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2. Let|u′
1/an}bracketri}ht,...,|u′
n/an}bracketri}htbe any orthonormal basis for P′,
and define |ur/an}bracketri}ht= secθP|u′
r/an}bracketri}ht. Eq. (305) then implies
P= sec2θn/summationdisplay
r=1P|u′
r/an}bracketri}ht/an}bracketle{tu′
r|P=n/summationdisplay
r=1|ur/an}bracketri}ht/an}bracketle{tur| (313)
Given any |ψ/an}bracketri}ht ∈Pwe have
|ψ/an}bracketri}ht=P|ψ/an}bracketri}ht=n/summationdisplay
r=1/an}bracketle{tur|ψ/an}bracketri}ht|ur/an}bracketri}ht (314)
Since dim( P) =nit follows that the |ur/an}bracketri}htare linearly independent. In particular
|ur/an}bracketri}ht=P|ur/an}bracketri}ht=n/summationdisplay
s=1/an}bracketle{tus|ur/an}bracketri}ht|us/an}bracketri}ht (315)
Since the |ur/an}bracketri}htare linearly independent this means
/an}bracketle{tus|ur/an}bracketri}ht=δrs (316)
So the|ur/an}bracketri}htare an orthonormal basis for P. It follows, that if |ψ′/an}bracketri}htis any vector in
P′, then
/vextenddouble/vextenddoubleP|ψ′/an}bracketri}ht/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay
r=1/an}bracketle{tu′
r|ψ′/an}bracketri}htP|u′
r/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay
r=1/an}bracketle{tu′
r|ψ′/an}bracketri}ht|ur/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble|ψ′/an}bracketri}ht/vextenddouble/vextenddouble(317)
implying that P′is uniformly inclined to Pat angleθ.
It will be convenient to summarise all this in the form of a lemma:
Lemma 15. LetP,P′be any two subspaces, real or complex, having the same
dimensionn. LetP,P′be the corresponding projectors. Then the following state-
ments are equivalent:
(a)P′is uniformly inclined to Pat angleθ.
(b)Pis uniformly inclined to P′at angleθ.
(c)
PP′P= cos2θP (318)
(d)
P′PP′= cos2θP′(319)
Suppose these conditions are satisfied for some θin the range 0< θ <π
2, and
let|u1/an}bracketri}ht,...|un/an}bracketri}htbe any orthonormal basis for P. Then there exists an orthonormal
basis|u′
1/an}bracketri}ht,...,|u′
n/an}bracketri}htforP′such that
P′|ur/an}bracketri}ht= cosθ|u′
r/an}bracketri}ht (320)
P|u′
r/an}bracketri}ht= cosθ|ur/an}bracketri}ht (321)
We are now in a position to state the main results of this section. Let Qr
(respectively ¯Qr) be the subspace onto which Qr(respectively QT
r) projects. We
then have39
Theorem 16. For each pair of distinct indices r,sthe subspaces Qr,¯Qrhave the
orthogonal decomposition
Qr=Q0
rs⊕Qrs (322)
¯Qr=¯Q0
rs⊕¯Qrs (323)
where
Q0
rs⊥Qrs dim(Q0
rs) = 1 dim( Qrs) =d−2
¯Q0
rs⊥¯Qrs dim(¯Q0
rs) = 1 dim( ¯Qrs) =d−2
We have
(a)Relation of the subspaces QrandQs:
(1)Q0
rs⊥QsrandQrs⊥Q0
sr.
(2)Q0
rsandQ0
srare inclined at angle cos−1/parenleftbig1
d+1/parenrightbig
.
(3)QrsandQsrare uniformly inclined at angle cos−1/parenleftig
1√d+1/parenrightig
.
(b)Relation of the subspaces ¯Qrand¯Qs:
(1)¯Q0
rs⊥¯Qsrand¯Qrs⊥¯Q0
sr.
(2)¯Q0
rsand¯Q0
srare inclined at angle cos−1/parenleftbig1
d+1/parenrightbig
.
(3)¯Qrsand¯Qsrare uniformly inclined at angle cos−1/parenleftig
1√d+1/parenrightig
.
(c)Relation of the subspaces Qrand¯Qs:
(1)Q0
rs⊥¯Qsr,Qrs⊥¯Q0
srandQrs⊥¯Qsr.
(2)Q0
rsand¯Q0
srare inclined at angle cos−1/parenleftbigd
d+1/parenrightbig
.
The relations between these subspaces are, perhaps, easier to a ssimilate if pre-
sented pictorially. In the following diagrams the line joining each pair of subspaces