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Computing Moore-Penrose Inverses of Ore PolynomialMatrices
Yang Zhang
Department of Mathematics
University of Manitoba, Canada
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
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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
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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).
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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]
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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
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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.
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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.
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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.
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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.
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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 .
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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.
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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.
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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∗.
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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.
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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∗.
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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].
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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′.
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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
).
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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
.
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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
.
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