/n ∥An∥1/n ≤ |||An|||1/n ≤ ∆1/n ∥An∥1/n

The limits exist and equal ρ for the ends of the above inequality. Hence, by the squeezing

theorem, ρ = limn→∞ |||An|||1/n. ■

Example 13.3.4 Consider

 9 −1 2

−2 8 4

1 1 8

 . Estimate the absolute value of the largest

eigenvalue.

A laborious computation reveals the eigenvalues are 5, and 10. Therefore, the right

answer in this case is 10. Consider∣∣∣∣A7

∣∣∣∣1/7 where the norm is obtained by taking themaximum of all the absolute values of the entries. Thus 9 −1 2

−2 8 4

1 1 8

7

=

 8015 625 −1984 375 3968 750

−3968 750 6031 250 7937 500

1984 375 1984 375 6031 250

and taking the seventh root of the largest entry gives

ρ (A) ≊ 8015 6251/7 = 9. 688 951 236 71.

Of course the interest lies primarily in matrices for which the exact roots to the characteristicequation are not known and in the theoretical significance.

13.4 Series and Sequences of Linear Operators

Before beginning this discussion, it is necessary to define what is meant by convergence inL (X,Y ) .