Computing Moore-Penrose Inverses of Ore Polynomial Matrices Yang Zhang

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1 Computing Moore-Penrose Inverses of Ore Polynomial Matrices Yang Zhang Department of Mathematics University of Manitoba, Canada

2 Outline History and motivation. Theorems and algorithms for quaternion polynomials Examples 1

3 History Applying algebraic methods to differential equations: 1930 s - 40 s. E. Noether O. Ore N. Jacobson J. Wedderburn First Question: differential operators? correspondence 2 rings

4 Differential Operators: Multiplication Given differential operators: D := d dt, td2 + 1 := t d2 dt over Q(t). Question: how to write D (td 2 + 1) in the form n i=0 a i D i? where a i Q(t).? True: D (td 2 + 1) = td 3 +D 3

5 Differential Operators: Multiplication Given differential operators: D := d dt, td2 + 1 := t d2 dt over Q(t). Question: how to write D (td 2 + 1) in the form n i=0 a i D i? where a i Q(t).? True: D (td 2 + 1) = td 3 +D (D (td 2 + 1))(t 3 ) = D ((td 2 + 1)(t 3 )) = 12t + 3t 2 (td 3 +D)(t 3 ) = (td 3 )(t 3 ) +D(t 3 ) = 6t + 3t 2 Need conditions to swap the positions of t and D! 4

6 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, R Ore polynomial ring R[x;σ,δ] over R is the set of usual polynomials in x over R, i.e., { r i x i r i R}, with usual + and xr = σ(r)x + δ(r) r R. δ(a + b) = δ(a) + δ(b), δ(ab) = σ(a)δ(b) + δ(a)b R Appeared in Noether and Schmeidler (1920). More discussion given in Ore (1933). 5

7 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

8 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 := d dt x and td2 + 1 := t d2 dt tx D (td 2 + 1) = x(tx 2 + 1) = xtx = (tx + 1)x = tx 3 + x = td 3 +D

9 Examples: Difference Operators Consider Q(t)[x; σ, δ]. Pure difference case: σ(t) = t + 1 shift mapping δ(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

10 History: Ore Polynomials R One of the main research fields in Ring Theory s 1950s: Noether, Ore, Jacobson, etc. R s present: Cohn, Goodearl, Lam, etc. One of the main research fields in Computer Algebra. R Applications: Differential (difference) equations, Model theory, Coding theory, Control theory, Cryptography. R Maple packages: Ore, Ore module. 9

11 Matrices over Ore Polynomial Rings Let R[x;σ,δ] be an Ore polynomial ring and R[x;σ,δ] n m be the set of 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 much more complex structure. Many of the algorithmic breakthroughs in computer algebra over the past three decades do not obviously apply in these domains. 10

12 Moore-Penrose Inverses for Quaternion Matrices In 1843 Sir Rowan Hamilton discovered the algebra H of real quaternion, which is a four-dimensional non-commutative algebra over R with canonical basis 1, i, j, k satisfying the conditions: that implies i 2 = j 2 = k 2 = ijk = 1, 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 ᾱ = a bi cj dk, and the norm α is α = αᾱ. 11

13 Quaternion Polynomials The study of quaternion polynomials may go back to Niven in the early 1940 s. A quaternion polynomial f (x) over H is defined as f (x) = a n x n + + a 1 x + a 0, where x commutes element-wise with H. a i H, i = 0,...,n, The conjugate of f (x) = a n x n + + a 0 H[x] is defined as f (x) = ā n x n + + ā 0, and has the following properties: Properties Let f,g H[x]. Then (i) f g = ḡ f (ii) f f = f f R[x], where R are reals (iii) If f g R[x], then f g = g f. 12

14 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 = (A i j ). If A = P + Qj with P, Q C[x] m n, then χ A = denotes the complex adjoint of A. ( ) P Q Q P C[x] 2m 2n Moreover, A T,A H[x] n m denote the transpose and the conjugate transpose of A, respectively. 13

15 Definitions A H[x] n m is called a Moore-Penrose inverse of A H[x] m n if it is 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 H m m is unitary if AA = A A = I m. 14

16 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 H m m is a unitary matrix, then (UA) = A U. 15

17 Properties Lemma If E H[x] m m is a symmetric projection, that is, E = E 2 = E, then E H m m. Lemma A H m m is hermitian, that is, A = A, if and only if there exists a unitary matrix U H m m such that U AU = diag(d 1,...,d m ), where d i are the eigenvalues of A. 16

18 Theorem Let A H[x] m n. Then A has the Moore-Penrose inverse A if and only if ( ) A1 A A = U with U H m m unitary and A 1 A 1 + A 2A 2 a unit in H[x]r r with r min{m, n}. Moreover, ( ) A A = 1 (A 1A 1 + A 2A 2 ) 1 0 A 2 (A 1A 1 + A 2A U. 2 )

19 Leverrier-Faddeev algorithm Given A H[x] m m. An element λ H is called an eigenvalue of A if there exists a vector X H m 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 f B (λ) = det(λi 2m χ B ) is called the characteristic polynomial of A. Theorem Let A H[x] m n and B = AA. Then f B (λ) = g(λ) 2 where g(λ) (R[x])[λ]. 18

20 Characteristic polynomials Let A H[x] m n, B = AA and f B (λ) = g(λ) 2. Then g(b) = 0. We will call g(λ) the generalized characteristic polynomial of A. Lemma Let A H[x] m n have the Moore-Penrose inverse A. Set B = AA. Then (i) B = (A ) A and B B = AA. (ii) B B = BB and (B B) 2 = B B. (iii) ( B ) k = ( B k ) and (B n k ) B n k = B B, for any k N. 19

21 Formula Theorem Let A H[x] m n have the Moore-Penrose inverse A and B = AA. Suppose the generalized characteristic polynomial of A is g(λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m, where a i R[x]. If k is the largest integer such that a k 0, then the generalized inverse of A is given by A = 1 a k A [ B k 1 + a 1 B k a k 1 I ]. If a i = 0 for all 1 i m, then A = 0. 20

22 Fadeev-Leverrier s method Lemma Let A H[x] m n have the Moore-Penrose inverse A and set B = AA. Then for 1 k m, tr [( B k + a 1 B k a k 1 B )] = ka k, where the a i arise from the generalized characteristic polynomial of A: g(λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m. 21

23 Laverrier-Faddeev algorithm Let A H[x] m n have the generalized inverse A and B = AA. Suppose the generalized characteristic polynomial of A is g(λ) = λ m + a 1 λ m a k λ m k + + a m 1 λ + a m, where a i R[x]. Define a 0 = 1. If p is the largest integer such that a p 0 and we construct the sequence A 0,, A p as follows: A 0 = 0 1 = q 0 B 0 = I A 1 = AA tra B = q 1 B 1 = A 1 q 1 I... A p 1 = AA tra B p 1 p 2 p 1 = q p 1 B p 1 = A p 1 q p 1 I A p = AA tra B p p 1 p = q p B p = A p q p I then q i (x) = a i (x), i = 0,, p. 22

24 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. B 0 I m, a for i = 1,...,m do A i AA B i 1, a i tra i i, B i A i + a i I m 3. Find the maximal index p such that a p Return A = { 1 a p A B p 1, p > 0, 0, p = 0. 23

25 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) 0. Then the evaluation of f (x) at x = r is defined 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 the roots of f lie in at most n conjugacy classes of H. 24

26 Interpolation Theorem Let c 1,, c n be n pairwise non-conjugate elements of H. Then there is a unique monic polynomial n H[x] of degree n such that g n (c 1 ) = = g n (c n ) = 0. Moreover, c 1,...,c n are the only roots (up to conjugacy classes) of g n in H. Theorem Let c 1,, c n+1 H be pairwise non-conjugate and let d 1,,d n+1 H. Then there exists a unique lowest degree polynomial f H[x], of degree p n, such that f (c i ) = d i for all 1 i n

27 Degree Bonds For a given A H[x] m n, the degree dega = max{deg(a i 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 c 1,, c k+1 H be pairwise non-conjugate and let A 1,, A k+1 H n m. Then there is a unique lowest degree matrix A H[x] n m of degree p k, such that A(c i ) = A i for all 1 i k

28 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 a p 0. We construct the sequence A 0,, A p as follows: A 0 = 0 1 = q 0 B 0 = I... A p 1 = AA tra B p 1 p 2 p 1 = q p 1 B p 1 = A p 1 q p 1 I A p = AA tra B p p 1 p = q p B p = A p q p I. 27

29 Algorithm Theorem In the above setting, let k = (2m 1)degA and c 1,, c k+1 R be k + 1 distinct real numbers such that q p (c s ) 0 for any 1 s k + 1. Let S = {1,, k + 1} \ {s }. Then where A(c s ) = and A = k+1 A(c s ) g S s =1 1 [ ] q p (c s ) A(c s ) B(c s ) p 1 q 1 (c s )B(c s ) p 2 q p 1 (c s )I g S (c α ) = { 0 α S, 1 α = s. 28

30 Example A = ( 14x i + 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 i + 2j + 8k 3x 3 + 3i 15j 12k i + 15j 15k 6j + 12k 3x i + 3j + 12k 4x 4 + 4i 20j 16k i + 20j 20k 8j + 16k 4x i + 4j + 16k ) H 4 3 [x] The upper bound of the degree of A is less than (2m 1)degA = (2 4 1) 1 = 7. Choose c 1 = 0 and c 2 = 1. ) A(c 1 ) = ( i + 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 i + 15j 15k 6j + 12k i + 3j + 12k 4 + 4i 20j 16k i + 20j 20k 8j + 16k i + 4j + 16k and A(c 2 ) = ( ) i + 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 i + 15j 15k 6j + 12k i + 3j + 12k i 20j 16k i + 20j 20k 8j + 16k i + 4j + 16k 29

31 we calculate and obtain: A(c 1 ) = A(0) = i 228j 342k i 96j + 81k i i + 426j 382k i 93j 149k i i 176j + 292k i + 68j + 194k i + 12j i + 228j + 342k i + 96j 81k i 12 and A(c 2 ) = A(1) = i 244j 330k i 8j + 15k i i + 406j 402k i + 17j 39k i i 160j + 300k i 20j + 150k i + 60j i + 244j + 330k i + 8j 15k i 78 30

32 A 2 = A(c s ) g S = A(0) (1 x) + A(1) x s =1 = ( i 16j + 12k)x i 228j 342k ( 66 55i + 8 ( i 20j 20k)x i + 426j 382k (44 88i (16j + 8k)x i 176j + 292k ( 88j ( 12 10i + 16j 12k)x i + 228j 342k ( i 88 ( i 48j + 36k)x i + 126j + 54k ( i 64j + ( i 60j 60k)x i 72j + 204k ( i 80 (48j + 24k)x i + 12j 204k (64j + 32 ( 36 30i + 48j 36k)x i 126j 54k ( 48 40i + 64j 31

33 Example Let A = ( ) 1 i + 2k 3. i 6 + j Then its Moore-Penrose inverse can be found by using our Maple package as follows: i j i k A = i j k i j k i k i k

Serge Ballif January 18, 2008

Serge Ballif January 18, 2008 ballif@math.psu.edu The Pennsylvania State University January 18, 2008 Outline Rings Division Rings Noncommutative Rings s Roots of Rings Definition A ring R is a set toger with two binary operations +