Computing Moore-Penrose Inverses of Ore PolynomialMatrices

Yang Zhang

Department of Mathematics

University of Manitoba, Canada

Outline

•History and motivation.

• Theorems and algorithms for quaternion polynomials

• Examples

1

History

Applying algebraic methods to differential equations: 1930’s - 40’s.

E. Noether O. Ore N. Jacobson J. Wedderburn

First Question:

differential operators ? correspondence←−−−−−−−−−−−−−−→ rings

2

Differential Operators: Multiplication

Given differential operators: D := ddt , tD2+1 := t d2

dt2 +1 over Q(t).

Question: how to write D ◦ (tD2+1) in the form ∑ni=0 aiD i ?

where ai ∈Q(t).

? True: D ◦ (tD2+1) = tD3+D

3

Differential Operators: Multiplication

Given differential operators: D := ddt , tD2+1 := t d2

dt2 +1 over Q(t).

Question: how to write D ◦ (tD2+1) in the form ∑ni=0 aiD i ?

where ai ∈Q(t).

? True: D ◦ (tD2+1) = tD3+D

(D ◦ (tD2+1))(t3) = D ◦ ((tD2+1)(t3)) = 12t +3t2

(tD3+D)(t3) = (tD3)(t3)+D(t3) = 6t +3t2

Need conditions to ”swap” the positions of t and D !

4

Ore Polynomials

R Let σ be an automorphism of a ring R, i.e., σ is 1-1 and

σ(a+b) = σ(a)+σ(b) σ(ab) = σ(a)σ(b) ∀a,b ∈ R.

R A σ-derivation δ of R is a mapping R→ R satisfying: ∀a,b ∈ R,

δ(a+b) = δ(a)+δ(b), δ(ab) = σ(a)δ(b)+δ(a)b

ROre polynomial ring R[x;σ,δ] over R is the set of usual polynomi-als in x over R, i.e., {∑rixi | ri ∈ R}, with usual ”+ ” and

xr = σ(r)x+δ(r) ∀r ∈ R.

R Appeared in Noether and Schmeidler (1920). More discussiongiven in Ore (1933).

5

Examples: Usual Polynomials

When σ = 1: identity, and δ = 0: 0-derivative, i.e.,

σ(r) = r and δ(r) = 0 for any r ∈ R

we have

xr = σ(r)x+δ(r) = rx+0 = rx, Commutative !

In this case,

Ore polynomial ring R[x;1,0]||

Usual polynomial ring R[x]

6

Examples: Differential operators

Consider Q(t)[x;σ,δ].

Pure differential case: σ(t) = t: identity mapping;δ(t) = 1: usual derivative.

Commutative Rule: xt =σ(t)x+δ(t)= tx+1. Non-commutative!

Set D := ddt ←→ x and tD2+1 := t d2

dt2 +1←→ tx2+1.

D ◦ (tD2+1) = x(tx2+1) = xtx2+1 = (tx+1)x2+1 = tx3+ x2+1

= tD3+D2+1

7

Examples: Difference Operators

Consider Q(t)[x;σ,δ].

Pure difference case: σ(t) = t +1 shift mappingδ(t) = 0 zero derivative

Commutative rule: ∀ f (t) ∈Q(t)

x f (t) = σ( f (t))x+δ( f (t)) = f (t +1)x

In particular, xt = σ(t)x = (t +1)x.

8

History: Ore Polynomials

ROne of the main research fields in Ring Theory.- 1920s – 1950s: Noether, Ore, Jacobson, etc.- 1960s – present: Cohn, Goodearl, Lam, etc.

ROne of the main research fields in Computer Algebra.

R Applications:Differential (difference) equations, Model theory, Coding theory,Control theory, Cryptography.

RMaple packages: Ore, Ore module.

9

Matrices over Ore Polynomial Rings

Let R[x;σ,δ] be an Ore polynomial ring and R[x;σ,δ]n×m be the setof all n×m matrices over R[x;σ,δ].

Questions: compute various generalized inverses in R[x;σ,δ]n×m:

{1}-inverse, {1,2}-inverse, Moore-Penrose inverse, etc.

Difficult points:Ore polynomials are noncommutative algebra, and have a muchmore complex structure.

Many of the algorithmic breakthroughs in computer algebra overthe past three decades do not obviously apply in these domains.

10

Moore-Penrose Inverses for Quaternion Matrices

In 1843 Sir Rowan Hamilton discovered the algebra H of real quater-nion, which is a four-dimensional non-commutative algebra over Rwith canonical basis 1, i, j,k satisfying the conditions:

i2 = j2 = k2 = ijk =−1,

that implies

ij =−ji = k, jk =−kj = i, and ki =−ik = j.

The elements in H can be written in a unique way:

α = a+bi+ cj+dk.The conjugate of α is defined as α = a−bi−cj−dk, and the norm|α| is |α|=

√αα.

11

Quaternion Polynomials

The study of quaternion polynomials may go back to Niven in theearly 1940’s.

A quaternion polynomial f (x) over H is defined as

f (x) = anxn+ · · ·+a1x+a0, ai ∈H, i = 0, . . . ,n,

where x commutes element-wise with H.

The conjugate of f (x) = anxn + · · ·+ a0 ∈ H[x] is defined as f (x) =anxn+ · · ·+ a0, and has the following properties:

Properties Let f ,g ∈ H[x]. Then (i) f g = g f (ii) f f = f f ∈ R[x],where R are reals (iii) If f g ∈ R[x], then f g = g f .

12

Definitions

Let H[x]m×n denote the set of all m×n matrices over H[x].

For A ∈H[x]m×n, the conjugate A of A is defined as A = (Ai j).

If A = P+Qj with P, Q ∈ C[x]m×n, then χA =

(P Q−Q P

)∈ C[x]2m×2n

denotes the complex adjoint of A.

Moreover, AT ,A∗ ∈H[x]n×m denote the transpose and the conjugatetranspose of A, respectively.

13

Definitions

A† ∈ H[x]n×m is called a Moore-Penrose inverse of A ∈ H[x]m×n if itis a solution of the following system of equations:

AXA = A, XAX = X , (AX)∗ = AX , (XA)∗ = XA.

Note that we require that A† must be in H[x]n×m.

A ∈Hm×m is unitary if AA∗ = A∗A = Im.

14

Properties

Let A ∈H[x]m×n and B ∈H[x]n×l. Then

(i) (AB)∗ = B∗A∗ and AA∗ = (AA∗)∗.

(ii) If A has a Moore-Penrose inverse A†, then (A∗)† =(A†)∗,

A†(A†)∗A∗ = A† = A∗

(A†)∗A† and A†AA∗ = A∗ = A∗AA†.

(iii) If A has a Moore-Penrose inverse A†, then A† is unique.

(iv) Let A have the Moore-Penrose inverse A†. If U ∈ Hm×m is a uni-tary matrix, then (UA)† = A†U∗.

15

Properties

Lemma If E ∈ H [x]m×m is a symmetric projection, that is, E = E2 =E∗, then E ∈Hm×m.

Lemma A ∈ Hm×m is hermitian, that is, A = A∗, if and only if thereexists a unitary matrix U ∈Hm×m such that

U∗AU = diag(d1, . . . ,dm),

where di are the eigenvalues of A.

16

Theorem

Let A ∈ H [x]m×n. Then A has the Moore-Penrose inverse A† if andonly if

A =U(

A1 A2

0 0

)with U ∈ Hm×m unitary and A1A∗1 +A2A∗2 a unit in H [x]r×r with r ≤min{m, n}. Moreover,

A† =

(A∗1 (A1A∗1+A2A∗2)

−1 0A∗2 (A1A∗1+A2A∗2)

−1 0

)U∗.

17

Leverrier-Faddeev algorithm

Given A ∈ H[x]m×m. An element λ ∈ H is called an eigenvalue of Aif there exists a vector X ∈Hm×1[x] such that AX = Xλ.

Lemma Let A ∈H[x]m×n. Then eigenvalues of AA∗ are real.

Let A ∈ H[x]m×n and set B = AA∗. Then fB (λ) = det(λI2m−χB) iscalled the characteristic polynomial of A.

Theorem Let A ∈ H [x]m×n and B = AA∗. Then fB (λ) = g(λ)2 whereg(λ) ∈ (R [x]) [λ].

18

Characteristic polynomials

Let A∈H [x]m×n, B = AA∗ and fB (λ) = g(λ)2. Then g(B) = 0. We willcall g(λ) the generalized characteristic polynomial of A.

Lemma Let A ∈ H[x]m×n have the Moore-Penrose inverse A†. SetB = AA∗. Then(i) B† = (A∗)† A† and B†B = AA†.

(ii) B†B = BB† and (B†B)2 = B†B.

(iii)(B†)k

=(Bk)† and (Bn−k)†Bn−k = B†B, for any k ∈ N.

19

Formula

Theorem Let A ∈ H [x]m×n have the Moore-Penrose inverse A† andB = AA∗. Suppose the generalized characteristic polynomial of A is

g(λ) = λm+a1λ

m−1+ · · ·+akλm−k + · · ·+am−1λ+am,

where ai ∈ R [x].If k is the largest integer such that ak 6= 0, then the generalizedinverse of A is given by

A† =− 1ak

A∗[Bk−1+a1Bk−2+ · · ·+ak−1I

].

If ai = 0 for all 1≤ i≤ m, then A† = 0.

20

Fadeev-Leverrier’s method

Lemma Let A ∈ H [x]m×n have the Moore-Penrose inverse A† andset B = AA∗. Then for 1≤ k ≤ m,

tr[(

Bk +a1Bk−1+ · · ·+ak−1B)]

=−kak,

where the ai arise from the generalized characteristic polynomialof A:

g(λ) = λm+a1λ

m−1+ · · ·+akλm−k + · · ·+am−1λ+am.

21

Laverrier-Faddeev algorithm

Let A ∈H [x]m×n have the generalized inverse A† and B = AA∗. Sup-pose the generalized characteristic polynomial of A is

g(λ) = λm+a1λ

m−1+ · · ·+akλm−k + · · ·+am−1λ+am,

where ai ∈ R [x]. Define a0 = 1. If p is the largest integer such thatap 6= 0 and we construct the sequence A0, · · · , Ap as follows:

A0 = 0 −1 = q0 B0 = IA1 = AA∗B0

trA11 = q1 B1 = A1−q1I

... ... ...Ap−1 = AA∗Bp−2

trAp−1p−1 = qp−1 Bp−1 = Ap−1−qp−1I

Ap = AA∗Bp−1trAp

p = qp Bp = Ap−qpI

then qi (x) =−ai (x) , i = 0, · · · , p.

22

Laverrier-Faddeev algorithm

Leverrier-Faddav algorithm for quaternion polynomial matrices

Input: A ∈H[x]m×n

Output: The Moore-Penrose inverse A† of A in H[x]n×m if exists.

1. B0← Im, a0← 1

2. for i = 1, . . . ,m doAi← AA∗Bi−1, ai←−trAi

i , Bi← Ai+aiIm

3. Find the maximal index p such that ap 6= 0.

4. Return A† =

{− 1

apA∗Bp−1, p > 0,

0, p = 0.

23

Finding Moore-Penrose inverses by interpolation

Let f , g and h ∈H [x], f = gh and r ∈H.

If h(r) = 0, then f (r) = 0.

Otherwise, set β = h(r) 6= 0. Then the evaluation of f (x) at x = r isdefined as

f (r) = g(βrβ

−1)h(r) .

In particular, if r is a root of f but not of h, then βrβ−1 is a root of g.

In 1965, Gordon proved that if f ∈ H [x] is of degree n, then theroots of f lie in at most n conjugacy classes of H.

24

Interpolation

Theorem Let c1, · · · , cn be n pairwise non-conjugate elements ofH. Then there is a unique monic polynomial n ∈ H [x] of degree nsuch that gn (c1) = · · ·= gn (cn) = 0.Moreover, c1, . . . ,cn are the only roots (up to conjugacy classes) ofgn in H.

Theorem Let c1, · · · , cn+1 ∈ H be pairwise non-conjugate and letd1, · · · ,dn+1 ∈H. Then there exists a unique lowest degree polyno-mial f ∈ H [x], of degree p ≤ n, such that f (ci) = di for all 1 ≤ i ≤n+1.

25

Degree Bonds

For a given A ∈H [x]m×n, the degree

degA = max{deg(Ai j) | 1≤ i≤ m, 1≤ j ≤ n} .

Lemma Let A ∈H [x]m×n have the Moore-Penrose inverse A†. Then

degA† ≤ (2m−1)degA.

Proposition Let c1, · · · , ck+1 ∈ H be pairwise non-conjugate and letA1, · · · , Ak+1 ∈ Hn×m. Then there is a unique lowest degree matrixA∈H [x]n×m of degree p≤ k, such that A(ci) = Ai for all 1≤ i≤ k+1.

26

Interpolation method

Let A ∈ H [x]m×n have the Moore-Penrose inverse A†, and set B =AA∗. Let p be the largest integer such that ap 6= 0. We constructthe sequence A0, · · · , Ap as follows:

A0 = 0 −1 = q0 B0 = I... ... ...

Ap−1 = AA∗Bp−2trAp−1

p−1 = qp−1 Bp−1 = Ap−1−qp−1IAp = AA∗Bp−1

trApp = qp Bp = Ap−qpI.

27

Algorithm

Theorem In the above setting, let k = (2m−1)degA and c1, · · · ,ck+1 ∈ R be k+1 distinct real numbers such that qp (cs′) 6= 0 for any1≤ s′ ≤ k+1. Let S = {1, · · · , k+1}\{s′}. Then

A† =k+1

∑s′=1

A(cs′)† gS

where

A(cs′)† =

1qp (cs′)

A(cs′)∗[B(cs′)

p−1−q1 (cs′)B(cs′)p−2−·· ·−qp−1 (cs′) I

]and

gS (cα) =

{0 α ∈ S,1 α = s′.

28

Example

A =

(14x+14+76i+70j+56k 56−28i−70j+70k 28j−56k 14x−56−8i−14j−56k−2x−2−43i−10j−8k −8+4i+10j−10k −4j+8k −2x+8−31i+2j+8k−3x−3+3i−15j−12k −12+6i+15j−15k −6j+12k −3x+12+21i+3j+12k−4x−4+4i−20j−16k −16+8i+20j−20k −8j+16k −4x+16+28i+4j+16k

)∈H4×3[x]

The upper bound of the degree of A† is less than (2m− 1)degA =(2×4−1) ·1 = 7. Choose c1 = 0 and c2 = 1.

A(c1) =

(14+76i+70j+56k 56−28i−70j+70k 28j−56k −56−8i−14j−56k−2−43i−10j−8k −8+4i+10j−10k −4j+8k 8−31i+2j+8k−3+3i−15j−12k −12+6i+15j−15k −6j+12k 12+21i+3j+12k−4+4i−20j−16k −16+8i+20j−20k −8j+16k 16+28i+4j+16k

)and

A(c2) =

(28+76i+70j+56k 56−28i−70j+70k 28j−56k −42−8i−14j−56k−4−43i−10j−8k −8+4i+10j−10k −4j+8k 6−31i+2j+8k−6+3i−15j−12k −12+6i+15j−15k −6j+12k 9+21i+3j+12k−8+4i−20j−16k −16+8i+20j−20k −8j+16k 12+28i+4j+16k

).

29

we calculate and obtain:

A(c1)† = A(0)† =

1230175

×140−560i−228j−342k 355+1730i−96j+81k −255−870i+126j+54k −340−1160i+168j+72k276+88i+426j−382k 282+416i−93j−149k −252−276i−72j+204k −336−368i−96j+272k32+16i−176j+292k −176−88i+68j+194k 96+48i+12j−204k 128+64i+16j−272k−140−122i+228j+342k −355+2021i+96j−81k 255−1176i−126j−54k 340−1568i−168j−72k

and

A(c2)† = A(1)† =

1230175

×152−550i−244j−330k 289+1675i−8j+15k −219−840i+78j+90k −292−1120i+104j+120k268+104i+406j−402k 326+328i+17j−39k −276−228i−132j+144k −368−304i−176j+192k

32+16i−160j+300k −176−88i−20j+150k 96+48i+60j−180k 128+64i+80j−240k−152−132i+244j+330k −289+2076i+8j−15k 219−1206i−78j−90k 292−1608i−104j−120k

.

30

A† =2

∑s′=1

A(cs′)† gS = A(0)† (1− x)+A(1)† x

=1

230175×

(12+10i−16j+12k)x+140−560i−228j−342k (−66−55i+88j−66k)x+355+1730i−96j+81k(−8+16i−20j−20k)x+276+88i+426j−382k (44−88i+110j+110k)x+282+416i−93j−149k

(16j+8k)x+32+16i−176j+292k (−88j−44k)x−176−88i+68j+194k(−12−10i+16j−12k)x−140−122i+228j−342k (66+55i−88j+66k)x−355+2021i+96j−81k(36+30i−48j+36k)x−255−870i+126j+54k (48+40i−64j+48k)x−340−1160i+168j+72k(−24+48i−60j−60k)x−252−276i−72j+204k (−32+64i−80j+80k)x−366−368i−96j+272k

(48j+24k)x+96+48i+12j−204k (64j+32k)x+128+64i+16j−272k(−36−30i+48j−36k)x+255−1176i−126j−54k (−48−40i+64j−48k)x+340−1568i−168j−72k

.

31

Example

Let A =

(1 i+2k 3i 6+ j 7

)2×3

.

Then its Moore-Penrose inverse can be found by using our Maplepackage as follows:

A† =

47

347 +21

694i+ 11694j − 21

694−11347i− 11

694k

− 63347−

28347i+ 21

694j− 101694k 61

694 +21694i− 6

347j+ 21347k

57347 +

49694i+ 77

694k 21347−

21694i− 33

694k

3×2

.

32