392 CHAPTER 14. ANALYSIS OF LINEAR TRANSFORMATIONS

Corollary 14.2.3 Let J be in Jordan canonical form

J =

J1

. . .

Js

where each Jk is a block diagonal having λ k on the main diagonal and strings of ones onthe super diagonal, as described earlier. Also let ρ ≡max{|λ i| : λ i ∈ σ (J)} . Then for anynorm ∥·∥

limn→∞∥Jn∥1/n = ρ

Proof: For convenience, take the norm to be given as ∥A∥ ≡ max{∣∣Ai j

∣∣ , i, j}

. Thenwith this norm,

∥Jn∥1/n = max{∥Jn

k ∥1/n ,k = 1, · · · ,s

}From Lemma 14.2.2,

limn→∞∥Jn

k ∥1/n = |λ k|

Therefore,

limn→∞∥Jn∥1/n = lim

n→∞

(max

{∥Jn

k ∥1/n ,k = 1, · · · ,s

})= max

k

(limn→∞∥Jn

k ∥1/n)= max

k|λ k|= ρ.

Now let the norm on the matrices be any other norm say |||·||| . By equivalence of norms,there are δ ,∆ such that

δ ∥A∥ ≤ |||A||| ≤ ∆∥A∥

for all matrices A. Therefore,

δ1/n ∥Jn∥1/n ≤ |||Jn|||1/n ≤ ∆

1/n ∥Jn∥1/n

and so, passing to a limit, it follows that, since limn→∞ δ1/n = limn→∞ ∆1/n = 1,

ρ = limn→∞|||Jn|||1/n ■

Theorem 14.2.4 (Gelfand) Let A be a complex p× p matrix. Then if ρ is the absolutevalue of its largest eigenvalue,

limn→∞||An||1/n = ρ.

Here ||·|| is any norm on L (Cn,Cn).

Proof: Let ||·|| be the operator norm on L (Cn,Cn). Then letting J denote the Jordanform of A,S−1AS = J and these two J,A have the same eigenvalues. Thus it follows fromCorollary 14.2.3

lim supn→∞

||An||1/n = lim supn→∞

∣∣∣∣SJnS−1∣∣∣∣1/n ≤ lim supn→∞

(∥S∥

∥∥S−1∥∥∥Jn∥)1/n

= limn→∞

(∥S∥

∥∥S−1∥∥)1/n ∥Jn∥1/n = ρ