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= (d+1)K2
st−d+1
dd2/summationdisplay
r=1K2
rsK2
rt
=d2δst−1
d+1(299)
from which it follows
d2/summationdisplay
r=1Qr=d2/summationdisplay
r=1QT
r=d2
d+1/parenleftbig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig
(300)
Eq. (290) follows from this and the fact that Rr=Qr+QT
r.
/square
8.Geometrical Considerations
In this section we show that there are some interesting geometrica l relationships
between the subspaces onto which the operators Qr,QT
rand¯Rrproject. The
original motivation for this work was an observation concerning the subspaces onto
which the ¯Rrproject. ¯Rris a real matrix, and so it deﬁnes a 2( d−2) subspace
inRd2, which we will denote Rr. We noticed that for each pair of distinct indices
randsthe intersection Rr∩Rsis a 1-dimensional line. This led us to speculate
that a set of hyperplanes parallel to the Rrmight be the edges of an interesting
polytope. We continue to think that this could be the case. Unfortu nately we have
not been able to prove it. However, it appears to us that the result s we obtained37
while trying to prove it have an interest which is independent of the tr uth of the
motivating speculation.
We will begin with some terminology. Let Pbe any projector (on either RN
orCN), letPbe the subspace onto which Pprojects, and let \|ψ/an}bracketri}htbe any non-zero
vector. Then we deﬁne the angle between \|ψ/an}bracketri}htandPin the usual way, to be
θ= cos−1/parenleftigg/vextenddouble/vextenddoubleP\|ψ/an}bracketri}ht/vextenddouble/vextenddouble
/vextenddouble/vextenddouble\|ψ/an}bracketri}ht/vextenddouble/vextenddouble/parenrightigg
(301)
(soθis the smallest angle between \|ψ/an}bracketri}htand any of the vectors in P).
Suppose, now, that P′is another projector, and let P′be the subspace onto
whichP′projects. We will say that P′is uniformly inclined to Pif every vector in
P′makes the same angle θwithP. Ifθ= 0 this means that P′⊆P, while ifθ=π
2
it means P′⊥P. Suppose, on the other hand, that 0 < θ <π
2. Let\|u′
1/an}bracketri}ht,...,\|u′
n/an}bracketri}ht
be any orthonormal basis for P′, and deﬁne \|ur/an}bracketri}ht= secθP\|u′
r/an}bracketri}ht. Then/an}bracketle{tur\|ur/an}bracketri}ht= 1
for allr. Moreover, if P,P′are complex projectors,
/an}bracketle{tu′
r+eiφu′
s\|P\|u′
r+eiφu′
s/an}bracketri}ht= 2cos2θ/parenleftig
1+Re/parenleftbig
eiφ/an}bracketle{tur\|us/an}bracketri}ht/parenrightbig/parenrightig
(302)
for allφandr/ne}ationslash=s. On the other hand it follows from the assumption that P′is
uniformly inclined to Pthat
/an}bracketle{tu′
r+eiφu′
s\|P\|u′
r+eiφu′
s/an}bracketri}ht= 2cos2θ (303)
for allφandr/ne}ationslash=s. Consequently
/an}bracketle{tur\|us/an}bracketri}ht=δrs (304)
for allr,s. It is easily seen that the same is true if P,P′are real projectors.
Suppose we now make the further assumption that dim( P′) = dim( P) =n. Then
\|u1/an}bracketri}ht,...,\|un/an}bracketri}htis an orthonormal basis for P, and we can write
P=n/summationdisplay
r=1\|ur/an}bracketri}ht/an}bracketle{tur\| (305)
P′=n/summationdisplay
r=1\|u′
r/an}bracketri}ht/an}bracketle{tu′
r\| (306)
Observe that
/an}bracketle{tu′
r\|us/an}bracketri}ht=/an}bracketle{tu′
r\|P\|us/an}bracketri}ht= cosθ/an}bracketle{tur\|us/an}bracketri}ht= cosθδrs (307)
for allr,s. Consequently
P′\|ur/an}bracketri}ht= cosθ\|ur/an}bracketri}ht (308)
for allr. It follows that
/vextenddouble/vextenddoubleP′\|ψ/an}bracketri}ht/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay
r=1cosθ/an}bracketle{tur\|ψ/an}bracketri}ht\|u′
r/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble\|ψ/an}bracketri}ht/vextenddouble/vextenddouble (309)
for all\|ψ/an}bracketri}ht ∈P. SoPis uniformly inclined to P′at the same angle θ.
It follows from Eqs. ( 305) and (306) that
PP′P= cos2θP (310)
P′PP′= cos2θP′(311)
Eq. (310), or equivalently Eq. ( 311), is not only necessary but also suﬃcient for
the subspaces to be uniformly inclined. In fact, let P,P′be any two subspaces38
which have the same dimension n, but which are not assumed at the outset to be
uniformly inclined, and let P,P′be the corresponding projectors. Suppose
PP′P= cos2θP (312)
for someθin the range 0 ≤θ≤π
2. It is immediate that P=P′ifθ= 0, and
P⊥P′ifθ=π
2. Either way, the subspaces are uniformly inclined. Suppose, on
the other hand, that 0 <θ<π

= (d+1)K2

st−d+1

dd2/summationdisplay

r=1K2

rsK2

rt

=d2δst−1

d+1(299)

from which it follows

d2/summationdisplay

r=1Qr=d2/summationdisplay

r=1QT

r=d2

d+1/parenleftbig

I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig

(300)

Eq. (290) follows from this and the fact that Rr=Qr+QT

r.

/square

8.Geometrical Considerations

In this section we show that there are some interesting geometrica l relationships

between the subspaces onto which the operators Qr,QT

rand¯Rrproject. The

original motivation for this work was an observation concerning the subspaces onto

which the ¯Rrproject. ¯Rris a real matrix, and so it deﬁnes a 2( d−2) subspace

inRd2, which we will denote Rr. We noticed that for each pair of distinct indices

randsthe intersection Rr∩Rsis a 1-dimensional line. This led us to speculate

that a set of hyperplanes parallel to the Rrmight be the edges of an interesting

polytope. We continue to think that this could be the case. Unfortu nately we have

not been able to prove it. However, it appears to us that the result s we obtained37

while trying to prove it have an interest which is independent of the tr uth of the

motivating speculation.

We will begin with some terminology. Let Pbe any projector (on either RN

orCN), letPbe the subspace onto which Pprojects, and let |ψ/an}bracketri}htbe any non-zero

vector. Then we deﬁne the angle between |ψ/an}bracketri}htandPin the usual way, to be

θ= cos−1/parenleftigg/vextenddouble/vextenddoubleP|ψ/an}bracketri}ht/vextenddouble/vextenddouble

/vextenddouble/vextenddouble|ψ/an}bracketri}ht/vextenddouble/vextenddouble/parenrightigg

(301)

(soθis the smallest angle between |ψ/an}bracketri}htand any of the vectors in P).

Suppose, now, that P′is another projector, and let P′be the subspace onto

whichP′projects. We will say that P′is uniformly inclined to Pif every vector in

P′makes the same angle θwithP. Ifθ= 0 this means that P′⊆P, while ifθ=π

2

it means P′⊥P. Suppose, on the other hand, that 0 < θ <π

2. Let|u′

1/an}bracketri}ht,...,|u′

n/an}bracketri}ht

be any orthonormal basis for P′, and deﬁne |ur/an}bracketri}ht= secθP|u′

r/an}bracketri}ht. Then/an}bracketle{tur|ur/an}bracketri}ht= 1

for allr. Moreover, if P,P′are complex projectors,

/an}bracketle{tu′

r+eiφu′

s|P|u′

r+eiφu′

s/an}bracketri}ht= 2cos2θ/parenleftig

1+Re/parenleftbig

eiφ/an}bracketle{tur|us/an}bracketri}ht/parenrightbig/parenrightig

(302)

for allφandr/ne}ationslash=s. On the other hand it follows from the assumption that P′is

uniformly inclined to Pthat

/an}bracketle{tu′

r+eiφu′

s|P|u′

r+eiφu′

s/an}bracketri}ht= 2cos2θ (303)

for allφandr/ne}ationslash=s. Consequently

/an}bracketle{tur|us/an}bracketri}ht=δrs (304)

for allr,s. It is easily seen that the same is true if P,P′are real projectors.

Suppose we now make the further assumption that dim( P′) = dim( P) =n. Then

|u1/an}bracketri}ht,...,|un/an}bracketri}htis an orthonormal basis for P, and we can write

P=n/summationdisplay

r=1|ur/an}bracketri}ht/an}bracketle{tur| (305)

P′=n/summationdisplay

r=1|u′

r/an}bracketri}ht/an}bracketle{tu′

r| (306)

Observe that

/an}bracketle{tu′

r|us/an}bracketri}ht=/an}bracketle{tu′

r|P|us/an}bracketri}ht= cosθ/an}bracketle{tur|us/an}bracketri}ht= cosθδrs (307)

for allr,s. Consequently

P′|ur/an}bracketri}ht= cosθ|ur/an}bracketri}ht (308)

for allr. It follows that

/vextenddouble/vextenddoubleP′|ψ/an}bracketri}ht/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay

r=1cosθ/an}bracketle{tur|ψ/an}bracketri}ht|u′

r/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble|ψ/an}bracketri}ht/vextenddouble/vextenddouble (309)

for all|ψ/an}bracketri}ht ∈P. SoPis uniformly inclined to P′at the same angle θ.

It follows from Eqs. ( 305) and (306) that

PP′P= cos2θP (310)

P′PP′= cos2θP′(311)

Eq. (310), or equivalently Eq. ( 311), is not only necessary but also suﬃcient for

the subspaces to be uniformly inclined. In fact, let P,P′be any two subspaces38

which have the same dimension n, but which are not assumed at the outset to be

uniformly inclined, and let P,P′be the corresponding projectors. Suppose

PP′P= cos2θP (312)

for someθin the range 0 ≤θ≤π

2. It is immediate that P=P′ifθ= 0, and

P⊥P′ifθ=π

2. Either way, the subspaces are uniformly inclined. Suppose, on

the other hand, that 0 <θ<π