Noncommutative Partial Fractions and Continued Fractions

Noncommutative Partial Fractions and Continued Fractions

Darlayne Addabbo

Professor Robert Wilson

Department of Mathematics

Rutgers University

July 16, 2010

Overview

1) Existence of partial fraction decompositions over division rings

2) Relation between Laurent series and quasideterminants

3) Generalization of Galois’s result that every periodic continued fraction satisfies a quadratic equation

Partial Fractions in C (complex numbers)

• If f(x) is a polynomial over C of degree n with distinct roots , then

for some in C.€

λ1,...,λ n

€

1

f (x)=

A1

x − λ1

+ ...+Anx − λ n

€

A1,...,An

€

Example

€

1

x 2 −1=

1

2x −1

−

1

2x +1

We may also write as

€

1

f (x)=

A1

x − λ1

+ ...+Anx − λ n

€

( f (x))−1 = (x − λ1)−1A1 + ...+ (x − λ n )

−1An

The two expressions are the same over C. However, over a division ring,

is ambiguous. (It could be equal to

or .)

To avoid ambiguity, we prefer the second notation.

€

Aix − λ i

€

(x − λ i)−1Ai

€

Ai(x − λ i)−1

Can we generalize the method of partial fractions to an arbitrary

division ring, D?• Recall that a division ring satisfies all of the axioms of

a field except that multiplication is not required to be commutative.

• Over a field, if f(x) is a monic polynomial of degree n, with n distinct roots ,

• But this doesn’t work in a division ring. It doesn’t even work in the quaternions.

€

λ1,...,λ n

€

f (x) = (x − λ1)...(x − λ n )

(Recall) The Algebra of Quaternions

• The algebra of quaternions is a four dimensional vector space over R (the real numbers), with basis 1, i, j, k and multiplication satisfying:

• ij=-ji=k• jk=-kj=i• ki=-ik=j • 1 is the multiplicative identity

An example

has roots i+1 and 1+i+jCheck:

Similarly,

€

f (x) = x 2 − (2 + j)x + (2 + j − k)

€

(i +1)2 − (2 + j)(i +1) + 2 + j − k

= −1+ 2i +1− 2i + k − 2 − j + 2 + j − k

= 0

€

(1+ i + j)2 − (2 + j)(1+ i + j) + 2 + j − k = 0.

But

Instead,

Note that each root corresponds to the rightmost factor of one of these expressions. (This is due to a result of Ore.)

€

f (x) ≠ (x − (1+ i + j))(x − (1+ i))

€

f (x) = (x − (1− i + j))(x − (1+ i)) = (x − (1− i))(x − (1+ i + j))

Using the above to solve partial fractions

where , are elements of the quaternions

€

(x 2 − (2 + j)x + (2 + j − k))−1 = (x − (1+ i))−1A1 + (x − (1+ i + j))−1A2

€

x 2 − (2 + j)x + (2 + j − k) = (x − y)(x − (1+ i))

= (x − z)(x − (1+ i + j))

€

y

€

z

So which gives

and (since we can write and in terms of 1+i and 1+i+j),

Thus

So we have generalized the method of partial fractions.

€

1 = (x − y)A1 + (x − z)A2

€

A1 + A2 = 0

€

(1+ i)A1 + (1+ i + j)A2 =1

€

y

€

z

€

1 1

1+ i 1+ i + j

⎡

⎣ ⎢

⎤

⎦ ⎥A1

A2

⎡

⎣ ⎢

⎤

⎦ ⎥=

0

1

⎡

⎣ ⎢

⎤

⎦ ⎥

Theorem

Let , with

, have distinct roots,

.

Then there exist elements, such that

€

f (x) = x n + B1xn−1 + ...+ Bn

€

B1,...,Bn ∈ D

€

λ1,...,λ n ∈ D

€

A1,...,An ∈ D

€

( f (x))−1 = (x − λ1)−1A1 + ...+ (x − λ n )

−1An

1) First prove inductively that

€

(x − yn )(x − yn−1)...(x − y1) =

€

x n + (−1)( y i)xn−1 + (−1)2( y j1 y j0 )

j1 > j0

∑i=1

n

∑ x n−2 + ...+ (−1)n ynyn−1...y1

2) By the Gelfand-Retakh Vieta Theorem, we can write

€

f (x) = x n + B1xn−1 + B2x

n−2 + ...+ Bn

€

=(x − a1,1)(x − a1,2)...(x − a1,n )

€

=(x − a2,1)(x − a2,2)...(x − a2,n )

€

=...

€

=(x − an,1)(x − an,2)...(x − an,n )

where each is an element of our division ring, D, and for all

€

ai, j ,1 ≤ i ≤ n,1 ≤ j ≤ n,

€

ai,n = λ i

€

i,1 ≤ i ≤ n

3) Obtaining our set of equations

Recall that we want to find such that

We multiply both sides of our equation by f(x). Notice that our terms cancel.

€

Ais

€

(x − λ1)−1A1 + (x − λ 2)−1A2 + ...+ (x − λ n )

−1An

€

=(x n + B1xn−1 + B2x

n−2 + ...+ Bn )−1

€

(x − λ i)−1

4) Obtaining our System of Equations

Comparing terms on both sides of our equation, we obtain

Using the Generalized Cramer’s Rule, we can solve this system of equations .

€

1 1 1 ... 1

λ1 λ 2 λ 3 ... λ nλ1

2 λ 22 λ 3

2 ... λ n2

M M M M M

λ1n−1 λ 2

n−1 λ 3n−1 ... λ n

n−1

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

A1

A2

A3

M

An

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

=

0

0

0

M

1

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

Continued Fractions

In the commutative case, periodic continued fractions are often written as follows

, a field.€

α =a0 +1

a1 +1

a2 +1

...an +1

α

€

a0,a1,...,an ∈ F

Continued Fractions

In the noncommutative case, we avoid ambiguity by writing our continued fractions using nested parentheses.

where , a division ring

€

α =a0 + (a1 + (...+ (an +α −1)−1...)−1

€

a0,a1,...,an ∈ D

There is a theorem by Galois which says that every periodic continued fraction is a solution to a quadratic equation. I generalized this to show that a periodic continued fraction over a division ring, D, satisfies

where

We can write in terms of , the repeating terms of our continued fraction

€

xAx + Bx + xC + E = 0

€

A,B,C,E ∈ D

€

a1,a2,...,an

€

A,B,C,E

Current Work

Galois showed that there is a relationship between the complex conjugate of a periodic continued fraction and the periodic continued fraction obtained when we write the repeating terms in reverse order. We want to generalize this to the non-commutative case.

References

Gelʹfand, I. M.; Retakh, V. S. Theory of noncommutative determinants, and characteristic functions of graphs. (Russian) Funktsional. Anal. i Prilozhen. 26 (1992), no. 4, 1--20, 96; translation in Funct. Anal. Appl. 26 (1992), no. 4, 231--246 (1993)

Holtz, Olga, and Mikhail Tyaglov. "Structured Matrices, Continued Fractions, and Root Localization of Polynomials.” http://www.cs.berkeley.edu/~oholtz/RF.pdf

Lauritzen, Neils. "Continued Fractions and Factoring.”http://www.dm.unito.it/~cerruti/ac/cfracfact.pdf

Wilson, Robert L. "Three Lectures on Quasideterminants." Lecture. http://www.mat.ufg.br/bienal/2006/mini/wilson.pdf