Matrices of nonnegative rank at most three Emil Horobet (with Rob - - PowerPoint PPT Presentation

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Matrices of nonnegative rank at most three Emil Horobet (with Rob H. Eggermont and Kaie Kubjas) Eindhoven University of Technology e.horobet@tue.nl Nonnegative rank A fictional study on Does watching football cause hair loss?

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Matrices of nonnegative rank at most three

Emil Horobet ¸ Eindhoven University of Technology e.horobet@tue.nl (with Rob H. Eggermont and Kaie Kubjas)

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Nonnegative rank

  • A fictional study on ”Does watching football cause hair

loss?”

  51 45 33 28 30 29 15 27 38   =: U

≤ 2 h 2 − 6 h ≥ 6 h plenty medium less

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SLIDE 3

Nonnegative rank

  • A fictional study on ”Does watching football cause hair

loss?”

  51 45 33 28 30 29 15 27 38   =: U

≤ 2 h 2 − 6 h ≥ 6 h plenty medium less

  • The 3 × 3 contingency table is of rank 2, so apparently they

are correlated.

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SLIDE 4

Nonnegative rank

  • A fictional study on ”Does watching football cause hair

loss?”

  51 45 33 28 30 29 15 27 38   =: U

≤ 2 h 2 − 6 h ≥ 6 h plenty medium less

  • The 3 × 3 contingency table is of rank 2, so apparently they

are correlated.

  • We get a much better understanding if we consider the

hidden variable gender.

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SLIDE 5

U =   3 9 15 4 12 20 7 21 35   +   48 36 18 24 18 9 8 6 3  

  • We get the sum of two rank one contingency tables:
  • So watching football and hair loss are independent given

the gender.

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SLIDE 6

U =   3 9 15 4 12 20 7 21 35   +   48 36 18 24 18 9 8 6 3  

  • We get the sum of two rank one contingency tables:
  • So watching football and hair loss are independent given

the gender.

Main motivation: Determine whether such an ”expanatory random variable” exists and what the number of its states is?

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Nonnegative rank

Given M ∈ Rm×n

≥0

its nonnegative rank is the smallest r such that there exist A ∈ Rm×r

≥0

and B ∈ Rr×n

≥0

with M = A · B.

  • Matrices of nonnegative rank at most r form a semialgebraic

set Mr

m×n.

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Nonnegative rank

Given M ∈ Rm×n

≥0

its nonnegative rank is the smallest r such that there exist A ∈ Rm×r

≥0

and B ∈ Rr×n

≥0

with M = A · B.

  • Matrices of nonnegative rank at most r form a semialgebraic

set Mr

m×n.

  • If the nonnegative rank is 1 or 2, then it equals the rank.
  • First interesting example is M3

m×n.

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  • We want to solve optimization problems on M3

m×n (e.g.

maximum likelihood estimation).

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  • We want to solve optimization problems on M3

m×n (e.g.

maximum likelihood estimation).

  • For this we need to understand the Zariski closure of the

boundary. topological boundary ∂(M3

m×n)

algebraic boundary ∂(M3

m×n)

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Algebraic boundary of M3

m×n

Let us denote the multiplication map by µ : Mm×3 × M3×n → Mm×n ((aik), (bkj)) → (xij).

Theorem (Kubjas-Robeva-Sturmfels)

The irreducible components of ∂(M3

m×n) are

  • components defined by xij = 0
  • components parametrized by

(xij) = A · B or (xij) = B · A, where A has four zeros in three columns and distinct rows, and B has three zeros in distinct rows and columns.

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Consider the following irreducible component Let A =                     ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ . . . . . . . . .                     and B =      ∗ ∗ · · · ∗ ∗ · · · ∗ ∗ · · ·      µ : Mm×3 × M3×n Mm×n A × B ⊆ µ(A × B) ⊆

X

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Main Theorem (Eggermont-H.-Kubjas)

The ideal of the variety X is generated by certain degree 6 polynomials and degree 4 determinants I(X) = (fi, detj,k)i,j,k. The following theorem was previously conjectured by Kubjas-Robeva-Sturmfels.

Generators of the ideal of X.

Moreover, these polynomials form a Gr¨

  • bner basis with respect

to the graded reverse lexicographic term order.

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Steps of the proof

GL3 A × B g · (A, B) = (Ag−1, gB)

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Steps of the proof

GL3 A × B g · (A, B) = (Ag−1, gB) GL3 R[Mm×3 × M3×n]

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SLIDE 16

Steps of the proof

GL3 A × B g · (A, B) = (Ag−1, gB) GL3 R[Mm×3 × M3×n] GL3 · (A × B) has codim. 1

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Steps of the proof

GL3 A × B g · (A, B) = (Ag−1, gB) GL3 R[Mm×3 × M3×n] GL3 · (A × B) has codim. 1 Pick f ∈ I(X) from K-R-S µ∗f = det Bi · f6,3

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Steps of the proof

GL3 A × B g · (A, B) = (Ag−1, gB) GL3 R[Mm×3 × M3×n] GL3 · (A × B) has codim. 1 Pick f ∈ I(X) from K-R-S µ∗f = det Bi · f6,3 I(GL3(A × B)) = (f6,3)

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Steps of the proof (cont.)

I(GL3(A × B)) = (f6,3)

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Steps of the proof (cont.)

I(GL3(A × B)) = (f6,3) GL3(A × B) dense in X µ∗I(X) = I(GL3(A × B) ∩ Imµ∗

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Steps of the proof (cont.)

I(GL3(A × B)) = (f6,3) GL3(A × B) dense in X µ∗I(X) = I(GL3(A × B) ∩ Imµ∗ Imµ∗ = R[Mm×3 × M3×n]GL3

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Steps of the proof (cont.)

I(GL3(A × B)) = (f6,3) GL3(A × B) dense in X µ∗I(X) = I(GL3(A × B) ∩ Imµ∗ Imµ∗ = R[Mm×3 × M3×n]GL3 µ∗I(X) = (f6,3)GL3 = {

i f6,3 · det Bi · hi, hi is GL3-inv.}

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Steps of the proof (cont.)

I(GL3(A × B)) = (f6,3) GL3(A × B) dense in X µ∗I(X) = I(GL3(A × B) ∩ Imµ∗ Imµ∗ = R[Mm×3 × M3×n]GL3 µ∗I(X) = (f6,3)GL3 = {

i f6,3 · det Bi · hi, hi is GL3-inv.}

I(X) = (fi, detj,k)i,j,k

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Stabilization of the boundary

  • For matrix rank, if m, n are large enough, then already a

submatrix has the given rank.

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Stabilization of the boundary

  • For matrix rank, if m, n are large enough, then already a

submatrix has the given rank.

  • When both m and n tend to infinity, this is not true for the
  • nonneg. rank.

Example of Moitra: A 3n × 3n matrix with nonneg. rank 4 and all its 3n × n submatrices have nonneg. rank 3.

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Stabilization of the boundary

  • For matrix rank, if m, n are large enough, then already a

submatrix has the given rank.

  • When both m and n tend to infinity, this is not true for the
  • nonneg. rank.

Example of Moitra: A 3n × 3n matrix with nonneg. rank 4 and all its 3n × n submatrices have nonneg. rank 3. Given M ∈ Rm×n

≥0

, let W = Span(M) ∩ ∆m−1 and V = Cone(M) ∩ ∆m−1.

Lemma

Let rank(M) = r, then M has nonneg. rank exactly r if and

  • nly if there exists a (r −1)-simplex ∆, such that V ⊆ ∆ ⊆ W.
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Example

Here M ǫ is a 12 × 12 matrix of rank 3 and nonnegative rank 4. Any 12 × 5 (i.e 3n × (⌈ 3

2n⌉ − 1)) submatrix has nonnegative

rank 3. M M ǫ

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  • We have seen that there is no stabilization for the

topological boundary.

  • What about the stabilization of the algebraic boundary?

Stabilization of the boundary

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  • We have seen that there is no stabilization for the

topological boundary.

  • What about the stabilization of the algebraic boundary?

Claim

For n > 4, let M ∈ ∂(M3

m×n). Then we can find a column i0

such that M i0 ∈ ∂(M3

m×n−1),

where M i0 is obtained by removing the i0-th column of M.

Stabilization of the boundary

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Conjecture

For given r ≥ 3, there exists n0 ∈ N, such that for all n ≥ n0, and all matrices M on ∂(Mr

m×n), there is a column i0 such

that M i0 ∈ ∂(Mr

m×n−1),

where M i0 is obtained by removing the i0-th column of M.

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Conjecture

For given r ≥ 3, there exists n0 ∈ N, such that for all n ≥ n0, and all matrices M on ∂(Mr

m×n), there is a column i0 such

that

Thank you!

M i0 ∈ ∂(Mr

m×n−1),

where M i0 is obtained by removing the i0-th column of M.