Even though it is in general impractical to compute the Jordan form, its existence is all that is needed in
order to prove an important theorem about something which is relatively easy to compute. This is the
spectral radius of a matrix.
Definition 15.2.1Define σ
(A)
to be the eigenvalues of A. Also,
ρ(A ) ≡ max (|λ| : λ ∈ σ(A))
The number, ρ
(A)
is known as the spectral radius of A.
Recall the following symbols and their meaning.
lim sup an, lim inf an
n→ ∞ n→∞
They are respectively the largest and smallest limit points of the sequence
{an}
where ±∞ is allowed in
the case where the sequence is unbounded. They are also defined as
lim ns→up∞ an ≡ nli→m∞ (sup{ak : k ≥ n}),
lim inf an ≡ lim (inf {ak : k ≥ n}).
n→ ∞ n→∞
Thus, the limit of the sequence exists if and only if these are both equal to the same real number.
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 J^{p} = 0 and so the limit is obviously 0. Therefore, assume
λ≠0.
( )
n ∑p n i n−i
J = i N λ
i=0
Then
( ) ( )
n ∑p n ∥∥ i∥∥ n−i n n∑p n −i
∥J ∥ ≤ i N |λ | = |λ| + C |λ| i |λ| (15.4)
i=0n n i=1
≤ |λ| (1+ Cnp ) ≤ |λ|C˜np (15.5)
lim ∥Jn∥1∕n = lim (max {∥Jn ∥1∕n,k = 1,⋅⋅⋅,s})
n→ ∞ n→∞ ( k )
= max lim ∥Jn∥1∕n = max|λk| = ρ.
k n→∞ k k
Now let the norm on the matrices be any other norm say
|||⋅|||
. By equivalence of norms, there are δ,Δ
such that
δ∥A ∥ ≤ |||A||| ≤ Δ ∥A ∥
for all matrices A. Therefore,
1∕n n 1∕n n 1∕n 1∕n n 1∕n
δ ∥J ∥ ≤ |||J ||| ≤ Δ ∥J ∥
and so, passing to a limit, it follows that, since lim_{n→∞}δ^{1∕n} = lim_{n→∞}Δ^{1∕n} = 1,
n 1∕n
ρ = nl→im∞ |||J ||| ■
Theorem 15.2.4(Gelfand) Let A be a complex p×p matrix. Then if ρ is the absolute value of its largesteigenvalue,
n 1∕n
nl→im∞ ||A || = ρ.
Here
||⋅||
is any norm on ℒ
(ℂn, ℂn)
.
Proof:Let
||⋅||
be the operator norm on ℒ
(ℂn,ℂn)
. Then letting J denote the Jordan form of
A,S^{−1}AS = J and these two J,A have the same eigenvalues. Thus it follows from Corollary 15.2.3
n 1∕n || n −1||1∕n ( ∥ −1∥ n )1∕n
lim snu→p∞ ||A || = lim ns→up∞ ||SJ S || ≤ lim nsu→p∞ ∥S ∥∥S ∥ ∥J ∥
( ∥ ∥)1∕n 1∕n
= nl→im∞ ∥S ∥∥S− 1∥ ∥Jn∥ = ρ
Letting λ be the largest eigenvalue of A,
|λ|
= ρ, and Ax = λx where
∥x∥
= 1,
∥An∥ ≥ ∥Anx ∥ = ρn
and so
lim inf ∥An∥1∕n ≥ ρ ≥ lim sup ∥An ∥1∕n
n→∞ n→∞
If follows that liminf _{n→∞}
||An||
^{1∕n} = limsup_{n→∞}
||An ||
^{1∕n} = lim_{n→∞}
||An||
^{1∕n} = ρ. As in
Corollary 15.2.3, there is no difference if any other norm is used because they are all equivalent.
■
I would argue that a better way to prove this theorem is to use the theory of complex analysis and tie it
in with a Laurent series. However, there is a non trivial issue related to the set of convergence of the
Laurent series which involves the theory of functions of a complex variable and this knowledge is not being
assumed here. Thus the above gives an algebraic proof which does not involve so much hard
analysis.
Example 15.2.5Consider
( 9 − 1 2 )
| |
( − 2 8 4 )
1 1 8
. Estimate the absolute value of the largesteigenvalue.
A laborious computation reveals the eigenvalues are 5 and 10. Therefore, the right answer in this case is
10. Consider
|||| 9||||
A
^{1∕9} where the norm is obtained by taking the maximum of all the absolute values of the
entries. Thus
and taking the seventh root of the largest entry gives
1∕9
ρ(A ) ≈ 800390625 = 9.7556.
Of course the interest lies primarily in matrices for which the exact roots to the characteristic equation are
not known and in the theoretical significance.