Alex Townsend Cornell Universitytownsend@cornell.edu

Vanni Noferini Sujit Rao

ALGEBRAIC-GEOMETRIC METHODS FOR MULTIDIMENSIONAL POLYNOMIAL ROOTFINDING

Joint work with:

MULTIDIMENSIONAL ROOTFINDING

• Polynomials are of maximal degree (degree in each variable)

• Number of isolated solutions (Bernstein’s Theorem)

n n

d!nd

Rootfinding problem: Find all the solutions to:

Assumptions (the algebraically boring situation):•Zero-dimensional polynomial system with well-conditioned roots•Finite and simple solutions (Jacobian is invertible at roots)

0

B@p1(x1, . . . , xd)

...pd(x1, . . . , xd)

1

CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd

TWO DIFFICULT BUT FEASIBLE REGIMES

Robotics (inverse kinematics):

Witness sets convert this to a zero dimensional solution set [Sommese & Wampler, 2005].

Small n, large d Large n, small d

Rootfinders: Resultants, Homotopy, Möller-Stetter, Gröbner bases, Newton's method, etc.

−1 −0.5 0 0.5 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

[Belyaev, 2015]

Random plane waves:

NUMERICAL STABILITY

Goal: A stable numerical algorithm for polynomial rootfinding.

Absolute condition number for rootfinding: In 1D:

x

⇤is a root, [J(x

⇤)]

jk

=

@pj

@xk(x

⇤)

A stable numerical method should compute a root with errorO(kJ(x⇤

)

�1k2✏), ✏ = unit roundo↵

(x⇤) =1

|p0(x⇤)|(x⇤) = kJ(x⇤)�1k2

[Higham, 2002]

INHERIT ROBUSTNESS FROM EIGENSOLVER

(�, A) =||v||2||w||2

|v⇤w|Av = �v w = left eigenvector

Algebraic-geometric

method

For a simple eigenvalue:

[Kressner, Peláez, & Moro, 09]

UNIVARIATE POLYNOMIAL ROOTFINDING

p(x) = a0 + a1x+ · · ·+ anxn

For example,

is the characteristic polynomial forthe companion matrix:

C =

2

6664

0 � a0an

1 � a1an

. . ....

1 �an�1

an

3

7775

Leads to a numerically stable algorithm for rootfinding on the unit circle. [Van Dooren & Dewilde, 83]

ROOTFINDING IN ONE VARIABLE

DEMO

PARTIAL SURVEY OF GLOBAL ROOTFINDERS

Rootfinding problem: Find all the solutions to:0

B@p1(x1, . . . , xd)

...pd(x1, . . . , xd)

1

CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd

TWO IDEAS FROM ALGEBRAIC GEOMETRY

1) Multidimensional resultants

2) Möller-Stetter matrices

[Noferini & T., 2015]

[Noferini, Rao, & T., 2018]

Michael Möller Hans Stetter

LINEAR EXAMPLE: CRAMER’S RULE

Cramer’s rule as an eigenproblem: A1v = �x1(A(:, 1)eT1 )v

HIDDEN-VARIABLE RESULTANTS FOR BIVARIATE POLYNOMIALS

p(x, y) = q(x, y) = 0

p(x, y) =nX

j,k=0

ajkxjy

kq(x, y) =

nX

j,k=0

bjkxjy

k

Rootfinding problem:

where

q[x](y) =nX

k=0

bk(x)yk

p[x](y) =nX

k=0

ak(x)yk

“Hide” one of the variables:

,

Bézout resultant:

The matrix R is singular, where

p(s)q(t)� p(t)q(s)

s� t=

n�1X

j,k=0

Rjksjtk

p(y) and q(y) have a common root

RESULTANTS WITH MATRIX POLYNOMIALS

In algebraic geometry they form the resultant: R(x) = det(R[x])

We avoid the determinant: [Nakatsukasa, Noferini, T., 2015]

CAYLEY RESULTANT MATRIX (TRYING TO AVOID THE EYE-WATERING TENSOR MANIPULATIONS)

0

B@p1(x1, . . . , xd)

...pd(x1, . . . , xd)

1

CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd

In higher dimensions, one can still use multidimensional resultants:

R(x1)v = 0Still, obtain a poly. eig. problem:

DEVASTATING EXAMPLE FOR CAYLEY RESULTANTS

Devastating example for absolute (and relative) conditioning: 0

B@p1(x1, . . . , xd)

.

.

.

pd(x1, . . . , xd)

1

CA =

0

B@x

21.

.

.

x

2d

1

CA+ uQ

0

B@x1.

.

.

xd

1

CA , Q = orthogonal

Cond. no. of rootfinding problem: kJ(x⇤)�1k2 =1

u

kQ�1k2 =1

u

Cond. no. of eigenproblem: (x⇤, R(x1)) =

1

det(J(x⇤))=

1

u

d

J(x⇤) = uQ

x

⇤ = 0

[Noferini & T., 2015]

A devastating example exists for any reasonable basis, any linear combination of solution components, and symbolic calculation of resultant. Also, devastating for relative conditioning.

A NEGATIVE RESULT WITH (HOPEFULLY) A POSITIVE IMPACT

BivariatePolynomial (in MAPLE), chebfun2v/roots (in Chebfun), & Roots (in Mathematica). Numerically unstable bivariate rootfinders:

ROOTFINDING IN TWO DIMENSIONS

DEMO

TWO IDEAS FROM ALGEBRAIC GEOMETRY

1) Multidimensional resultants

2) Möller-Stetter matrices

[Noferini & T., 2015]

[Noferini, Rao, & T., 2018]

Michael Möller Hans Stetter

BACK TO THE COMPANION MATRIX

p(x) = 0, p(x) = a0 + a1x+ · · ·+ anxn

2

6664

0 � a0an

1 � a1an

. . ....

1 �an�1

an

3

7775

2

6664

b0b1...

bn�1

3

7775=

2

66664

� bn�1a0

an

b0 � bn�1a1

an

...

bn�2 � bn�1an�1

an

3

77775

Companion matrix as Euclidean division:q(x) = b0 + b1x+ . . .+ bn�1x

n�1

xq(x) = b0x+ b1x2 + . . .+ bn�1x

n

Euclidean division: xq(x) = c(x)p(x) + r(x), deg(r) < n

c(x) = bn�1

an, r(x) = xq(x)� bn�1

anp(x)

C : Pn�1 ! Pn�1, C(q) = mod(xq, p)

MÖLLER-STETTER MATRICES

One can construct a matrix

0

B@p1(x1, . . . , xd)

...pd(x1, . . . , xd)

1

CA = 0Suppose that has N solutions.

M1 2 CN⇥N

[Lazard, 81], [Möller, 93], [Möller-Stetter, 95]whose eigs are the x1

component of the solutions.

M1 =

2

664

0 0 0 01 �u 0 �u2

0 �u 0 �u2

0 0 1 �u

3

775

For example: p(x, y) = x

2 + u(x+ y)q(x, y) = y

2 + u(x� y)

represents multivariable Euclidean division wrtM1 {1, x, y, xy}

u =p22

DEVASTATING EXAMPLE FOR MÖLLER-STETTER

Devastating example for absolute (and relative) conditioning: 0

B@p1(x1, . . . , xd)

.

.

.

pd(x1, . . . , xd)

1

CA =

0

B@x

21.

.

.

x

2d

1

CA+ uQ

0

B@x1.

.

.

xd

1

CA , Q = orthogonal

A devastating example exists for any reasonable basis, any linear combination of solution components, and symbolic calculation of the Möller-Stetter matrix.

[Noferini, Rao, & T., 2018]

Cond. no. of rootfinding problem: kJ(x⇤)�1k2 =1

u

kQ�1k2 =1

u

Cond. no. of eigenproblem:

J(x⇤) = uQ

x

⇤ = 0

(x⇤,M1) �

1

u

d

THANK YOU

Thanks to: Anthony Austin, Daniel Bates, John Boyd, Martin Lotz, Nick Higham, Tyler Jarvis, Gregorio Malajovich, Yuji Nakatsukasa, Bor Plestenjak, and Andrew Sommese, and Mike Stillman.

Is it possible to design a practical and stable multidimensional rootfinder based on eigensolvers?

Promising ongoing work by:

Bor Plestenjak Marc Van BarelTyler Jarvis

SYLVESTER RESULTANT MATRIX