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for gl(d,C) means the lrcannot all be zero. So we must in fact have l>0. Using
this result in Eq. ( 188) we find
β+d2l2γ=dl2(216)
while Eq. ( 214) implies
dβ+dl2γ=d−1+l2(217)
This gives us a pair of simultaneous equations in βandγ. Solving them we obtain
β=d
d+1(218)
γ=1
dl2/parenleftbigg
l2−1
d+1/parenrightbigg
(219)
Substituting these expressions into Eqs. ( 186) and (187) we deduce Eqs. ( 210)
and (211). /square
The next lemma shows that each Lris a linear combination of a rank-1projector
and the identity:
Lemma 11. LetLbe any Hermitian matrix ∈gl(d,C)which is not a multiple of
the identity. Then
rank(ad L)≥2(d−1) (220)
The lower bound is achieved if and only if Lis of the form
L=ηI+ξP (221)
wherePis a rank- 1projector and η,ξare any pair of real numbers. The eigenvalues
ofadLare then ±ξ(each with multiplicity d−1) and0(with multiplicity d2−2d+2).
Proof.Letλ1≥λ2≥ ··· ≥λdbe the eigenvalues of Larranged in decreasing
order, and let |b1/an}bracketri}ht,|b2/an}bracketri}ht,...,|bd/an}bracketri}htbe the correspondingeigenvectors. We mayassume,
without loss of generality, that the |br/an}bracketri}htare orthonormal. We have
adL/parenleftbig
|br/an}bracketri}ht/an}bracketle{tbs|/parenrightbig
=/bracketleftbig
L,|br/an}bracketri}ht/an}bracketle{tbs|/bracketrightbig
= (λr−λs)|br/an}bracketri}ht/an}bracketle{tbs| (222)
So the eigenvalues of ad Lareλr−λs. SinceLis not a multiple of the identity we
must haveλr/ne}ationslash=λr+1for somerin the range 1 ≤r≤d−1. We then have that
λs−λt/ne}ationslash= 0 if either s≤r<tort≤r<s. There are 2 r(d−r) such pairs s,t. So
rank(ad L)≥2r(d−r)≥2(d−1) (223)
Suppose, now that the lower bound is achieved. Then r(d−r) =d−1, implying
thatr= 1 ord−1. Also we must have λs=λs+1for alls/ne}ationslash=r. So either
L=λ2I+(λ1−λ2)|b1/an}bracketri}ht/an}bracketle{tb1| (224)
or
L=λd−1I−(λd−1−λd)|bd/an}bracketri}ht/an}bracketle{tbd| (225)
Either way Land the spectrum of ad Lare as described. /square
The final ingredient needed to complete the proof is29
Lemma 12. LetLr,CrstandCrbe as defined in the statement of Theorem 7.
Suppose that the Crhave the spectral decomposition
Cr=Pr−PT
r (226)
wherePris a rankd−1projector which is orthogonal to its own transpose. Let l,
ǫ′
rbe as in the statement of Corollary 10. Then there is a fixed sign ǫ=±1such
that
Πr=ǫǫ′
rLr−ǫl−1
dI (227)
is a rank- 1projector for all r.
Proof.Define
L′
r=ǫ′
rLr−l−1
dI (228)
Then it follows from Corollary 10that
Tr(L′
r) = 1 (229)
for allr,
Tr(L′
rL′
s) =dδrs+1
d+1(230)
for allr,s, and
d2/summationdisplay
r=1L′
r=dI (231)
It is also easily seen that if we define C′
rst=ǫ′
rǫ′
sǫ′
tCrstthen
[L′
r,L′
s] =d2/summationdisplay
t=1C′
rstL′
t (232)
and
C′
r=P′
r−P′
rT(233)
whereP′
ris a rank-1 projector which is orthogonal to its own transpose (se e the
first part of the proof of Theorem 7). In particular
rank/parenleftbig
adL′r/parenrightbig
= 2(d−1) (234)
and the eigenvalues of ad L′rall equal to ±1 or 0. So, taking account of the fact
that Tr(L′
r) = 1, we can use Lemma 11to deduce that there is a family of rank-1
projectors Π′
rand signsξr=±1 such that
L′