positive semidefinite rank
TRANSCRIPT
Positive Semidefinite Rank
Joint work with:
Hamza Fawzi (MIT)João Gouveia (U Coimbra)
Pablo Parrilo (MIT)Richard Robinson (UW)
Rekha R.Thomas University of Washington, Seattle
arXiv:1407.4095
Definition: Given M 2 Rp⇥q+
rkpsd(M) = min
8<
:k : 9A1, . . . , Ap 2 Sk
+
B1, . . . , Bq 2 (Sk+)
⇤
s.t. Mij = hAi, Bji
9=
;
Definition: Given M 2 Rp⇥q+
rkpsd(M) = min
8<
:k : 9A1, . . . , Ap 2 Sk
+
B1, . . . , Bq 2 (Sk+)
⇤
s.t. Mij = hAi, Bji
9=
;
M =
2
40 1 11 0 11 1 0
3
5
A1 =
1 00 0
�A2 =
0 00 1
�A3 =
1 �1
�1 1
�
B1 =
0 00 1
�B2 =
1 00 0
�B3 =
1 11 1
�
S2+-factorization of M :
rk(M) = 3
rkpsd(M) = 2
Definition: Given M 2 Rp⇥q+
rkpsd(M) = min
8<
:k : 9A1, . . . , Ap 2 Sk
+
B1, . . . , Bq 2 (Sk+)
⇤
s.t. Mij = hAi, Bji
9=
;
M =
2
40 1 11 0 11 1 0
3
5
A1 =
1 00 0
�A2 =
0 00 1
�A3 =
1 �1
�1 1
�
B1 =
0 00 1
�B2 =
1 00 0
�B3 =
1 11 1
�
S2+-factorization of M :
rk(M) = 3
rkpsd(M) = 2
rkpsd(I3) = 3
Factorizations of matrices/operators:
(i) factoring through subspaces:
rk(M) = min {k : M = AB, A 2 Rp⇥k, B 2 Rk⇥q}
i.e., 9 a1, . . . , ap 2 Rk, b1, . . . , bq 2 (Rk)⇤ s.t. Mij = hai, bji
Factorizations of matrices/operators:
C = {Ci} “closed’’ family of closed convex cones
cone rank of M GPT 2013
(i) factoring through subspaces:
rk(M) = min {k : M = AB, A 2 Rp⇥k, B 2 Rk⇥q}
(ii) factoring through cones:
C = {Rk+}
C = {Sk+}
nonnegative rank
psd rank
rk+(M)
rkpsd(M)
i.e., 9 a1, . . . , ap 2 Rk, b1, . . . , bq 2 (Rk)⇤ s.t. Mij = hai, bji
rkC(M) = min
⇢k : 9 a1, . . . , ap 2 Ck, b1, . . . , bq 2 (Ck)⇤
s.t. Mij = hai, bji
�
rk(M) = 1 , rkpsd(M) = 1 rk(M) = 2 , rkpsd(M) = 2
1
2
p1 + 8 rk(M)� 1
2 rkpsd(M) rk+(M) min(p, q)
rk(M) = 1 , rkpsd(M) = 1 rk(M) = 2 , rkpsd(M) = 2
1
2
p1 + 8 rk(M)� 1
2 rkpsd(M) rk+(M) min(p, q)
Mabc =
2
4a b cc a bb c a
3
5 , a, b, c � 0
rk = rkpsd = 1 , a = b = c
else, rk = 3 ) rkpsd = 2, 3
rkpsd = 2 ,a2 + b2 + c2 2(ab+ bc+ ac)
rk(M) = 1 , rkpsd(M) = 1 rk(M) = 2 , rkpsd(M) = 2
1
2
p1 + 8 rk(M)� 1
2 rkpsd(M) rk+(M) min(p, q)
Mabc =
2
4a b cc a bb c a
3
5 , a, b, c � 0
rk = rkpsd = 1 , a = b = c
else, rk = 3 ) rkpsd = 2, 3
rkpsd = 2 ,a2 + b2 + c2 2(ab+ bc+ ac)
Mp⇥qrkpsd=2 ⇢ Mp⇥q
rk=3 Kubjas-Robeva-Robinson 2014
Square root rank
rkp(M) := min {rkpM}
pM Hadamard square root of M
Square root rank
rkp(M) := min {rkpM}
pM Hadamard square root of M
pM =
2
41 2 33 �1 22 �3 �1
3
5 , rkpM = 2
M =
2
41 4 99 1 44 9 1
3
5 rkp(M) = 2
rk
p(M) 2
,(a, b, c) on this surface
Square root rank
rkp(M) := min {rkpM}
pM Hadamard square root of M
pM =
2
41 2 33 �1 22 �3 �1
3
5 , rkpM = 2
M =
2
41 4 99 1 44 9 1
3
5 rkp(M) = 2
rk
p(M) 2
,(a, b, c) on this surface
rk
p(M) = min {k : 9 Sk
+-factorization of M with rank one factors}
rkpsd(M) rkp(M)
M 2 {0, 1}p⇥q ) rkpsd(M) rk(M)Corollaries:
rk rk+ rkpsd rkp
rk = ⌧ ⌧, > ⌧, >rk+ � = � ⌧,�rkpsd <,� ⌧ = ⌧rkp <,� ⌧,� � =
Euclidean distance matrices:
Mij = (i� j)2rk(M) = 3, rkpsd(M) = 2
rk+(M) = ⇥(log2 n) Hrubes 2012
Cannot assume all factors have rank one to study psd rank.
Geometric Interpretation
⇡ linear map, L a�ne space
Def: P ✓ Rn(convex set) has a psd lift of size k if P = ⇡(Sk
+ \ L)
Geometric Interpretation
⇡ linear map, L a�ne space
Def: P ✓ Rn(convex set) has a psd lift of size k if P = ⇡(Sk
+ \ L)
P = conv(p1, . . . , pv) ⇢ Rn
Q = {x 2 Rn : aTj x �j , j = 1, . . . , f}
SP,Q 2 Rv⇥f , (SP,Q)ij := �j � a>j pislack matrix:
P ✓ Q
Geometric Interpretation
⇡ linear map, L a�ne space
Def: P ✓ Rn(convex set) has a psd lift of size k if P = ⇡(Sk
+ \ L)
P = conv(p1, . . . , pv) ⇢ Rn
Q = {x 2 Rn : aTj x �j , j = 1, . . . , f}
SP,Q 2 Rv⇥f , (SP,Q)ij := �j � a>j pislack matrix:
rkpsd(SP,Q) = min {k : 9 ⇡, L s.t. P ✓ ⇡(Sk+ \ L) ✓ Q}
rkpsd(SP,P ) = min {k : s.t. P = ⇡(Sk+ \ L)}
Thm:
Yannakakis 91:
GPT 2013: cone-lifts, P convex set, rkC(slack operators)
polyhedral-lifts, P polytope, rk+(SP,P )
SP =
2
664
1 1 0 00 1 1 00 0 1 11 0 0 1
3
775 , rkpsd = 3
P = [�1, 1]2 =
8<
:(x, y, z) 2 R3 :
2
41 x y
x 1 z
y z 1
3
5 ⌫ 0
9=
;
SP =
2
664
1 1 0 00 1 1 00 0 1 11 0 0 1
3
775 , rkpsd = 3
P = [�1, 1]2 =
8<
:(x, y, z) 2 R3 :
2
41 x y
x 1 z
y z 1
3
5 ⌫ 0
9=
;
Q = [�1, 1]2, P = [�a, a]⇥ [�b, b]
�a a
�b
b
0
0�1�1
1
1
SP,Q =
2
664
1 + a 1 + b 1� a 1� b1� a 1 + b 1 + a 1� b1� a 1� b 1 + a 1 + b1 + a 1� b 1� a 1 + b
3
775
rkpsd =
8><
>:
3 if a2 + b2 > 1
2 if 0 < a2 + b2 1
1 if a = b = 0
PSD ranks of polytopes
Briet-Dadush-Pokutta 2013: not all {0, 1}-polytopes in Rn
have small psd lifts
no concrete 0/1-polytope family so far with large psd rank
favorite candidate: COR(n) = conv(aa> : a 2 {0, 1}n)
PSD ranks of polytopes
Briet-Dadush-Pokutta 2013: not all {0, 1}-polytopes in Rn
have small psd lifts
no concrete 0/1-polytope family so far with large psd rank
no family so far with large gap between psd and nonnegative rank
GPT 13: rkpsd(P ) = k ) P has at most kO(k2n)facets
eg. psd rank of n-gons grow to infinity with n
rkpsd(generic n-gon) = ⌦(
4pn)GRT 13:
favorite candidate: COR(n) = conv(aa> : a 2 {0, 1}n)
Equivariant psd lifts of polytopes:
(lifts that respect the symmetries of the polytope -- a la Yannakakis)
FSP 2014:
Fawzi-Parrilo-Saunderson 2013:
Lee-Raghavendra-Steurer-Tan 2013
GPT 13:
provides exponential gap in sizes of LP and PSD equivariant lifts
n prime power
rk
eq+ (regular n-gon) = n
rk
eqpsd(regular n-gon) = ⇥(log n)
rkeqpsd(CUT(n)) �✓
n
dn/4e
◆
Lower bounds on psd rank
rkpsd(M2) rk(M)support not powerful: Lee-Theis 12
Barvinok 12:M has k distinct entries ) rkpsd(M)
✓k � 1 + rk(M)
k � 1
◆
Lower bounds on psd rank
rkpsd(M2) rk(M)support not powerful: Lee-Theis 12
Barvinok 12:M has k distinct entries ) rkpsd(M)
✓k � 1 + rk(M)
k � 1
◆
A lower bounding technique using results of Renegar:
(1) L \ Sk+ is bounded by polynomials of degree k
(2) bound poly describing algebraic boundary of ⇡(L \ Sk+) = P
(3) degree of this boundary poly � # facets of P
rkpsd(n-gon) = ⌦
slog n
log log n
!
GPT 13:e.g. ... deWolf-Lee-Wei 14
Polytopes of min psd rank Gouveia-Robinson-T. 13:
rk(SP ) = n+ 1 rkpsd(P )
equality holds , rk
p(SP ) = n+ 1
Theorem:
Polytopes of min psd rank Gouveia-Robinson-T. 13:
rk(SP ) = n+ 1 rkpsd(P )
equality holds , rk
p(SP ) = n+ 1
Theorem:
Theorem: rkpsd(STAB(G)) = n+ 1 , G perfect
in R2 :
in R3 : and their polars
biplanar
psd minimal matroid polytopesGrande-Sanyal 14:
a corollary ....
Fawzi-Gouveia-Robinson 14
M is the slack matrix of a biplanar cuboid
Open for nonnegative rank
2
66666666664
0 0 2 1 0 11 0 0 2 0 10 1 2 0 0 11 2 0 0 0 10 0 2 1 1 01 0 0 2 1 00 1 2 0 1 01 2 0 0 1 0
3
77777777775
9M : rkQpsd(M) > rkRpsd(M)
PSD rank of convex sets
Helton-Nie Conjecture: Every convex semi-algebraic set has finite psd rank.
Long history of constructing psd lifts of interesting convex sets in optimization/control/real algebraic geometry ....
i.e. show rkpsd(C) < 1
Scheiderer 12: true for convex hulls of algebraic curves
C ⇢ Rn ( rkpsd(C) = rkpsd(slack operator of C)
Complexity issues:
Robinson 14: Square root rank is NP-hard to compute.
2
666664
1 0 · · · 0 a210 1 · · · 0 a22...
...0 0 · · · 1 a2n1 1 · · · 1 0
3
777775
Complexity issues:
Robinson 14: Square root rank is NP-hard to compute.
Is psd rank NP-hard to compute?Vavasis 2009: nonnegative rank is NP-hard
2
666664
1 0 · · · 0 a210 1 · · · 0 a22...
...0 0 · · · 1 a2n1 1 · · · 1 0
3
777775
Complexity issues:
Robinson 14: Square root rank is NP-hard to compute.
Is psd rank NP-hard to compute?Vavasis 2009: nonnegative rank is NP-hard
For fixed k, is it NP-hard to decide if rkpsd(M) k?
poly-time decidable for nonnegative rankArora-Ge-Kannan-Moitra 12, Moitra 13:
2
666664
1 0 · · · 0 a210 1 · · · 0 a22...
...0 0 · · · 1 a2n1 1 · · · 1 0
3
777775
Some of the things I did not talk about ...
• Space of psd factorizations• Quantum information interpretation of psd rank• Approximate factorizations and approximations of polytopes • Symmetric psd factorizations