Matrix rigidity and elimination theory
Abhinav Kumarjoint work with Satya Lokam, Vijay Patankar and Jalayal Sarma M.N.
MIT
April 25, 2012
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 1 / 1
Rigidity of a matrix
Definition
Let A be an n × n matrix with entries in a field F . The rigidity Rig(A, r)of A for target rank r is the smallest number of entries of A that need tobe changed to make the rank at most r .
Most of the time we’ll work with F = C. If F is a subfield of C (e.g. Q)we may want to consider changes in some finite extension field L/F , anddefine Rig(A, r , L).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 2 / 1
Rigidity of a matrix
Definition
Let A be an n × n matrix with entries in a field F . The rigidity Rig(A, r)of A for target rank r is the smallest number of entries of A that need tobe changed to make the rank at most r .
Most of the time we’ll work with F = C. If F is a subfield of C (e.g. Q)we may want to consider changes in some finite extension field L/F , anddefine Rig(A, r , L).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 2 / 1
Rigidity of a matrix
Definition
Let A be an n × n matrix with entries in a field F . The rigidity Rig(A, r)of A for target rank r is the smallest number of entries of A that need tobe changed to make the rank at most r .
Most of the time we’ll work with F = C. If F is a subfield of C (e.g. Q)we may want to consider changes in some finite extension field L/F , anddefine Rig(A, r , L).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 2 / 1
Valiant’s theorem
Valiant defined matrix rigidity in his study of lower bounds on complexityfor computing linear forms in n variables, for circuits using gates which cancompute linear combinations of two variables.
To compute one such linear combination, we can use a binary tree to get acircuit of depth log2(n) and size n.
But: say we want to use such a circuit to simultaenously compute n linearforms in n variables, given by the entries of the vector Ax , wherex = (x1, . . . , xn) is the column vector of inputs.
Valiant showed that this cannot be done with circuits which are bothshallow (of depth O(log n)) and small (of size O(n)).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 3 / 1
Valiant’s theorem
Valiant defined matrix rigidity in his study of lower bounds on complexityfor computing linear forms in n variables, for circuits using gates which cancompute linear combinations of two variables.
To compute one such linear combination, we can use a binary tree to get acircuit of depth log2(n) and size n.
But: say we want to use such a circuit to simultaenously compute n linearforms in n variables, given by the entries of the vector Ax , wherex = (x1, . . . , xn) is the column vector of inputs.
Valiant showed that this cannot be done with circuits which are bothshallow (of depth O(log n)) and small (of size O(n)).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 3 / 1
Valiant’s theorem
Valiant defined matrix rigidity in his study of lower bounds on complexityfor computing linear forms in n variables, for circuits using gates which cancompute linear combinations of two variables.
To compute one such linear combination, we can use a binary tree to get acircuit of depth log2(n) and size n.
But: say we want to use such a circuit to simultaenously compute n linearforms in n variables, given by the entries of the vector Ax , wherex = (x1, . . . , xn) is the column vector of inputs.
Valiant showed that this cannot be done with circuits which are bothshallow (of depth O(log n)) and small (of size O(n)).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 3 / 1
Valiant’s theorem II
More precisely,
Theorem
Let A1,A2, . . . be an infinite family of matrices, where An is an n × n real
matrix and for some κ, c , ǫ > 0, we have Rig(A, κn) ≥ cn1+ǫ. Then given
any fixed c1, c2 > 0, there does not exist a family of straight line programs
for the corresponding sets of linear forms that achieve size c1n and depth
c2 log n simultaneously for all n.
Remarks:
The proof uses some graph theoretic property of the circuit (namely,that it’s a grate.
We’ll see that most matrices have quadratic rigidity, so asking forsuper-linear rigidity is perhaps not so bad.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 4 / 1
Valiant’s theorem II
More precisely,
Theorem
Let A1,A2, . . . be an infinite family of matrices, where An is an n × n real
matrix and for some κ, c , ǫ > 0, we have Rig(A, κn) ≥ cn1+ǫ. Then given
any fixed c1, c2 > 0, there does not exist a family of straight line programs
for the corresponding sets of linear forms that achieve size c1n and depth
c2 log n simultaneously for all n.
Remarks:
The proof uses some graph theoretic property of the circuit (namely,that it’s a grate.
We’ll see that most matrices have quadratic rigidity, so asking forsuper-linear rigidity is perhaps not so bad.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 4 / 1
Valiant’s theorem II
More precisely,
Theorem
Let A1,A2, . . . be an infinite family of matrices, where An is an n × n real
matrix and for some κ, c , ǫ > 0, we have Rig(A, κn) ≥ cn1+ǫ. Then given
any fixed c1, c2 > 0, there does not exist a family of straight line programs
for the corresponding sets of linear forms that achieve size c1n and depth
c2 log n simultaneously for all n.
Remarks:
The proof uses some graph theoretic property of the circuit (namely,that it’s a grate.
We’ll see that most matrices have quadratic rigidity, so asking forsuper-linear rigidity is perhaps not so bad.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 4 / 1
Valiant’s theorem II
More precisely,
Theorem
Let A1,A2, . . . be an infinite family of matrices, where An is an n × n real
matrix and for some κ, c , ǫ > 0, we have Rig(A, κn) ≥ cn1+ǫ. Then given
any fixed c1, c2 > 0, there does not exist a family of straight line programs
for the corresponding sets of linear forms that achieve size c1n and depth
c2 log n simultaneously for all n.
Remarks:
The proof uses some graph theoretic property of the circuit (namely,that it’s a grate.
We’ll see that most matrices have quadratic rigidity, so asking forsuper-linear rigidity is perhaps not so bad.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 4 / 1
Easy upper bound
Proposition
Any n × n matrix X has rigidity at most (n − r)2 for target rank r .
Proof.
If the rank less than r , rigidity is zero. Else assume w.lo.g. that the topleft r × r block is nonsingular. Write X as
(
A B
C D
)
.
Modify D to CA−1B .
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 5 / 1
Easy upper bound
Proposition
Any n × n matrix X has rigidity at most (n − r)2 for target rank r .
Proof.
If the rank less than r , rigidity is zero. Else assume w.lo.g. that the topleft r × r block is nonsingular. Write X as
(
A B
C D
)
.
Modify D to CA−1B .
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 5 / 1
Generic rigidity
An extension of this argument shows that generic matrices have maximalrigidity, i.e. the subset of matrices of rigidity less than (n − r)2 iscontained in a proper Zariski-closed subset.
Proof: if rigidity is k < (n − r)2, then the matrix is in the image of a mapfrom a variety of matrices of the form
(
A B
C CA−1B
)
.
(which has dimension ≤ n2 − (n − r)2) times Ak . Sincek + n2 − (n − r)2 < n2 = dimMatn, the image is contained in a properclosed subset.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 6 / 1
Generic rigidity
An extension of this argument shows that generic matrices have maximalrigidity, i.e. the subset of matrices of rigidity less than (n − r)2 iscontained in a proper Zariski-closed subset.
Proof: if rigidity is k < (n − r)2, then the matrix is in the image of a mapfrom a variety of matrices of the form
(
A B
C CA−1B
)
.
(which has dimension ≤ n2 − (n − r)2) times Ak . Sincek + n2 − (n − r)2 < n2 = dimMatn, the image is contained in a properclosed subset.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 6 / 1
Questions
All this leads to the natural question:
Construct an explicit family of reasonably natural matrices of super-linearrigidity for target rank linear in the dimension n.
Remarks:
Or even better, quadratic rigidity.
Or better still, maximal rigidity (n − r)2.
Note that these extensions will not give a better lower complexity bound,but they are natural questions from the algebraic point of view.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 7 / 1
Questions
All this leads to the natural question:
Construct an explicit family of reasonably natural matrices of super-linearrigidity for target rank linear in the dimension n.
Remarks:
Or even better, quadratic rigidity.
Or better still, maximal rigidity (n − r)2.
Note that these extensions will not give a better lower complexity bound,but they are natural questions from the algebraic point of view.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 7 / 1
Questions
All this leads to the natural question:
Construct an explicit family of reasonably natural matrices of super-linearrigidity for target rank linear in the dimension n.
Remarks:
Or even better, quadratic rigidity.
Or better still, maximal rigidity (n − r)2.
Note that these extensions will not give a better lower complexity bound,but they are natural questions from the algebraic point of view.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 7 / 1
Questions
All this leads to the natural question:
Construct an explicit family of reasonably natural matrices of super-linearrigidity for target rank linear in the dimension n.
Remarks:
Or even better, quadratic rigidity.
Or better still, maximal rigidity (n − r)2.
Note that these extensions will not give a better lower complexity bound,but they are natural questions from the algebraic point of view.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 7 / 1
Questions
All this leads to the natural question:
Construct an explicit family of reasonably natural matrices of super-linearrigidity for target rank linear in the dimension n.
Remarks:
Or even better, quadratic rigidity.
Or better still, maximal rigidity (n − r)2.
Note that these extensions will not give a better lower complexity bound,but they are natural questions from the algebraic point of view.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 7 / 1
Examples I
One guess for highly rigid matrices: totally regular matrices, i.e. all minorsare nonsingular.
This guess is not correct! Valiant shows:
Proposition
For each n there is an n × n totally regular matrix A such that
Rig(A, n(log log log n)/(log log n)) ≤ n1+O(1/ log log n).
Remarks:
That is, one can bring rank down to o(n) by changing o(n1+ǫ) entries.
Proof uses superconcentrators, and is also non-explicit.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 8 / 1
Examples I
One guess for highly rigid matrices: totally regular matrices, i.e. all minorsare nonsingular.
This guess is not correct! Valiant shows:
Proposition
For each n there is an n × n totally regular matrix A such that
Rig(A, n(log log log n)/(log log n)) ≤ n1+O(1/ log log n).
Remarks:
That is, one can bring rank down to o(n) by changing o(n1+ǫ) entries.
Proof uses superconcentrators, and is also non-explicit.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 8 / 1
Examples I
One guess for highly rigid matrices: totally regular matrices, i.e. all minorsare nonsingular.
This guess is not correct! Valiant shows:
Proposition
For each n there is an n × n totally regular matrix A such that
Rig(A, n(log log log n)/(log log n)) ≤ n1+O(1/ log log n).
Remarks:
That is, one can bring rank down to o(n) by changing o(n1+ǫ) entries.
Proof uses superconcentrators, and is also non-explicit.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 8 / 1
Examples I
One guess for highly rigid matrices: totally regular matrices, i.e. all minorsare nonsingular.
This guess is not correct! Valiant shows:
Proposition
For each n there is an n × n totally regular matrix A such that
Rig(A, n(log log log n)/(log log n)) ≤ n1+O(1/ log log n).
Remarks:
That is, one can bring rank down to o(n) by changing o(n1+ǫ) entries.
Proof uses superconcentrators, and is also non-explicit.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 8 / 1
Examples I
One guess for highly rigid matrices: totally regular matrices, i.e. all minorsare nonsingular.
This guess is not correct! Valiant shows:
Proposition
For each n there is an n × n totally regular matrix A such that
Rig(A, n(log log log n)/(log log n)) ≤ n1+O(1/ log log n).
Remarks:
That is, one can bring rank down to o(n) by changing o(n1+ǫ) entries.
Proof uses superconcentrators, and is also non-explicit.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 8 / 1
Examples II
Nevertheless, here are some natural familes.
Vandermonde matrices: row j is 1, xj , x2j , . . . , x
nj .
Discrete Fourier transform matrices: n = p prime say, Vandermondewith xj = e2π
√−1j/p.
Generalized Hadamard matrices H: entries complex numbers hij ofabsolute value 1, and such that HH† = nIn.
Circulant matrices: each row is a shift of the previous one.
Cauchy matrix Cij = 1/(i + j − 1).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 9 / 1
Examples II
Nevertheless, here are some natural familes.
Vandermonde matrices: row j is 1, xj , x2j , . . . , x
nj .
Discrete Fourier transform matrices: n = p prime say, Vandermondewith xj = e2π
√−1j/p.
Generalized Hadamard matrices H: entries complex numbers hij ofabsolute value 1, and such that HH† = nIn.
Circulant matrices: each row is a shift of the previous one.
Cauchy matrix Cij = 1/(i + j − 1).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 9 / 1
Examples II
Nevertheless, here are some natural familes.
Vandermonde matrices: row j is 1, xj , x2j , . . . , x
nj .
Discrete Fourier transform matrices: n = p prime say, Vandermondewith xj = e2π
√−1j/p.
Generalized Hadamard matrices H: entries complex numbers hij ofabsolute value 1, and such that HH† = nIn.
Circulant matrices: each row is a shift of the previous one.
Cauchy matrix Cij = 1/(i + j − 1).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 9 / 1
Examples II
Nevertheless, here are some natural familes.
Vandermonde matrices: row j is 1, xj , x2j , . . . , x
nj .
Discrete Fourier transform matrices: n = p prime say, Vandermondewith xj = e2π
√−1j/p.
Generalized Hadamard matrices H: entries complex numbers hij ofabsolute value 1, and such that HH† = nIn.
Circulant matrices: each row is a shift of the previous one.
Cauchy matrix Cij = 1/(i + j − 1).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 9 / 1
Examples II
Nevertheless, here are some natural familes.
Vandermonde matrices: row j is 1, xj , x2j , . . . , x
nj .
Discrete Fourier transform matrices: n = p prime say, Vandermondewith xj = e2π
√−1j/p.
Generalized Hadamard matrices H: entries complex numbers hij ofabsolute value 1, and such that HH† = nIn.
Circulant matrices: each row is a shift of the previous one.
Cauchy matrix Cij = 1/(i + j − 1).
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 9 / 1
Progress so far
Matrices Ω(.) References
Vandermonde n2
rRazborov ’89, Pudlak ’94Shparlinsky ’97, Lokam ’99
Hadamard n2
rKashin-Razborov ’98
Parity Check n2
rlog(n
r) Friedman ’93
Pudlak-Rodl ’94
Totally regular n2
rlog(n
r) Shokrallahi-Spielmann-Stemann ’97√
pij n(n − 16r) Lokam ’06
ζij = e2π
√
−1pij (n − r)2 This work.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 10 / 1
Statement of our result
We construct matrices with maximal rigidity, but over a number field ofrelatively high degree.
Theorem
Let ∆(n) = 2n2n2and let pij > ∆(n) be distinct primes for 1 ≤ i , j ≤ n.
Let A(n) have (i , j) entry ζij = e2π√−1/pij . Then Rig(A(n), r) = (n − r)2.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 11 / 1
Determinantal ideal
Recall that the variety of matrices of rank at most r is defined by thedeterminantal ideal I (n, r) with generators all (r + 1)× (r + 1) minors of
x11 x12 · · · x1nx21 x22 · · · x2n...
......
xn1 xn2 · · · xnn
It’s an irreducible variety of dimension n2 − (n − r)2.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 12 / 1
Elimination ideals
Let π be a pattern (Valiant calles it a mask) of positions where we allowchanges, of cardinality k strictly less than (n− r)2 say. The set of matriceswhose rank can be brought down to at most r by changing entries in π liein the image of a variety of dimension n2 − (n − r)2 + k .
Its closure is defined by the elimination ideal I (n, r , π) = I (n, r) ∩ Q[xπ]where π means positions not in π.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 13 / 1
Elimination ideals
Let π be a pattern (Valiant calles it a mask) of positions where we allowchanges, of cardinality k strictly less than (n− r)2 say. The set of matriceswhose rank can be brought down to at most r by changing entries in π liein the image of a variety of dimension n2 − (n − r)2 + k .
Its closure is defined by the elimination ideal I (n, r , π) = I (n, r) ∩ Q[xπ]where π means positions not in π.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 13 / 1
Sketch of proof
As a generic matrix has maximal rigidity, these elimination ideals I (n, r , π)are all non-zero, as π runs over all patterns of size strictly less than(n − r)2.
So the ideal I (n, r , π) is nonzero. We’ll use effective Nullstellensatz boundsto show that there’s a multivariate polynomial of low enough degree, andthen use Galois theory to show that it cannot vanish on the roots of unityconstructed. Therefore, the matrix with entries Aij = e2π
√−1/pij does not
lie in V (I (n, r , π)) for any π, and therefore has maximal rigidity.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 14 / 1
Sketch of proof
As a generic matrix has maximal rigidity, these elimination ideals I (n, r , π)are all non-zero, as π runs over all patterns of size strictly less than(n − r)2.
So the ideal I (n, r , π) is nonzero. We’ll use effective Nullstellensatz boundsto show that there’s a multivariate polynomial of low enough degree, andthen use Galois theory to show that it cannot vanish on the roots of unityconstructed. Therefore, the matrix with entries Aij = e2π
√−1/pij does not
lie in V (I (n, r , π)) for any π, and therefore has maximal rigidity.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 14 / 1
Hypersurfaces
In fact, we can show pretty easily that the corresponding union ofsubvarieties V (I (n, r , π)) is a (reducible) hypersurface.
Explicit equations for these, or understanding of their geometry, might beuseful in trying to understand why some families of matrices (e.g.Vandermonde) seem to have maximal rigidity.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 15 / 1
Hypersurfaces
In fact, we can show pretty easily that the corresponding union ofsubvarieties V (I (n, r , π)) is a (reducible) hypersurface.
Explicit equations for these, or understanding of their geometry, might beuseful in trying to understand why some families of matrices (e.g.Vandermonde) seem to have maximal rigidity.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 15 / 1
Effective Nullstellensatz
We’ll use the following bounds, which rely on the Bezout inequality fordegrees proved by Heintz (1983)
Theorem (Dickenstein, Fitchas, Giusti, Sessa ’91)
Let I = 〈f1, . . . , fs〉 be an ideal in the polynomial ring F [Y ] over an infinite
field F , where Y = y1, . . . , ym. Let dmax be the maximum total degree
of a generator fi . Let Z = yi1 , . . . , yiℓ ⊂ Y be a subset of indeterminates
of Y . If I ∩ F [Z ] 6= (0) then there exists a non-zero polynomial
g ∈ I ∩ F [Z ] such that g =∑s
i=1 gi fi , with gi ∈ F [Y ] anddeg(gi fi ) ≤ dm(dm + 1), where d = max(dmax , 3).
Applying it to our situation, where d = r + 1 and m = n2, we get a boundof ∆(n) = 2n2n
2for the total degree of g .
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 16 / 1
Effective Nullstellensatz
We’ll use the following bounds, which rely on the Bezout inequality fordegrees proved by Heintz (1983)
Theorem (Dickenstein, Fitchas, Giusti, Sessa ’91)
Let I = 〈f1, . . . , fs〉 be an ideal in the polynomial ring F [Y ] over an infinite
field F , where Y = y1, . . . , ym. Let dmax be the maximum total degree
of a generator fi . Let Z = yi1 , . . . , yiℓ ⊂ Y be a subset of indeterminates
of Y . If I ∩ F [Z ] 6= (0) then there exists a non-zero polynomial
g ∈ I ∩ F [Z ] such that g =∑s
i=1 gi fi , with gi ∈ F [Y ] anddeg(gi fi ) ≤ dm(dm + 1), where d = max(dmax , 3).
Applying it to our situation, where d = r + 1 and m = n2, we get a boundof ∆(n) = 2n2n
2for the total degree of g .
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 16 / 1
Roots of unity
Finally, we have the following lemma.
Lemma
Let N be a positive integer, and θ1, . . . , θm be algebraic numbers such that
Q(θi ) is Galois over Q and such that
[Q(θi ) : Q] ≥ N and Q[θi ] ∩Q(θ1, . . . , θi−1, θi+1, . . . , θm) = Q for all i .
Let g(x1, . . . , xm) ∈ Q[x1, . . . , xm] be a nonzero polynomial such that
deg(g) < N. Then g(θ1, . . . , θm) 6= 0.
The proof is easy by induction, using the linear disjointness property.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 17 / 1
Conclusion of proof
To finish the proof of the theorem, we note that choosing distinct primespij for 1 ≤ i , j ≤ n, and setting θij = ζpij := e2π
√−1/pij , the linear
disjointness property is satisfied. So we just need to make sure thatpij − 1 > ∆(n) := 2n2n
2.
If we want real matrices, we can take θij = ζpij + ζ−1pij
, these generate thetotally real subfields of the cyclotomic fields, so we just need to ensure,(pij − 1)/2 > ∆(n).
Of course, this falls far short of our desired goal, since the roots of unityhave very high degree. What we would like is to have rational entries, andpreferably some systematic family of these.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 18 / 1
Conclusion of proof
To finish the proof of the theorem, we note that choosing distinct primespij for 1 ≤ i , j ≤ n, and setting θij = ζpij := e2π
√−1/pij , the linear
disjointness property is satisfied. So we just need to make sure thatpij − 1 > ∆(n) := 2n2n
2.
If we want real matrices, we can take θij = ζpij + ζ−1pij
, these generate thetotally real subfields of the cyclotomic fields, so we just need to ensure,(pij − 1)/2 > ∆(n).
Of course, this falls far short of our desired goal, since the roots of unityhave very high degree. What we would like is to have rational entries, andpreferably some systematic family of these.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 18 / 1
Conclusion of proof
To finish the proof of the theorem, we note that choosing distinct primespij for 1 ≤ i , j ≤ n, and setting θij = ζpij := e2π
√−1/pij , the linear
disjointness property is satisfied. So we just need to make sure thatpij − 1 > ∆(n) := 2n2n
2.
If we want real matrices, we can take θij = ζpij + ζ−1pij
, these generate thetotally real subfields of the cyclotomic fields, so we just need to ensure,(pij − 1)/2 > ∆(n).
Of course, this falls far short of our desired goal, since the roots of unityhave very high degree. What we would like is to have rational entries, andpreferably some systematic family of these.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 18 / 1
Semicontinuity?
One approach to constructing rigid matrices over the rationals is byconstructing rigid matrices over some totally real extension (as we havejust done), and then approximating these real numbers by rationalnumbers, hoping that rigidity does not change.
This will work reasonably well in our situation, since we’ve actuallyconstructed a matrix with a Zariski neighborhood (and therefore Euclideanneighborhood) disjoint from the less-than-maximally-rigid locus.
But one might wonder whether some kind of semicontinuity property holdsfor rigidity in general, as it holds for the rank function of matrices. Wewould like it if Rig(A, r) ≥ ℓ, that the same held true in a neighborhood ofA.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 19 / 1
Semicontinuity?
One approach to constructing rigid matrices over the rationals is byconstructing rigid matrices over some totally real extension (as we havejust done), and then approximating these real numbers by rationalnumbers, hoping that rigidity does not change.
This will work reasonably well in our situation, since we’ve actuallyconstructed a matrix with a Zariski neighborhood (and therefore Euclideanneighborhood) disjoint from the less-than-maximally-rigid locus.
But one might wonder whether some kind of semicontinuity property holdsfor rigidity in general, as it holds for the rank function of matrices. Wewould like it if Rig(A, r) ≥ ℓ, that the same held true in a neighborhood ofA.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 19 / 1
Semicontinuity?
One approach to constructing rigid matrices over the rationals is byconstructing rigid matrices over some totally real extension (as we havejust done), and then approximating these real numbers by rationalnumbers, hoping that rigidity does not change.
This will work reasonably well in our situation, since we’ve actuallyconstructed a matrix with a Zariski neighborhood (and therefore Euclideanneighborhood) disjoint from the less-than-maximally-rigid locus.
But one might wonder whether some kind of semicontinuity property holdsfor rigidity in general, as it holds for the rank function of matrices. Wewould like it if Rig(A, r) ≥ ℓ, that the same held true in a neighborhood ofA.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 19 / 1
Examples I
Unfortunately, this expectation is false. Let a, b, c , d , e be non-zerorational numbers. Consider
A =
a b c
d 0 0e 0 0
We have rank(A) = 2; Rig(A, 1) = 2.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 20 / 1
Examples II
Now, for any ǫ > 0 let
A(δ) =
a b c
d bdδ cdδe beδ ceδ
Change a to 1δ , rank of the matrix goes down to 1. Rig(A(δ), 1) = 1.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 21 / 1
Examples III
One can even produce such counterexamples which are maximally rigid.For instance,
A =
a b c
d e 0e 0 i
has Rig(A, 1) = 4 = (3− 1)2.
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 22 / 1
Examples IV
But
A(δ) =
a b c
d e cdδe bgδ i
has rigidity Rig(A(δ), 1) = 3, since one can change the diagonal entries toget
B =
1/δ b c
d bdδ cdδe bgδ cgδ
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 23 / 1
Some more thoughts, questions
Perhaps we can try to use similar techniques to show that the eliminationideals do not contain the ideal of the locus of Vandermonde matrices.Rational normal curves?
Can we systematically improve the bounds for effective Nullstellensatz,with some hypothesis on the degrees or type of generators. For example,for elimination ideals of determinantal ideals.
Coming back to the Valiant example of totally regular matrices with lowrigidity, can we find explicit examples of these, using algebraic geometryrather than graph theory?
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 24 / 1
Some more thoughts, questions
Perhaps we can try to use similar techniques to show that the eliminationideals do not contain the ideal of the locus of Vandermonde matrices.Rational normal curves?
Can we systematically improve the bounds for effective Nullstellensatz,with some hypothesis on the degrees or type of generators. For example,for elimination ideals of determinantal ideals.
Coming back to the Valiant example of totally regular matrices with lowrigidity, can we find explicit examples of these, using algebraic geometryrather than graph theory?
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 24 / 1
Some more thoughts, questions
Perhaps we can try to use similar techniques to show that the eliminationideals do not contain the ideal of the locus of Vandermonde matrices.Rational normal curves?
Can we systematically improve the bounds for effective Nullstellensatz,with some hypothesis on the degrees or type of generators. For example,for elimination ideals of determinantal ideals.
Coming back to the Valiant example of totally regular matrices with lowrigidity, can we find explicit examples of these, using algebraic geometryrather than graph theory?
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 24 / 1
Reference: “Using Elimination Theory to construct Rigid Matrices”,Abhinav Kumar, Satya V. Lokam, Vijay M. Patankar and Jayalal SarmaM.N. arXiv:0910.5301.
Thank you!
Abhinav Kumar (MIT) Matrix rigidity and elimination theory April 25, 2012 25 / 1