Datasets:
alx-ai
/
arxiv_papers

Dataset card Data Studio Files Community
text
stringlengths
0
44.4k
s/bardbler/an}bracketri}ht/an}bracketri}ht=d
2(d+1)/parenleftbig
3K2
rs+1/parenrightbig
(367)
where in deriving Eq. ( 364) we used the fact that GraGasGsbGbr= (d+1)TrasTrsb
(in view of the fact that r/ne}ationslash=s). Substituting these expressions into Eqs. ( 358)
and (359) we deduce Eqs. ( 356) and (357). /square
Now define the rank d−1 projectors
Qrs=Qr−/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (368)
QT
rs=QT
r−/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗
rs/bardbl (369)
andletQ0
rs,Qrs,¯Q0
rsand¯Qrsbe, respectively, the subspacesontowhich /bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl,
Qrs,/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗
rs/bardblandQ∗
rsproject. It is immediate that we have the orthogonal
decompositions
Qr=Q0
rs⊕Qrs (370)
¯Qr=¯Q0
rs⊕¯Qrs (371)
Using Lemma 18we find
Qsr/bardblfrs/an}bracketri}ht/an}bracketri}ht=Qrs/bardblfsr/an}bracketri}ht/an}bracketri}ht= 0 (372)45
implying that Q0
rs⊥QsrandQrs⊥Q0
sr, and
/vextendsingle/vextendsingle/an}bracketle{t/an}bracketle{tfrs/bardblfsr/an}bracketri}ht/an}bracketri}ht/vextendsingle/vextendsingle=1
d+1(373)
implying that Q0
rsandQ0
srare inclined at angle cos−1/parenleftbig1
d+1/parenrightbig
. Using Lemma 18
together with Lemma 19we find
QrsQsrQrs=QrsQsQrs
=QrQsQr−/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardblQsQr−QrQs/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl
+/an}bracketle{t/an}bracketle{tfrs/bardblQs/bardblfrs/an}bracketri}ht/an}bracketri}ht/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl
=1
d+1Qr−1
d+1/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl
=1
d+1Qrs (374)
which in view of Lemma 15implies that QrsandQsrare uniformly inclined at angle
cos−1/parenleftbig1√d+1/parenrightbig
. This proves part (a) of the theorem. Parts (b) and (c) are prov ed
similarly.
Proof of Theorem 17.Define
/bardblgrs/an}bracketri}ht/an}bracketri}ht=1√
2/parenleftbig
/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht+/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightbig
(375)
/bardbl¯grs/an}bracketri}ht/an}bracketri}ht=i√
2/parenleftbig
/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht−/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightbig
(376)
By construction the components of /bardblgrs/an}bracketri}ht/an}bracketri}ht,/bardbl¯grs/an}bracketri}ht/an}bracketri}htin the standard basis are real, so
we can regard them as ∈Rd2. They are orthonormal:
/an}bracketle{t/an}bracketle{tgrs/bardblgrs/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{t¯grs/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 1 and /an}bracketle{t/an}bracketle{tgrs/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 0 (377)
It is also readily verified, using Lemma 18, that
¯Rr/bardblgrs/an}bracketri}ht/an}bracketri}ht=/bardblgrs/an}bracketri}ht/an}bracketri}ht (378)
¯Rr/bardbl¯grs/an}bracketri}ht/an}bracketri}ht=/bardbl¯grs/an}bracketri}ht/an}bracketri}ht (379)
So
Rrs=¯Rr−/bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl−/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardbl (380)
is a rank 2 d−4 projector. If we define R0
rs,R1
rsandRrsto be, respectively,
the subspaces onto which /bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl,/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardblandRrsproject we have the
orthogonal decomposition
Rr=R0
rs⊕R1
rs⊕Rrs (381)
It follows from Eqs. ( 333) and (334) that
/bardblgrs/an}bracketri}ht/an}bracketri}ht=−/bardblgsr/an}bracketri}ht/an}bracketri}ht (382)
implying that R0
rs=R0
srfor allr/ne}ationslash=s. It is also easily verified, using Lemma 18,
that/vextendsingle/vextendsingle/an}bracketle{t/an}bracketle{t¯grs/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht/vextendsingle/vextendsingle=d−1
d+1(383)
from which it follows that R1
rsandR1
srare inclined at angle cos−1/parenleftbigd−1
d+1/parenrightbig
. We next
observe that
Rrs=Qrs+QT
rs (384)46
Using Lemma 18once again we deduce
Rrs/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht=Rsr/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 0 (385)
from which it follows that R1
rs⊥RsrandRrs⊥R1
sr. Finally, we know from
Theorem 16thatQT