and so for p large enough, this term on the right in the above inequality is less than ε. Since ε is arbitrary,
this shows the partial sums of 15.6 are a Cauchy sequence. Therefore by Corollary 11.5.4 it follows that
these partial sums converge. As to the last claim,
for any t,s. Also note that all of this works
with no change if A ∈ℒ
(X, X )
where X is a Hilbert space, possibly not finite dimensional. In fact you
don’t even need a Hilbert space. It would work fine with a Banach space, and you would replace the inner
product with the pairing with the dual space but this requires more functional analysis than what is
considered here.
As a special case, suppose λ ∈ ℂ and consider
∑∞ tkλk-
k!
k=0
where t ∈ ℝ. In this case, A_{k} =
tkλk-
k!
and you can think of it as being in ℒ
(ℂ,ℂ )
. Then the following
corollary is of great interest.
Corollary 15.3.5Let
∑∞ tkλk ∑∞ tkλk
f (t) ≡ k! ≡ 1+ k!
k=0 k=1
Then this function is a well defined complex valued function and furthermore, it satisfies the initial valueproblem,
y′ = λy,y(0) = 1
Furthermore, if λ = a + ib,
|f|(t) = eat.
Proof:The first part is a special case of the above theorem. Then
and so ze^{−ct} equals a constant which must be 1 because of the initial condition z
(0)
= 1.
■
Definition 15.3.6The function in Corollary 15.3.5given by that power series is denotedas
λt
exp(λt) or e .
The next lemma is normally discussed in advanced calculus courses but is proved here for the
convenience of the reader. It is known as the root test.
Definition 15.3.7For
{an}
any sequence of real numbers
lim sup an ≡ lim (sup {ak : k ≥ n})
n→ ∞ n→ ∞
Similarly
lim inf an ≡ lim (inf{ak : k ≥ n})
n→∞ n→∞
In case A_{n}is an increasing (decreasing) sequence which is unbounded above (below) then it isunderstood that lim_{n→∞}A_{n} = ∞(−∞) respectively. Thus either of limsup or liminf can equal
+∞ or −∞. However, the important thing about these is that unlike the limit, these alwaysexist.
It is convenient to think of these as the largest point which is the limit of some subsequence of
{an}
and the smallest point which is the limit of some subsequence of
{an }
respectively. Thus lim_{n→∞}a_{n} exists
and equals some point of
[− ∞, ∞ ]
if and only if the two are equal.
Lemma 15.3.8Let
{ap}
be a sequence of nonnegative terms and let
1∕p
r = lim ps→up∞ap .
Then if r < 1, it follows the series,∑_{k=1}^{∞}a_{k}converges and if r > 1, then a_{p}fails to converge to 0 so theseries diverges. If A is an n × n matrix and
r = lim sup ||Ap ||1∕p , (15.7)
p→ ∞
(15.7)
then if r > 1, then∑_{k=0}^{∞}A^{k}fails to converge and if r < 1 then the series converges. Notethat the series converges when the spectral radius is less than one and diverges if the spectralradius is larger than one. In fact, limsup_{p→∞}
Proof:Suppose r < 1. Then there exists N such that if p > N,
a1p∕p< R
where r < R < 1. Therefore, for all such p, a_{p}< R^{p} and so by comparison with the geometric series,
∑R^{p}, it follows ∑_{p=1}^{∞}a_{p} converges.
Next suppose r > 1. Then letting 1 < R < r, it follows there are infinitely many values of p at
which
R < a1p∕p
which implies R^{p}< a_{p}, showing that a_{p} cannot converge to 0 and so the series cannot converge
either.
To see the last claim, if r > 1, then
||Ap ||
fails to converge to 0 and so
{∑m Ak}
k=0
_{
m=0}^{∞} is not a
Cauchy sequence. Hence ∑_{k=0}^{∞}A^{k}≡ lim_{m→∞}∑_{k=0}^{m}A^{k} cannot exist. If r < 1, then for all n large
enough,
∥An∥
^{1∕n}≤ r < 1 for some r so
∥An ∥
≤ r^{n}. Hence ∑_{n}
∥An ∥
converges and so by Lemma 15.3.2, it
follows that ∑_{k=1}^{∞}A^{k} also converges. ■
Now denote by σ
(A )
^{p} the collection of all numbers of the form λ^{p} where λ ∈ σ
(A )
.
Lemma 15.3.9σ
(Ap )
= σ
(A)
^{p}≡
{λp : λ ∈ σ (A )}
.
Proof:In dealing with σ
p
(A )
, it suffices to deal with σ
p
(J )
where J is the Jordan form of A because
J^{p} and A^{p }are similar. Thus if λ ∈ σ
p
(A )
, then λ ∈ σ
p
(J )
and so λ = α where α is one of the entries on
the main diagonal of J^{p}. These entries are of the form λ^{p} where λ ∈ σ
(A )
. Thus λ ∈ σ
(A )
^{p} and this shows
σ
p
(A )
⊆ σ
(A )
^{p}.
Now take α ∈ σ
(A )
and consider α^{p}.
αpI − Ap = (αp−1I + ⋅⋅⋅+ αAp −2 + Ap−1)(αI − A)
and so α^{p}I − A^{p} fails to be one to one which shows that α^{p}∈ σ