18.3 Invariant Subspaces
where the Ki
are disjoint and compact. Let δi
Then letting Ui ≡ Ki + B
it follows that the Ui
are disjoint open sets. Let Γj
be a oriented cycle
such that for
and if z ∈ Γj∗, then
Let Γ be the union of these oriented cycles. Thus n
= 0 if
and Ui ⊇
Thus fi is analytic on Ω. Using the above definition,
Then by the spectral mapping theorem, Theorem 18.2.6,
Note that for λ ∈ ρ
as can be seen by multiplying both sides by
and observing that the result is
on both sides.
is one to one, the identity follows. Now let
be the linear transformation
From Lemma 18.1.1,
With these introductory observations, the following is the main result about invariant subspaces. First
is some notation.
Definition 18.3.1 Let X be a vector space and let Xk be a subspace. Then X = ∑
k=1nXk means that
every x ∈ X can be written in the form x = ∑
k=1nxk,xk ∈ Xk. We write
if whenever 0 = ∑
kxk, it follows that each xk = 0. In other words, we use the new notation when there is a
unique way to write each vector in X as a sum of vectors in the Xk. When this uniqueness holds, the sum
is called a direct sum. In case AXk ⊆ Xk, we say that Xk is A invariant and Xk is an invariant subspace.
Theorem 18.3.2 Let σ
∪k=1nKk where Kj ∩ Ki
= ∅, each Kj being compact. There exist
for each k
,n such that
- I = ∑
- PiPj = 0 if i≠j
- Pi2 = Pi for each i
- X = k=1nXk where Xk = PkX and each Xk is a Banach space.
- AXk ⊆ Xk which says that Xk is A invariant.
- Pkx = x if x ∈ Xk. If x ∈ Xj, then Pkx = 0 if k≠j.
Proof: Consider 1.
. Consider 2. Let λ
be on Γi
. Then from Theorem 18.2.4
Now from 1.,
and so, multiplying both sides by Pi,
This shows 3.
Consider 4. Note that from 1.
where Xk ≡ PkX. However, this is a direct sum because if 0 = ∑
kPkxk, then doing Pj to both sides and
using part 2.,
and so the summands are all 0 since j is arbitrary. As to Xk being a Banach space, suppose Pkxn → y. Is
y ∈ Xk? Pkxn = Pk
and so, by continuity of
this converges to Pky ∈ Xk.
Thus Pkxn → y
Pkxn → Pky
and y ∈ Xk
. Thus Xk
is a closed subspace of a Banach space and must therefore
be a Banach space itself.
5. follows from 18.9. If Pkx ∈ Xk, then APkx = PkAx ∈ Xk. Hence A : Xk → Xk.
Finally, suppose x ∈ Xk. Then x = Pky. Then Pkx = Pk2y = Pky = x so for x ∈ Xk,Pkx = x and Pk
restricted to Xk is just the identity. If Pjx is a vector of Xj, and k≠j, then PkPjx = 0x = 0.
From the spectral mapping theorem, Theorem 18.2.6, σ
only these two values. Then the following is also obtained
Theorem 18.3.3 Let n > 1,
where the Kk are compact and disjoint. Let Pk be the projection map defined above and Xk ≡ PkX. Then
define Ak ≡ APk. The following hold
- Ak : Xk → Xk,Akx = Ax for all x ∈ Xk so Ak is just the restriction of A to Xk.
- σ =
- A = ∑
- If we regard Ak as a mapping A : Xk → Xk, then σ =
Proof: Letting fk
be the function in the above theorem,
and so, by the spectral mapping theorem,
because the possible values for fk
λ ∈ σ
. This shows 2. Part 1. is obvious from
. So is Part 3. Consider the last claim about Ak
then in all of the above, Uk
could have excluded μ.
Assume this is the case. Thus
is analytic on
. Therefore, using Theorem 18.2.4
applied to the Banach space
and so μ
Therefore, Kk ⊇ σ
fails to be onto, then this must be the case for some
. Here is why. If y
then there is no x
If each Ak
is onto, then there would be xk ∈ Xk
Recall that Pkxk
Summing this on k,
and it would be the case that
is onto. Thus if
is not onto, then
: Xk → Xk
not onto for some k.
If μI − A
fails to be one to one, then there exists x≠
0 such that
where xk ∈ Xk.
Then, since Ak
is just the restriction of A
restriction of I
where I refers to Xk. Now recall that this is a direct sum. Hence
If each μI −Ak is one to one, then each xk = 0 and so it follows that x = 0 also, it being the sum of the xk.
It follows that σ
and so you cannot have σ
proper inclusion, for any k
since otherwise, the above could not hold.
It might be interesting to compare this with the algebraic approach to the same problem in the
appendix, Section A.11. That approach has the advantage of dealing with arbitrary fields of scalars and is
based on polynomials, the division algorithm, and the minimal polynomial where this is limited to the field
of complex numbers. However, the approach in this chapter based on complex analysis applies to arbitrary
Banach spaces whereas the algebraic methods only apply to finite dimensional spaces. Isn’t it interesting
how two totally different points of view lead to essentially the same result about a direct sum of invariant
Review of Linear Algebra