336 CHAPTER 13. NORMS

Then ∥∥∥∥ 1

ρnJnk

∥∥∥∥ ≤p∑

i=0

(n

i

)∥∥N ik

∥∥ ∣∣λn−ik

∣∣ρn−i

1

ρi(13.9)

Now there are p numbers∥∥N i

k

∥∥ so you could pick the largest, C. Also∣∣λn−ik

∣∣ρn−i

≤∣∣λn−p

k

∣∣ρn−p

so 13.9 is dominated by

≤ Cnp∣∣λn−p

k

∣∣ρn−p

p∑i=0

1

ρi≡ Ĉ

∣∣λn−pk

∣∣ρn−p

The ratio or root test shows that this converges to 0 as n→ ∞.What happens when |λk| = ρ?

1

ρnJnk = ωnI +

p∑i=1

(n

i

)N i

kωn−i 1

ρi

where |ω| = 1.1

ρn∥Jn

k ∥ ≤ 1 + npC

where C = max{∥∥N i

k

∥∥ , i = 1, · · · , p, k = 1..., s}∑p

i=11ρi . Thus

1

ρn∥Jn∥ ≤ 1

ρn

s∑k=1

∥Jnk ∥ ≤ s (1 + npC) = snpC

(1

npC+ 1

)and so

1

ρlim sup

n→∞∥Jn∥1/n ≤ lim sup

n→∞s1/n (npC)

1/n

(1

npC+ 1

)1/n

= 1

lim supn→∞

∥Jn∥1/n ≤ ρ

Next let x be an eigenvector for λ, |λ| = ρ and let ∥x∥ = 1. Then

ρn = ρn ∥x∥ = ∥Jnx∥ ≤ ∥Jn∥

and soρ ≤ ∥Jn∥1/n

Henceρ ≥ lim sup

n→∞∥Jn∥1/n ≥ lim inf

n→∞∥Jn∥1/n ≥ ρ ■

The following theorem is due to Gelfand around 1941.

Theorem 13.3.3 (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).