Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 ·...
Transcript of Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 ·...
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The Solution of the Kadison-Singer Problem
Adam Marcus (Crisply, Yale) Daniel Spielman (Yale)
Nikhil Srivastava (MSR India)
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Outline
Disclaimer The Kadison-Singer Problem, defined. Restricted Invertibility, a simple proof. Break Kadison-Singer, outline of proof.
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The Kadison-Singer Problem (‘59)
A posi've solu'on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich'nger Conjecture (‘05) Many others
Implied by:
Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS2 Conjecture
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A posi've solu'on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich'nger Conjecture (‘05) Many others
Implied by:
Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS2 Conjecture
The Kadison-Singer Problem (‘59)
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A posi've solu'on is equivalent to: Anderson’s Paving Conjectures (‘79, ‘81) Bourgain-‐Tzafriri Conjecture (‘91) Feich'nger Conjecture (‘05) Many others
Implied by:
Akemann and Anderson’s Paving Conjecture (‘91) Weaver’s KS2 Conjecture
The Kadison-Singer Problem (‘59)
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Let be a maximal Abelian subalgebra of , the algebra of bounded linear operators on Let be a pure state. Is the extension of to unique?
A
⇢ : A ! C⇢
See Nick Harvey’s Survey or Terry Tao’s Blog
B(`2(N))`2(N)
B(`2(N))
The Kadison-Singer Problem (‘59)
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Anderson’s Paving Conjecture ‘79
For all there is a k so that for every n-by-n symmetric matrix A with zero diagonals, there is a partition of into
kA(Sj , Sj)k ✏ kAk for j = 1, . . . , k
✏ > 0
Recall kAk = max
kxk=1kAxk
{1, ..., n} S1, ..., Sk
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0
@
1
A
Anderson’s Paving Conjecture ‘79
For all there is a k so that for every n-by-n symmetric matrix A with zero diagonals, there is a partition of into
kA(Sj , Sj)k ✏ kAk for j = 1, . . . , k
✏ > 0
Recall kAk = max
kxk=1kAxk
{1, ..., n} S1, ..., Sk
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kAk = supkxk=1
kAxk
Anderson’s Paving Conjecture ‘79
For all there is a k so that for every self-adjoint bounded linear operator A on , there is a partition of into
kA(Sj , Sj)k ✏ kAk for j = 1, . . . , k
✏ > 0`2
N S1, ..., Sk
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Anderson’s Paving Conjecture ‘79
For all there is a k so that for every n-by-n symmetric matrix A with zero diagonals, there is a partition of into
kA(Sj , Sj)k ✏ kAk for j = 1, . . . , k
✏ > 0
{1, ..., n} S1, ..., Sk
Is equivalent if restrict to projection matrices. [Casazza, Edidin, Kalra, Paulsen ‘07]
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Anderson’s Paving Conjecture ‘79
There exist an and a k so that for such that and then exists a partition of into k parts s.t.
✏ > 0
kvik2 1/2
{1, . . . , n}
Equivalent to [Harvey ‘13]:
v1, ..., vn 2 Cd
Pviv⇤i = I
���P
i2Sjviv⇤i
��� 1� ✏
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Moments of Vectors The moment of vectors in the direction of a unit vector is
v1, ..., vnu
v1
v2
v1
v2
v1
v2
u
X
i
(vTi u)2
2.5
u
u1 4
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Moments of Vectors The moment of vectors in the direction of a unit vector is
v1, ..., vnu
X
i
(vTi u)2 =
X
i
uT (vivTi )u
= uT
X
i
vivTi
!u
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Vectors with Spherical Moments
v1
�v1�v2
�v3
�v4
v4
v3
v2
For every unit vector uX
i
(vTi u)2 = 1u
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Vectors with Spherical Moments
v1
�v1�v2
�v3
�v4
v4
v3
v2
For every unit vector uX
i
(vTi u)2 = 1
X
i
vivTi = I
u
Also called isotropic position
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Partition into Approximately ½-Spherical Sets S1 S2
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Partition into Approximately ½-Spherical Sets
1/4 P
i2Sj(vTi u)
2 3/4
S1 S2
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Partition into Approximately ½-Spherical Sets
1/4 P
i2Sj(vTi u)
2 3/4
1/4 eigs(P
i2SjvivTi ) 3/4
S1 S2
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Partition into Approximately ½-Spherical Sets
1/4 P
i2Sj(vTi u)
2 3/4
1/4 eigs(P
i2SjvivTi ) 3/4
S1 S2
eigs(P
i2SjvivTi ) 3/4
X
i2S1
vivTi = I �
X
i2S2
vivTibecause
()
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Partition into Approximately ½-Spherical Sets
1/4 P
i2Sj(vTi u)
2 3/4
1/4 eigs(P
i2SjvivTi ) 3/4
S1 S2
X
i2S1
vivTi = I �
X
i2S2
vivTibecause
()���P
i2SjvivTi
��� 3/4
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S1
S2
Big vectors make this difficult
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S1 S2
Big vectors make this difficult
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Weaver’s Conjecture KS2
There exist positive constants and so that if all then exists a partition into S1 and S2 with
↵ ✏
kvik ↵
eigs(P
i2Sjviv⇤i ) 1� ✏
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Weaver’s Conjecture KS2
There exist positive constants and so that if all then exists a partition into S1 and S2 with
↵ ✏
Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer
kvik ↵
eigs(P
i2Sjviv⇤i ) 1� ✏
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eigs(P
i2Sjviv⇤i ) 1
2 + 3↵
Main Theorem
For all if all then exists a partition into S1 and S2 with
↵ > 0
kvik ↵
Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer
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A Random Partition? Works with high probability if all
(by Tropp ‘11, variant of Matrix Chernoff, Rudelson) kvik2 O(1/ log d)
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A Random Partition? Works with high probability if all
(by Tropp ‘11, variant of Matrix Chernoff, Rudelson)
Troublesome case: each is a scaled axis vector
kvik = ↵
are of each 1/↵2
kvik2 O(1/ log d)
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A Random Partition? Works with high probability if all
(by Tropp ‘11, variant of Matrix Chernoff, Rudelson)
Troublesome case: each is a scaled axis vector
kvik = ↵
are of each 1/↵2
chance that all in one direction land in same set is 2�1/↵2
kvik2 O(1/ log d)
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A Random Partition? Works with high probability if all
(by Tropp ‘11, variant of Matrix Chernoff, Rudelson)
Troublesome case: each is a scaled axis vector
kvik = ↵
are of each 1/↵2
chance that all in one direction land in same set is 2�1/↵2
kvik2 O(1/ log d)
Chance there exists a direction in which all land in same set is
1�⇣1� 2�1/↵2
⌘d! 1
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The Graphical Case
From a graph G = (V,E) with |V| = n and |E| = m Create m vectors in n dimensions:
va,b(c) =
8><
>:
1 if c = a
�1 if c = b
0 otherwise
X
(a,b)2E
va,bvTa,b = LG
If G is a good d-regular expander, all eigs close to d very close to spherical
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Partitioning Expanders
Can partition the edges of a good expander to obtain two expanders.
Broder-Frieze-Upfal ‘94: construct random partition guaranteeing degree at least d/4, some expansion Frieze-Molloy ‘99: Lovász Local Lemma, good expander Probability is works is low, but can prove non-zero
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Interlacing Families of Polynomials
A new technique for proving existence from very low probabilities
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Pi viv
⇤i = I
Restricted Invertibility (Bourgain-Tzafriri)
Special case: For with for every there is a so that
v1, ..., vn 2 Cd
k d S ⇢ {1, ..., n} , |S| = k
�k
�Pi2S viv⇤i
��
✓1�
qkd
◆2dn
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Restricted Invertibility (Bourgain-Tzafriri)
Special case: For with for every there is a so that
v1, ..., vn 2 Cd
k d S ⇢ {1, ..., n} , |S| = k
�k
�Pi2S viv⇤i
��
✓1�
qkd
◆2dn
Is far from singular on the span of {vi}i2S
Pi viv
⇤i = I
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�1(vv⇤) = v⇤v = kvk2 ⇡ dn
Restricted Invertibility (Bourgain-Tzafriri)
For with for every there is a so that
v1, ..., vn 2 Cd
k d S ⇢ {1, ..., n} , |S| = k
�k
�Pi2S viv⇤i
��
✓1�
qkd
◆2dn
For says , while
k = 1 �1(vv⇤) & dn
Similar bound for k a constant fraction of d!
Pi viv
⇤i = I
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Method of proof
Let be chosen uniformly from {v1, ..., vn}r1, ..., rk
1. is real rooted E�
hXrjr
⇤j
i(x)
the characteris*c polynomial in the variable x of the matrix inside the brackets
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Method of proof
Let be chosen uniformly from {v1, ..., vn}r1, ..., rk
1. is real rooted
2.
the k-‐th root of the polynomial
�k
⇣E�
hXrjr
⇤j
i(x)⌘� 1�
rk
d
!2d
n
E�
hXrjr
⇤j
i(x)
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Method of proof
Let be chosen uniformly from {v1, ..., vn}r1, ..., rk
1. is real rooted
2.
3. With non-zero probability
�k
⇣�
hXrjr
⇤j
i(x)
⌘� �k
⇣E�
hXrjr
⇤j
i(x)
⌘
�k
⇣E�
hXrjr
⇤j
i(x)⌘� 1�
rk
d
!2d
n
E�
hXrjr
⇤j
i(x)
Because is an interlacing family of polynomials
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Method of proof
Let be chosen uniformly from {v1, ..., vn}r1, ..., rk
1. is real rooted
2.
3. With non-zero probability
�k
⇣�
hXrjr
⇤j
i(x)
⌘� �k
⇣E�
hXrjr
⇤j
i(x)
⌘
�k
⇣E�
hXrjr
⇤j
i(x)⌘� 1�
rk
d
!2d
n
E�
hXrjr
⇤j
i(x)
Because is an interlacing family of polynomials
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Rank-1 updates of characteristic polynomials
As ,
Lemma: For a symmetric matrix A,
E�
⇥A+ r
j
r
⇤j
⇤(x) = (1� 1
n
@
x
)�[A](x)
E rjr⇤j = 1nI
Pi viv
⇤i = I
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Rank-1 updates of characteristic polynomials
As ,
Lemma: For a symmetric matrix A,
Proof: follows from rank-1 update for determinants:
det(A+ uu⇤) = det(A)(1 + u⇤A�1u)
E�
⇥A+ r
j
r
⇤j
⇤(x) = (1� 1
n
@
x
)�[A](x)
E rjr⇤j = 1nI
Pi viv
⇤i = I
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The expected characteristic polynomial
Lemma: For a symmetric matrix A,
Corollary:
E�
⇥A+ r
j
r
⇤j
⇤(x) = (1� 1
n
@
x
)�[A](x)
E�
hPk
j=1 rjr⇤j
i(x) = (1� 1
n
@
x
)kxd
![Page 43: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/43.jpg)
Real Roots Lemma: if is real rooted, so is
p(x) (1� c@
x
)p(x)
p(x)
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Real Roots Lemma: if is real rooted, so is
p(x) (1� c@
x
)p(x)
p(x)p
0(x)
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Real Roots Lemma: if is real rooted, so is
p(x) (1� c@
x
)p(x)
p(x)p
0(x)
�p
0(x)
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Real Roots Lemma: if is real rooted, so is
p(x) (1� c@
x
)p(x)
p(x)
�p
0(x)
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Real Roots Lemma: if is real rooted, so is
p(x) (1� c@
x
)p(x)
p(x)
�p
0(x)
+ + -‐ -‐ +
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Real Roots
Lemma: if is real rooted, so is p(x) (1� c@
x
)p(x)
E�
hPk
j=1 rjr⇤j
i(x) = (1� 1
n
@
x
)kxd
So, is real rooted E�
hPkj=1 rjr
⇤j
i(x)
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Lower bound on the kth root
E�
hPk
j=1 rjr⇤j
i(x) = (1� 1
n
@
x
)kxd
a scaled associated Laguerre polynomial
= x
d�k(1� 1n
@
x
)dxk
= x
d�kL
d�kk (nx)
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Lower bound on the kth root
E�
hPk
j=1 rjr⇤j
i(x) = (1� 1
n
@
x
)kxd
a scaled associated Laguerre polynomial
Let R be a k-‐by-‐d matrix of independent N (0, 1/n)
= x
d�kL
d�kk (nx)
E�[RR
T ] = L
d�kk (nx)
E�[RTR] = x
d�kL
d�kk (nx)
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Lower bound on the kth root
E�
hPk
j=1 rjr⇤j
i(x) = (1� 1
n
@
x
)kxd
a scaled associated Laguerre polynomial
�k
⇣E�
hXrjr
⇤j
i(x)⌘� 1�
rk
d
!2d
n
(Krasikov ‘06)
= x
d�kL
d�kk (nx)
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Proof: the polynomials form an interlacing family.
3. With non-zero probability
�k
⇣�
hXrjr
⇤j
i(x)
⌘� �k
⇣E�
hXrjr
⇤j
i(x)
⌘
pi1,i2,...,ik(x) = �
⇥vi1v
⇤i1 + · · ·+ vikv
⇤ik
⇤(x)
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Interlacing
Polynomial
interlaces
if
q(x) =Qd�1
i=1 (x� �i)
p(x) =Qd
i=1(x� ↵i)
↵1 �1 ↵2 · · ·↵d�1 �d�1 ↵d
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Interlacing
Polynomial
interlaces
if
q(x) =Qd�1
i=1 (x� �i)
p(x) =Qd
i=1(x� ↵i)
↵1 �1 ↵2 · · ·↵d�1 �d�1 ↵d
For example, interlaces p(x)@
x
p(x)
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Interlacing
Polynomial
interlaces
if
q(x) =Qd�1
i=1 (x� �i)
p(x) =Qd
i=1(x� ↵i)
↵1 �1 ↵2 · · ·↵d�1 �d�1 ↵d
If generalize to allow same degree,
Cauchy’s interlacing theorem says
interlaces �[A](x)�[A+ vv
⇤](x)
�d
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p1(x)
Common Interlacing
and have a common interlacing if can partition the line into intervals so that each contains one root from each polynomial
p2(x)
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
![Page 57: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/57.jpg)
p1(x)
Common Interlacing
and have a common interlacing if can partition the line into intervals so that each contains one root from each polynomial
p2(x)
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
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If have a common interlacing, then
Common Interlacing p1(x), p2(x), . . . , pn(x)
min
j�k(pj) �k
✓Ejpj
◆ max
j�k(pj)
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
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If have a common interlacing, then
Common Interlacing p1(x), p2(x), . . . , pn(x)
min
j�k(pj) �k
✓Ejpj
◆ max
j�k(pj)
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
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If have a common interlacing, then
Common Interlacing p1(x), p2(x), . . . , pn(x)
min
j�k(pj) �k
✓Ejpj
◆ max
j�k(pj)
�1
�2
�3
�d�1
) ) ) ) ( ( ( ( All are non-‐nega*ve here
All are non-‐posi*ve here
So, the average has a root between the smallest and largest kth roots
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Without a common interlacing
(x+ 1)(x+ 2) (x� 1)(x� 2)
![Page 62: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/62.jpg)
Without a common interlacing
(x+ 1)(x+ 2) (x� 1)(x� 2)
x
2 + 4
![Page 63: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/63.jpg)
Without a common interlacing (x+ 4)(x� 1)(x� 8)
(x+ 3)(x� 9)(x� 10.3)
![Page 64: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/64.jpg)
Without a common interlacing (x+ 4)(x� 1)(x� 8)
(x+ 3)(x� 9)(x� 10.3)
(x+ 3.2)(x� 6.8)(x� 7)
![Page 65: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/65.jpg)
If have a common interlacing, then
Common Interlacing p1(x), p2(x), . . . , pn(x)
min
j�k(pj) �k
✓Ejpj
◆ max
j�k(pj)
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
![Page 66: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/66.jpg)
is an interlacing family
if its members can be placed on the leaves of a tree so that when every node is labeled with the average of leaves below, siblings have common interlacings
Ei [ p2,i ]Ei [ p1,i ]
p2,2p2,1p1,1 p1,2
Ei,j [ pi,j ]
Interlacing Family of Polynomials {p�(x)}�
Ei [ p3,i ]
p2,3p1,3 p3,1 p3,2 p3,3
![Page 67: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/67.jpg)
is an interlacing family
For , set
Ei [ p2,i ]
p2,2p2,1p1,1 p1,2
Ei,j [ pi,j ]
Interlacing Family of Polynomials {p�(x)}�
Ei [ p3,i ]
p2,3p1,3 p3,1 p3,2 p3,3
� 2 {1, .., n}k pi1,...,ih = Eih+1,...,ik pi1,...,ik
Ei [ p1,i ]
![Page 68: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/68.jpg)
is an interlacing family
For , set
Interlacing Family of Polynomials {p�(x)}�
� 2 {1, .., n}k pi1,...,ih = Eih+1,...,ik pi1,...,ik
p2,2p2,1p1,1 p1,2 p2,3p1,3 p3,1 p3,2 p3,3
p1 p2
p;
p3
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Interlacing Family of Polynomials Theorem: There is an so that
p2,2p2,1p1,1 p1,2 p2,3p1,3 p3,1 p3,2 p3,3
p1 p2
p;
p3
i1, ..., ik
�k(pi1,...,ik) � �k (p;)
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It remains to prove that the polynomials form an interlacing family.
Will imply that with non-zero probability
�k
⇣�
hXrjr
⇤j
i(x)
⌘� �k
⇣E�
hXrjr
⇤j
i(x)
⌘
pi1,i2,...,ik(x) = �
⇥vi1v
⇤i1 + · · ·+ vikv
⇤ik
⇤(x)
Interlacing Family of Polynomials
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Common interlacings
have a common interlacing iff for every convex combination
p1(x), p2(x), . . . , pn(x)
Pj µj = 1, µj � 0
Pj µjpj(x)
is real rooted
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
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Common interlacings
have a common interlacing iff for every distribution on
p1(x), p2(x), . . . , pn(x)
is real rooted
�1
�2
�3
�d�1
) ) ) ) ( ( ( (
µ {1, ..., n}
Ej⇠µ pj(x)
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Common interlacings
have a common interlacing iff for every distribution on
p1(x), p2(x), . . . , pn(x)
is real rooted
µ {1, ..., n}
Ej⇠µ pj(x)
Proof: by similar picture. (Dedieu ‘80, Fell ‘92, Chudnovsky-Seymour ‘07)
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An interlacing family
pi1,i2,...,ik(x) = �
⇥vi1v
⇤i1 + · · ·+ vikv
⇤ik
⇤(x)
Nodes on the tree are labeled with
pi1,...,ih(x) = Eih+1,...,ik pi1,...,ik(x)
We need to show that for each the polynomials i1, ..., ih
pi1,...,ih,j(x) have a common interlacing
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An interlacing family
pi1,i2,...,ik(x) = �
⇥vi1v
⇤i1 + · · ·+ vikv
⇤ik
⇤(x)
Nodes on the tree are labeled with
pi1,...,ih(x) = Eih+1,...,ik pi1,...,ik(x)
We need to show that for each the polynomials i1, ..., ih
pi1,...,ih,j(x) have a common interlacing
Proof: = (1� 1
n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
![Page 76: Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR …cs- · 2014-05-19 · Adam Marcus (Crisply, Yale) Daniel Spielman (Yale) Nikhil Srivastava (MSR India)](https://reader035.fdocuments.in/reader035/viewer/2022081611/5f08e60c7e708231d4244259/html5/thumbnails/76.jpg)
An interlacing family We need to show that for each the polynomials i1, ..., ih
pi1,...,ih,j(x) have a common interlacing
Proof:
Cauchy’s interlacing theorem implies All interlace
�
⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
= (1� 1n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
�
⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih
⇤(x)
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An interlacing family We need to show that for each the polynomials i1, ..., ih
pi1,...,ih,j(x) have a common interlacing
Proof:
Cauchy’s interlacing theorem implies have a common interlacing
�
⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
= (1� 1n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
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An interlacing family Proof:
Cauchy’s interlacing theorem implies have a common interlacing. So, for all distributions is real rooted.
�
⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
Ej⇠µ �⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
µ
= (1� 1n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
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An interlacing family Proof:
Cauchy’s interlacing theorem implies have a common interlacing. So, for all distributions is real rooted. And, is also real rooted for every distribution .
�
⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
Ej⇠µ �⇥vi1v
⇤i1 + · · ·+ vihv
⇤ih + vjv
⇤j
⇤(x)
µ
Ej⇠µ
(1� 1n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
= (1� 1n
@
x
)k�h�1�
⇥v
i1v⇤i1+ · · ·+ v
ihv⇤ih
+ v
j
v
⇤j
⇤(x)
µQED
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Method of proof
Let be chosen uniformly from {v1, ..., vn}r1, ..., rk
1. is real rooted
2.
3. With non-zero probability
�k
⇣�
hXrjr
⇤j
i(x)
⌘� �k
⇣E�
hXrjr
⇤j
i(x)
⌘
�k
⇣E�
hXrjr
⇤j
i(x)⌘� 1�
rk
d
!2d
n
E�
hXrjr
⇤j
i(x)
Because is an interlacing family of polynomials
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In part 2
Will use the same approach to prove Weaver’s conjecture, and thereby Kadison-Singer But, employ multivariate analogs of these arguments and a direct bound on the roots of the polynomials.
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Main Theorem
For all if all and then exists a partition into S1 and S2 with
↵ > 0
kvik ↵
Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer
eigs(P
i2Sjviv⇤i ) 1
2 + 3↵
Pviv⇤i = I
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eigs
0
BB@
X
i2S1
viv⇤i 0
0X
i2S2
viv⇤i
1
CCA 12 + 3↵
We want
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We want
roots
0
BB@�
2
664
X
i2S1
viv⇤i 0
0
X
i2S2
viv⇤i
3
775 (x)
1
CCA 12 + 3↵
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We want
Consider expected polynomial with a random partition.
roots
0
BB@�
2
664
X
i2S1
viv⇤i 0
0
X
i2S2
viv⇤i
3
775 (x)
1
CCA 12 + 3↵
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Method of proof
1. Prove expected characteristic polynomial has real roots
2. Prove its largest root is at most
3. Prove is an interlacing family, so exists a partition whose polynomial
has largest root at most
1/2 + 3↵
1/2 + 3↵
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The Expected Polynomial Indicate choices by : �1, ...,�n i 2 S�i
p�1,...,�n(x) = �
2
664
X
i:�i=1
viv⇤i 0
0X
i:�i=2
viv⇤i
3
775 (x)
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0
BB@
X
i2S1
viv⇤i 0
0X
i2S2
viv⇤i
1
CCA =X
i
aia⇤i
The Expected Polynomial
ai =
✓vi0
◆for i 2 S1 ai =
✓0
vi
◆for i 2 S2
�i = 1 �i = 2
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Mixed Characteristic Polynomials
For independently chosen random vectors
is their mixed characteristic polynomial.
a1, ..., an
Theorem: It only depends on and, is real-rooted
E�[P
i aia⇤i ]
Ai = E aia⇤i
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Ai = E aia⇤i
Mixed Characteristic Polynomials
For independently chosen random vectors
is their mixed characteristic polynomial.
a1, ..., an
E�[P
i aia⇤i ] = µ(A1, ..., An)
Theorem: It only depends on and, is real-rooted
Tr (Ai) = Tr (E aia⇤i ) = ETr (aia⇤i ) = E kaik2
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Mixed Characteristic Polynomials
For independently chosen random vectors
is their mixed characteristic polynomial.
a1, ..., an
The constant term is the mixed discriminant of A1, . . . , An
E�[P
i aia⇤i ] = µ(A1, ..., An)
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The constant term When diagonal and , is a matrix permanent. d = n cd
0
@
1
A
0
@
1
A
0
@
1
A
0
@
1
A
A1 A2 Ad
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The constant term
is minimized when
Proved by Egorychev and Falikman ‘81. Simpler proof by Gurvits (see Laurent-Schrijver)
If and P
Ai = I Tr(Ai) = 1
When diagonal and , is a matrix permanent. Van der Waerden’s Conjecture becomes
d = n cd
cd Ai =1
nI
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The constant term
is minimized when
This was a conjecture of Bapat.
If and P
Ai = I Tr(Ai) = 1
For Hermitian matrices, is the mixed discriminant Gurvits proved a lower bound on : cd
cd
cd
Ai =1
nI
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Other coefficients One can generalize Gurvits’s results to prove lower bounds on all the coefficients.
is minimized when
If and P
Ai = I Tr(Ai) = 1
Ai =1
nI|ck|
But, this does not imply useful bounds on the roots
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Real Stable Polynomials
A multivariate generalization of real rootedness.
p 2 IR[z]Complex roots of come in conjugate pairs.
So, real rooted iff no roots with positive complex part.
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Real Stable Polynomials
Isomorphic to Gårding’s hyperbolic polynomials Used by Gurvits (in his second proof)
is real stable if
it has no roots in the upper half-plane
for all i
implies
p 2 IR[z1, . . . , zn]
imag(zi) > 0p(z1, . . . , zn) 6= 0
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Real Stable Polynomials
Isomorphic to Gårding’s hyperbolic polynomials Used by Gurvits (in his second proof)
is real stable if
it has no roots in the upper half-plane
for all i
implies
p 2 IR[z1, . . . , zn]
imag(zi) > 0p(z1, . . . , zn) 6= 0
See surveys of Pemantle and Wagner
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Real Stable Polynomials
Borcea-Brändén ‘08: For PSD matrices
is real stable
det[z1A1 + · · ·+ znAn]
A1, . . . , An
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Real Stable Polynomials
implies is real rooted
real stable p(z1, ..., zn)
p(x, x, ..., x)
implies is real stable
real stable p(z1, ..., zn)
(Lieb Sokal ‘81)
(1� @zi) p(z1, ..., zn)
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nY
i=1
1� @zi
!det
nX
i=1
zi
Ai
!���z1=···=zn=x
Real Roots
µ(A1, . . . , An)(x) =
So, every mixed characteristic polynomial is real rooted.
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p�1,...,�n(x) = �
2
664
X
i:�i=1
viv⇤i 0
0X
i:�i=2
viv⇤i
3
775 (x)
Our Interlacing Family Indicate choices by : �1, ...,�n i 2 S�i
p�1,...,�k(x) = E�k+1,...,�n
[ p�1,...,�n ] (x)
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Interlacing
p1(x) and have a common interlacing iff p2(x)
�p1(x) + (1� �)p2(x) is real rooted for all 0 � 1
We need to show that is real rooted.
�p�1,...,�k,1(x) + (1� �)p�1,...�k,2(x)
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Interlacing
p1(x) and have a common interlacing iff p2(x)
�p1(x) + (1� �)p2(x) is real rooted for all 0 � 1
We need to show that is real rooted.
�p�1,...,�k,1(x) + (1� �)p�1,...�k,2(x)
It is a mixed characteristic polynomial, so is real-rooted.
�k+1 = 1Set with probability Keep uniform for
��i i > k + 1
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An upper bound on the roots
Theorem: If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x)) (1 +p✏)2
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An upper bound on the roots
Theorem: If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x)) (1 +p✏)2
An upper bound of 2 is trivial (in our special case). Need any constant strictly less than 2.
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An upper bound on the roots
Theorem: If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x)) (1 +p✏)2
µ(A1, . . . , An)(x) =
nY
i=1
1� @zi
!det
nX
i=1
zi
Ai
!���z1=···=zn=x
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An upper bound on the roots
Define: is an upper bound on the roots of if
(w1, ..., wn)p(z1, ..., zn)
p(z1, ..., zn) > 0 for (z1, ..., zn) � (w1, ..., wn)
−8 −6 −4 −2 0 2 4 6 8 10−8
−6
−4
−2
0
2
4
6
8
10
w
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An upper bound on the roots
Define: is an upper bound on the roots of if
(w1, ..., wn)p(z1, ..., zn)
p(z1, ..., zn) > 0 for (z1, ..., zn) � (w1, ..., wn)
−8 −6 −4 −2 0 2 4 6 8 10−8
−6
−4
−2
0
2
4
6
8
10
w
Eventually set z1, ..., zn = x
so want w1 = · · · = wn
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−8 −6 −4 −2 0 2 4 6 8 10−8
−6
−4
−2
0
2
4
6
8
10
Action of the operators
p(x, y)
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Action of the operators
−8 −6 −4 −2 0 2 4 6 8 10 12−8
−6
−4
−2
0
2
4
6
8
10
p(x, y)(1� @y)p(x, y)
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Action of the operators
−8 −6 −4 −2 0 2 4 6 8 10 12−8
−6
−4
−2
0
2
4
6
8
10
p(x, y)(1� @y)p(x, y)
(1� ✏@y)p(x, y) ⇡ p(x, y � ✏)
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The roots of
Define:
(1� @
x
)p(x)
↵-max(�1, ...,�n) = max
n
u :
P 1u��i
= ↵o
↵-max(p(x)) = ↵-max(roots(p))
Theorem (Batson-S-Srivastava): If is real rooted and
p(x) ↵ > 0
↵-max((1� @
x
)p(x)) ↵-max(p(x)) +
1
1� ↵
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The roots of (1� @
x
)p(x)
Theorem (Batson-S-Srivastava): If is real rooted and
p(x) ↵ > 0
↵-max((1� @
x
)p(x)) ↵-max(p(x)) +
1
1� ↵
Proof: Define Set , so
u = ↵-max(p(x)) �p(u) = ↵
Suffices to show for all
�p�p0(u+ �) �p(u)
� � 11�↵
�p(u) =p0(u)
p(u)=
X
i
1
u� �i= @u log p(u)
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The roots of (1� @
x
)p(x)
Define �p(u) =p0(u)
p(u)=
X
i
1
u� �i
Set , so u = ↵-max(p(x)) �p(u) = ↵
Suffices to show for all
�p�p0(u+ �) �p(u)
� � 11�↵
�p(u)� �p(u+ �) ���0
p(u+ �)
1� �p(u+ �)
(algebra)
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The roots of (1� @
x
)p(x)
Define
�p(u)� �p(u+ �) ���0
p(u+ �)
1� �p(u+ �)
�p(u) convex for implies u > max(p(x))
�p(u)� �p(u+ �) � �(��0p(u+ �))
Monotone decreasing implies only need
� � 1
1� �p(u+ �)
�p(u) =p0(u)
p(u)=
X
i
1
u� �i
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The roots of (1� @
x
)p(x)
�p(u) convex for implies u > max(p(x))
�p(u)� �p(u+ �) � �(��0p(u+ �))
Monotone decreasing implies only need
� � 1
1� �p(u+ �)
and that � � 1
1� ↵=
1
1� �p(u) suffices.
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The roots of (1� @
x
)p(x)
�p(u) convex for implies u > max(p(x))
�p(u)� �p(u+ �) � �(��0p(u+ �))
Monotone decreasing implies only need
� � 1
1� �p(u+ �)
and that � � 1
1� ↵=
1
1� �p(u) suffices.
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The roots of (1� @
x
)p(x)
Theorem (Batson-S-Srivastava): If is real rooted and
p(x) ↵ > 0
↵-max((1� @
x
)p(x)) ↵-max(p(x)) +
1
1� ↵
Gives a sharp upper bound on the roots of associated Laguerre polynomials. The analogous argument with the min gives the lower bound that we claimed before.
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An upper bound on the roots
Theorem: If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x)) (1 +p✏)2
µ(A1, . . . , An)(x) =
nY
i=1
1� @zi
!det
nX
i=1
zi
Ai
!���z1=···=zn=x
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A robust upper bound
Define: is an -upper bound on if it is an -max, in each , and
(w1, ..., wn)p(z1, ..., zn)
↵
↵i zi ↵i ↵
Theorem: If is an -upper bound on , then ↵ pw
w + �ej is an -upper bound on , ↵ p� @zjp
� � 11�↵ for
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A robust upper bound Theorem: If is an -upper bound on , and ↵ pw
w + �ej is an -upper bound on , ↵ p� @zjp
� � 11�↵
Proof: Same as before, but need to know that
is decreasing and convex in , above the roots zj
@zip(z1, . . . , zn)
p(z1, . . . , zn)
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A robust upper bound Proof: Same as before, but need to know that
is decreasing and convex in , above the roots zj
@zip(z1, . . . , zn)
p(z1, . . . , zn)
Follows from a theorem of Helton and Vinnikov ’07:
Every bivariate real stable polynomial can be written
det(A+Bx+ Cy)
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A robust upper bound Proof: Same as before, but need to know that
is decreasing and convex in , above the roots zj
@zip(z1, . . . , zn)
p(z1, . . . , zn)
Or, as pointed out by Renegar, from a theorem Bauschke, Güler, Lewis, and Sendov ’01
Or, by a theorem of Brändén ‘07.
Or, see Terry Tao’s blog for a (mostly) self-contained proof
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An upper bound on the roots
Theorem: If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x)) (1 +p✏)2
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A probabilistic interpretation
For independently chosen random vectors with finite support
a1, ..., an
such that and ���E
⇥aia
Ti
⇤��� ✏E⇥ P
i aiaTi
⇤= I
then Pr
"�����X
i
aiaTi
����� (1 +p✏)2
#> 0
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Main Theorem
For all if all then exists a partition into S1 and S2 with
↵ > 0
kvik ↵
Implies Akemann-Anderson Paving Conjecture, which implies Kadison-Singer
eigs(P
i2SjvivTi ) 1
2 + 3↵
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Anderson’s Paving Conjecture ‘79
Reduction by Casazza-Edidin-Kalra-Paulsen ’07 and Harvey ’13:
There exist an and a k so that if all and then exists a partition of into k parts s.t.
✏ > 0
PvivTi = I
{1, . . . , n}
eigs(P
i2SjvivTi ) 1� ✏
kvik2 1/2
Can prove using the same technique
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A conjecture
If and then
PAi = I Tr(Ai) ✏
max-root (µ(A1, ..., An)(x))
is largest when Ai =✏dI
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Questions
How do operations that preserve real rootedness move the roots and the Stieltjes transform?
Can the partition be found in polynomial time?
What else can one construct this way?
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