14.2. THE SPECTRAL RADIUS 391

Recall the following symbols and their meaning. limsupn→∞ an, liminfn→∞ an. Theyare respectively the largest and smallest limit points of the sequence {an} where ±∞ isallowed in the case where the sequence is unbounded. They are also defined as

lim supn→∞

an ≡ limn→∞

(sup{ak : k ≥ n}) ,

lim infn→∞

an ≡ limn→∞

(inf{ak : k ≥ n}) .

Thus, the limit of the sequence exists if and only if these are both equal to the same realnumber.

Lemma 14.2.2 Let J be a p× p Jordan block

J =

λ 1 0

λ. . .. . . 1

0 λ

Then limn→∞ ||Jn||1/n = |λ |

Proof: The norm on matrices can be any norm. It could be operator norm for example.If λ = 0, there is nothing to show because Jp = 0 and so the limit is obviously 0. Therefore,assume λ ̸= 0.

Jn =p

∑i=0

(ni

)Ni

λn−i

Then

∥Jn∥ ≤p

∑i=0

(ni

)∥∥Ni∥∥ |λ |n−i = |λ |n +C |λ |np

∑i=1

(ni

)|λ |−i (14.4)

≤ |λ |n (1+Cnp)≤ |λ |n C̃np (14.5)

where the C depends on ∑pi=1 |λ |

−i and the∥∥Ni∥∥. Therefore,

lim supn→∞

∥Jn∥1/n ≤ |λ | lim supn→∞

(C̃np)1/n

= |λ |

Next let x be an eigenvector for λ such that ∥x∥ = 1, the norm being whatever norm isdesired. Then Jx= λx. It follows that Jnx= λ

nx. Thus ∥Jn∥≥ ∥Jnx∥= |λ |n ∥x∥= |λ |n .It follows that liminfn→∞ ∥Jn∥1/n ≥ |λ | . Therefore,

|λ | ≤ lim infn→∞∥Jn∥1/n ≤ lim sup

n→∞

∥Jn∥1/n ≤ |λ |

which shows that limn→∞ ∥Jn∥1/n = |λ |. The same conclusion holds for any other norm.Indeed, if |||·||| were another norm, there are constants δ ,∆ such that

δ1/n ∥Jn∥1/n ≤ |||Jn|||1/n ≤ ∆

1/n ∥Jn∥1/n

Then since limn→∞ δ1/n = limn→∞ ∆1/n = 1, the squeezing theorem from calculus implies

that limn→∞ |||Jn|||1/n = ρ . ■