This section discusses continua of fixed points—what they look like, and some properties. SinceφC,Dis (generally) only densely defined, it is not immediate that its fixed point
set is closed in MnC. However, it happens to be true.
LemmaA.1. Suppose that{Zn} is a sequence of fixed points ofφ≡φC,D, and that{Zn} converges toZ. ThenZis in the domain ofφand is a fixed point.
Proof. We note that for alln,Z−1
n =I−CZnD; thus{Zn−1} →I−CZD. Thus I=limZnZ−n1 =limZn(I−CZD)=Z(I−CZD). In particular, I−CZDis invertible, so thatZis in the
domain ofφ, and obviously it is a fixed point.
Although the following was deduced previously—whenCDis invertible—it is useful to give a more general and elementary proof.
CorollaryA.2. IfZis a limit point of fixed points ofφ≡φC,D, thenᏹZC,DZ has1as an eigenvalue.
Proof. Suppose that limZn=Zwhere theZnare each fixed points. SetTn=(Zn−Z)/Zn −Z(where · is any fixed algebra norm on MnC). Then{Tn}is a sequence of matrices
of norm one, so it has a subsequence which converges to another element of norm one. By reducing to a subsequence, we may assume that limTn=T exists (and is not zero),
and moreover thatZn−Z<1/(2Z · C · D). Of course, I−CZD=Z−1and I− CZnD=Zn−1. SetVn=ZC(Zn−Z)D, so thatVn<1/2, and thus
∞ i=0Vni<2. Now I−CZnD=I−CZD−C Zn−Z D=(I−CZD)I−ZCZn−Z D, (A.1)
so Zn= I−CZnD −1 =I−ZCZn−Z D−1·Z = I+Vn+Vn2 ∞ i=0 Vi n ·Z=Z+VnZ+Vn2Yn, (A.2) whereYn<2Z. Thus Tn= Zn−Z !!Zn−Z!!=ZC Zn−Z DZ !!Zn−Z!! + V 2 nYn !!Zn−Z!!. (A.3) SinceV2 n =OOO Zn−Z2
, the rightmost summand isOOO(Zn−Z), so on taking lim-
its, we deduce thatT=ZCTDZ. ThusTis an eigenvector ofᏹZC,DZ with eigenvalue 1.
The converse fails—we can have exactly one fixed point even though 1 is the only eigenvalue ofᏹZC,DZ.
IfφC,Dhas three fixed points in an (affine) line, then the entire line consists of fixed
points; another way to put this is that any one-dimensional affine subspace (i.e., translate of a subspace) containing three fixed points must be composed entirely of fixed points. Fork-dimensional affine subspaces, it is easy to construct examples containing exactly 2kfixed points (viaCandDdiagonal and with a mild assumption on their eigenvalues),
but whether 2k+ 1 is the critical value (i.e., if it contains 2k+ 1 fixed points, it contains a
continuum or better still, an affine line, of fixed points) is still unclear.
LemmaA.3 (Three in a row). Suppose that C,D,X,Z aren×nmatrices with {X,Z}
linearly independent. For a complex numberα, defineXα=αX+ (1−α)Z. Letφdenote φC,D.
(a)If there exist three distinct complex values ofαsuch thatXαis a fixed point ofφ, then Xαis a fixed point ofφfor all values ofα.
(b)If for allα,Xαis a fixed point ofφ, thenM:=Z−Xsatisfies the following:
(i) (MX−1)2=000,
(ii)Mis an eigenvector ofᏹXC,DXfor the eigenvalue1,
(iii)MCMD=000.
(c)IfXis a fixed point ofφandMis an eigenvector ofᏹXC,DXfor the eigenvalue1and
(MX−1)2=000, then for all complexα,X+ (1−α)Mis a fixed point ofφ.
Proof. (a) Since{Xα} is an affine line in MnC, we can assume (by relabelling) thatXα
is a fixed point forα=0, 1, and some other value,β. ThusXandZare fixed points, so
X−1=I−CXDandZ−1=I−CZD. Thus
βX+ (1−β)Z−1=Xβ−1=I−CXβD=β(I−CXD) + (1−β)(I−CZD)
and thus
I=βX+ (1−β)ZβX−1+ (1−β)Z−1
=β2+ (1−β)2I +β(1−β)ZX−1+XZ−1. (A.5)
We may divide the last equation byβ(1−β) (which is nonzero), and thus obtainZX−1+
XZ−1=2I. Multiply this byγ(1−γ); by reversing the implications, we find thatγX+ (1−
γ)Zis a fixed point for everyγ.
(b) SetE=ZX−1; from the argument in (a), we haveE+E−1=2I. Thus (E−I)2=000. AsZ=EX−1, it follows thatM=(E−I)X, and thusMX−1=E−I, yielding (i).
Now setN=E−I, so thatN2=000,M=NX, andZ=(I +N)X. Asφ(Z)=Z, we have thatCZD=I−Z−1, and thusC(X+NX)D=I−X−1(I−N)=I−X−1N−X−1. Since
CXD=I−X−1, we inferCNXD=X−1N. Pre- and post-multiplying byX, we obtain
XC·NX·DX=NX, which is the conclusion of (ii).
AsCNXD=X−1D=X−1N, on premultiplying by M=NX, we deduceMCMD=
N2=000 (iii).
(c) SetZ=M+X. It suffices to show, as in the proof of (a), thatZX−1+XZ−1=2I. SetN=MX−1so thatZ=(I +N)X, and thusZ−1=X−1(I−N). HenceZX−1+XZ−1=
I +N+ I−N=2I.
Forλa nonzero complex number, consider the one-parameter family of mapsφλ:X→
(λI−CXD)−1. We note that eachφ
λ is strongly conjugate toφλ:X→(I−λ−2CXD)−1
via the functionΨλ: MnC→MnCdefined byΨ(Z)=λZ, that is,Ψ◦φλ=φλ◦Ψ. Since φλ=φ
C/λ,D/λ, eachφλcomes within the purview of this article, and we expect, for exam-
ple,2nnfixed points generically. Assume thatCDis invertible. Asλ→0, it is clear that
{φλ}converges to the functionρC,D :X→(−CXD)−1. This is obviously defined only on
GL(n,C). As a limiting case of the maps studied here, we expect it to have similar generic properties. It turns out that this map is rather special—it behaves like the commutative case discussed earlier.
PropositionA.4. ForCDinvertible, the fixed points ofρC,D:X→ −(CXD)−1satisfy the following.
(a)IfD−1Cis nonderogatory and has exactlykJordan blocks in its normal form, then
there are exactly2kfixed points, and they all commute with each other andD−1C. (b)IfD−1Chas a multiple eigenvector, then there is an affine line of fixed points.
Proof. LetZbe a fixed point, so thatZCZD=I. SetY=ZC, so we obtain the equation
Y2C−1D= −I, yieldingY2= −D−1C; conversely, for anyY satisfying such an equation,
Y C−1is a fixed point.
IfD−1Cis nonderogatory, it is straightforward that all the square roots of−D−1Cin MnClie in the algebra generated byD−1C, hence commute with each other andD−1C.
PuttingD−1Cin Jordan form and noting that it is invertible, it is readily verifiable that each block contributes exactly two square roots in the block, and thus there are 2ksquare
On the other hand, ifD−1Chas a multiple eigenvector, we obtain a line of fixed square roots by noticing that a piece of the Jordan form looks like (λ00λ), and off-diagonal square
roots of this abound.
The mappingsX→ −(CXD)−1have no attractive fixed points—fromCXD= −X−1, it follows that XCXD= −I, so that XC= −(XD)−1, and thusρ((XC)−1)=ρ(XD)=
ρ(DX). However,Ᏸφ(X)(Y)= −(DX)−1Y(CX)−1=ᏹ
−(DX)−1,(CX)−1(Y). IfXwas an at-
tractive fixed point, thenρ(Ᏸφ(X))<1; however, this is justρ((DX)−1)·ρ((CX)−1)=
ρ((DX)−1)·ρ(DX), which is always at least one. Since the inverse of the map isX→
−(DXC)−1, that is, of the same form, neither can have any repulsive fixed points either.