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is labelled with the cosine of the angle between them. In particular a 0 o n the line
joining two subspaces indicates that they are orthogonal.
Q0
rs Qrs
Q0
sr Qsr0
01
d+11√d+1
0❅
❅
❅
❅
❅
❅❅0¯Q0
rs¯Qrs
¯Q0
sr¯Qsr0
01
d+11√d+1
0❅
❅
❅
❅
❅
❅❅0
Q0
rs Qrs
¯Q0
sr¯Qsr0
0d
d+10
0❅
❅
❅
❅
❅
❅❅040
Wewillprovethistheorembelow. Beforedoingso,however,letusst atetheother
mainresult ofthis section. Let Rrbe the subspace ontowhichthe ¯Rrproject. Since
¯Rris a real matrix we regard Rras a subspace of Rd2. We have
Theorem 17. For each pair of distinct indices r,sthe subspace Rrhas the decom-
position
Rr=R0
rs⊕R1
rs⊕Rrs (324)
whereR0
rs,R1
rs,Rrsare pairwise orthogonal and
dim(R0
rs) = 1 dim( R1
rs) = 1 dim( Rrs) = 2d−4 (325)
We have
(1)R0
rs=R0
sr.
(2)R1
rs⊥RsrandRrs⊥R1
sr.
(3)R1
rsandR1
srare inclined at angle cos−1/parenleftbigd−1
d+1/parenrightbig
.
(4)RrsandRsrare uniformly inclined at angle cos−1/parenleftig/radicalig
1
d+1/parenrightig
In particular, the subspaces ¯Rrand¯Rsintersect in a line.
In diagrammatic form the relations between these subspaces are
R0
rs=R0
srR1
rs Rrs
R1
sr Rsr0
0d−1
d+1/radicalig
1
d+1
0❅
❅
❅
❅
❅
❅❅0✟✟✟✟✟0
❍❍❍❍❍00
0
where, as before, each line is labelled with the cosine of the angle betw een the two
subspaces it connects.
Proof of Theorem 16.Let/bardbl1/an}bracketri}ht/an}bracketri}ht,...,/bardbld2/an}bracketri}ht/an}bracketri}htbethestandardbasisfor Hd2, asdefined
by Eq. (117). For each pair of distinct indices r,sdefine
/bardblfrs/an}bracketri}ht/an}bracketri}ht=i√
d+1Qr/bardbls/an}bracketri}ht/an}bracketri}ht (326)
/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht=−i√
d+1QT
r/bardbls/an}bracketri}ht/an}bracketri}ht (327)