Linear Algebra, Theory And Applications, 2012a

14.3. THE SPECTRAL RADIUS 349
⎛
= ρ
⎜
⎝
∣∣
( ) n ⎞
λ1
ρ
I + 1 ρ N 1
. .. ⎟
( ) n
⎠
λ2
ρ
I + 1 ρ N 2 ∣∣
1/n
(14.8)
From the definition of ρ, at least one of the λ k /ρ has absolute value equal to 1. Therefore,
⎛ ( ) n ⎞
λ 1
ρ
I + 1 ρ N 1/n
1
⎜
. .. ⎟
− 1 ≡ e ⎝
( ) n
⎠
n ≥ 0
λ
∣∣
2
ρ
I + 1 ρ N 2 ∣∣
because each N k has only zero terms on the main diagonal. Therefore, some term in the
matrix has absolute value at least as large as 1. Now also, since N p k
=0, the norm of
the matrix in the above is dominated by an expression of the form Cn p where C is some
constant which does not depend on n. This is because a typical block in the above matrix
is of the form
p∑
( )( ) n−i n λk
Nk
i i ρ
and each |λ k |≤ρ.
It follows that for n>p+1,
and so
i=1
Cn p ≥ (1 + e n ) n ≥
( ) n
e p+1
n
p +1
( 1/(p+1)
Cn
)) p
≥ e n ≥ 0
( n
p+1
Therefore, lim n→∞ e n =0. It follows from (14.8) that the expression in the norms in this
equation converges to 1 and so
lim ||J n || 1/n = ρ.
n→∞
In case ρ = 0 so that all the eigenvalues equal zero, it follows that J n =0foralln>p.
Therefore, the limit still exists and equals ρ.
The following theorem is due to Gelfand around 1941.
Theorem 14.3.3 (Gelfand) Let A be a complex p × p matrix. Then if ρ is the absolute
value of its largest eigenvalue,
lim
n→∞ ||An || 1/n = ρ.
Here ||·|| is any norm on L (C n , C n ).
Proof: First assume ||·|| is the special norm of the above lemma. Then letting J denote
the Jordan form of A, S −1 AS = J, it follows from Lemma 14.3.2
lim sup ||A n || 1/n = lim sup
∣ ∣ ∣SJ n S −1∣ ∣ ∣ 1/n
n→∞
n→∞
≤
((
lim sup p
2 ) ||S|| ∣ ∣ ∣S −1∣ ∣ ∣ ) 1/n
||J n || 1/n = ρ
n→∞