PIT problems in the light of and the noncommutative rank algorithm

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1 PIT problems in the light of and the noncommutative rank algorithm Gábor Ivanyos MTA SZTAKI Optimization, Complexity and Invariant Theory, IAS, June 4-8, 2018.

2 PIT problems in this talk Determinant: det(x 0 A 0 + x 1 A x k A k ) 0 exists a non-singular matrix in A = A 0,..., A 1 What is the rk A (commutative) rank of (= max rank in) A Constructive version (rank optimization): Find a matrix of max rank in A We assume square case most problems reducible to that

3 Overview Common block triangular forms of matrices Behavior of Wong sequences Module problems: from easy to hard If time left: Spaces spanned by unknown rank one matrices

4 Some notation M n (F ) = M n n (F ) Block matrices, holes in matrices: ( A B = C) A B 0 C Block (upper) triangular matrices: Matrix sets: ( {( } A A B = : B, Carbitrary ) C) ( A B C), A and C square

5 Notation (2) Product of sets: AU = {Au : A A, u U} (subspace when either A or U is a subspace) AB = {AB : A A, B B} (similarity): in the same orbit of conjugation by GL, changing the basis ( GL GL ): in the same orbit of (independent) left-right multiplication by GL changing the two bases independently

6 Oil and Vinegar signature schemes (Patarin (1997),...) Public key: P = (P 1,..., P k ) F [x] k x = (x 1,..., x n), deg P = 2 Message: a F k Valid signature: a solution of P(x) = a Private key (hidden structure): easy system: P s.t. P (y) = a P = P A, A GL n (F ) a linear change of variables easiness : P is linear in the first o variables: no terms x i x j with i, j {1,..., o} by a random substitution for x j (j = o + 1,..., n) we have a solvable linear system (with good chance) x 1,..., x o : oil variables ; x o+1,..., n vinegar variables

7 Oil and Vinegar (2) Key generation: choose such P randomly, and A randomly Tuning: choose the parameters k, o, n: P easy to solve hard to break Balanced O & V (Patarin 1997): n = 2o (and k o) Breaking Balanced O & V (Kipnis & Shamir 1998): P i = Q i + linear P i = Q i + linear, Q i = A T Q i A pick Q 0 = α i Q i random invertible with h.p. (for most P) Q 0 = A T Q 0 A Q 0 = α i Q i ) R i := Q 1 0 Q i R i = Q 1 0 Q i

8 Breaking Balanced O & V Q 0 = A T Q 0 A (Q 0 = α i Q i ) R i := Q0 1 Q i (R i = Q 1 0 Q i ) key property: R i = A 1 R i A easier than Q i = A T Q i A Proof. R i = A 1 Q 0 A T A T Q i A = A 1 R i ( ) ( ) ( ) A ( ) ( ) I Q i =, Q I i 1 I, I ( ) ( ) 2 ( ) ( ) R i = Q 0 1 I Q i = I

9 Breaking Balanced O & V (2) R i A 1 ( ) A unique common block triangular form of R i for most P up to lin. changes of the O and V variables separately do not disturb easiness find common block triangularization of R i O & V decomposition of Q i e.g. use the MeatAxe Kipnis & Shamir: simpler direct method (exploits specialties of the setting) Unbalanced O & V (Kipnis & Patarin 1999) better hardness : Bulygin, Petzoldt & Buchmann (2010)

10 Block triangular forms ( ) GAH, n t t zero lower left block reduces many problems to the diag. blocks e.g, finding full rk. A A; Find B A with invertible upper left block, B A with invertible lower block, λb + C will be invertible except for a few λs instability

11 Block triangular forms (2) The full (commutative) rank case: A 0 A invertible use A 0 as a bijection between the domain and range a prefect matchings: bipartite graphs digraphs New matrix space: A 1 0 A = {A 1 0 A : A A} I n, A0 1 A = A 1 0 AI GL GL A, inherits block triang. (GA 0 H) 1 GAH = H 1 A 1 0 G 1 GAH = H 1 A 1 0 AH natural action on A 1 0 A: conjugation X H 1 XH = two-sided action of GL GL preserving I

12 Block triangular forms (3) I n A, H 1 AH ( ), n t t zero block First t basis vectors span an H 1 AH-invariant subspace U U = H 1 U t-dim A-invariant subspace Remark: if I n A and dim AU dim U then AU = U. nontrivial strong components in digraphs Env(A) enveloping (matrix) algebra closure of A w.r.t. lin. comb. and multiplications transitive closure of digraphs A-invariant subspace: submodule for Env(A) (or for the free algebra)

13 Finding common invariant subspaces Quite well studied/understood Many of the methods: based on structure of Env(A) one-sided ideals, zero divisors for A = I, A 0 : factors of the minimum polynomials of A 0 general A: zero div. factoring min. pol. of good A Env(A) over algebraically closed fields: almost easy Depends on the computational model Representation size explosion? E.g. huge (composite) extensions 2 2 M 2n (Q) A... p n p n

14 Finding invariant subspaces - rationality issues over non-closed base fields (extensions not allowed) over finite fields: only randomized methods (in large char), factoring polynomials MeatAxe for group representations over Q only a partial decompositions Hardness of distinguishing full matrix algebras from division algebras over Q (Rónyai 1987): in some generalizations of the quaternions existence of zero divisors quadratic residuousity mod composite numbers finding zero divisors: factoring integers a motivation in conjecturing the regularity

15 Using blowups for block triangularization A = A M d (F ) (on F n F d ). Property A = (I n M d (F ))A = A (I n M d (F ) (this characterizes blowups) A U V = A (I n M d (F ))U (I n M d (F ))V I n M d (F )-invariant subspaces of F n F d : (I M d (F ))U = U U = U F d U = {u F n : u v U for some 0 v F d } Computing U: v 1,..., v d : basis for F d, u 1,..., u k: basis for U u i = u ij v j u ij F n U is spanned by u ij (i = 1,..., k j = 1,..., d)

16 Using blowups (2) Lower left zero blocks in blowups: A = A M d (F ) U, V (I M d (F ))-invariant subsp. U = U M d (F ), V = V M d (F ) A U V AU AV L.L.Z.B. in A L.L.Z.B. in A block triang forms of A block triang forms of A Application of constructive ncrank:find singular block triang of some blowup A block triang of A or an invertible element in some blowup A block triang A a block triang of A More serious applications in the next talk (?)

17 The Wong sequence Given A 0, A 1,..., A k M n (F ) A = A 0,..., A k rk A 0 = r < n, c = n r (co-rank of A 0 ) Idealistic goal: find case (1) A A s.t. rk A > r or case (2) U F n s.t. dim AU dim U c U 0 = (0), U j = A 1 0 U j, U j+1 = AU j (A 1 0 W : full inverse image of W at A 0) U 0 U 1... U l, U 0 U 1... U l..., U j,... stops inside im A 0 case (2) otherwise escapes from im A 0 : AU j im A 0 for some j length of the (escaping) Wong sequence l = min{j : AU j im A 0 }

18 Length 1 Wong sequence l = 1; basic case n = r + 1, A 0 = I r, r > 0 A ker A 0 im A 0 i: A i ker A 0 im A ( 0 B A i + λa 0 ) + λi b ( B ) + λi (b 0) b has rank > r if λ and is not an eigenvalue of B (F large enough) Blind algorithm compute rk (A i + λa 0 ) (i = 1,..., k, λ = λ 1,..., λ r+1 )

19 Length 1 - some examples Examples (long Wong sequences): A 0 = , k = 1, A 1 = i: rk (A 0 + A 1 ) > rk A 0 k = n > 1, A i = E ii : rk (A 0 + A i ) = rk A 0 Length one a nice property: independent of the basis for A preserved by GLn GL n preserved by base filed extension F = R; A 0, A i pos. szemidef. v ker A 0 \ ker A i = (im A 0 ) \ ker A i 0 v T A i v, but v T A 0 w = 0 for every w

20 Length 1 - examples (2) A i diagonal (i = 0,..., k) i, v: im A 0 ker A 0 = (0) ker A 0, A i -invariant (because A i A 0 = A 0 A i ) A i ker A 0 im A 0 ker A 0 ker A i Application: simplicity of finite extensions of Q: L: field extension of F = Q, L : F = n a L: F [a] = subring (=subfield) generated by F and a Task: find a s.t. L = F [a] Matrix representation of L a M a = matrix of x ax on L (n n) identify a with M a ; Facts: M a are simultaneously diagonalizable over C F [a] : F = #distinct eigenvalues of a

21 simplicity of extensions (2) a Ad Ma = matrix of X M a X XM a A := {Ad Ma : a L} n-dim subspace of M n 2(F ) = δ 1... δ n Ad diagonal in M n 2: Ad E ij = E ij E ij = (δ i δ j )E ij max. rank is n 2 n when δ i δ j for i j generalizes to direct sums of field extensions (over perfect base fields)

22 Short Wong sequences Key observation of Bläser, Jindal & Pandey (2017) A = A 0, A 1,..., A k A = A 0, A 1 (over F (x 1,..., x k )) A 1 = x 1A , x k A k A 0 not of max rank in A (F sufficiently large) A 0 not of max rank in A (rk = ncrk for pair A ) A 1 U l im A 0 U 1,..., U l Wong sequence for A 0 in A

23 Short Wong sequences (2) Assume basic case A 0 = ( Ir ), n = r + 1 A 1 U l = A 1l ker A 0 lower right entry of A 1 : nonzero degree l polynomial in x 1,..., x k has term a x i1... x il A 0 is not of max rank in A = A 0, x i1 A i x il A il A 0 is not of max rank in A = A 0, A i1,..., A il. Assume l length of Wong seq. for A. Then A 0 is of max rank in A 0, A 1,..., A k A 0 is of max rank in A 0, A i1,..., A il for every subset {i 1,..., i l } {1,..., k}

24 Short Wong sequences (3) Algorithm (Bläser, Jindal & Pandey (2017)) Input: A 0, A 1,..., A k and l k Output: A 0 A of rank > rk A 0 or: l IS TOO SMALL for every subset {i 1,..., i l } {1,..., k} try A 0 + l t=1 ω ta it for all (ω 1,..., ω l ) Ω l ( Ω = n) complexity (kn) l poly

25 Progress of Wong sequences Wong sequence U 0 = (0), U j = AA 1 0 A 0 U j im A 0 (j = 0,..., l 1) Lemma (BJP17 for case k = 2) Assume that rk A 0 = r < ncrk A. Then for every 1 j < l, dim U j dim U j 1 + ncrk A r. sufficient ( to ) prove for n = ncrk A ( basic case ) Ir A 0 = ; s = ncrk A, take an s s window of full ncrk containing the upper left r by r

26 Progress of Wong (2) ( ) Ir A 0 = F n = im A 0 ker A 0, block structure using (0) = U 0 < U 1 <... < U l 1 im A 0 B 1 B 12 B 17 B 18 B 19 B 21 B 2 B 27 B 28. A A = B 76 B 7 B 78 B 87 B 8 B 97 B 98 B jj square (I for A 0 ); B l+2,l 0 cyclically shift by n r (7 = l)

27 Progress of Wong (3) diagonal shifted by ncrk r to the right B 19 B 1 B 12 B 17 B 18 B 21 B 2 B 27 B 28. A B 76 B 7 B 78 B 87 B 8 B 97 B 98 B jj has ncrk A r columns by def of ncrk

28 Progress of Wong (3) A formal proof the lemma for 1 j < l: U j 1 = A 1 0 U j 1 = U j 1 ker A 0 for A 0 = ( Ir ) ; dim U j = dim AU j 1 dim U j 1 (full ncrank) = dim U j 1 + dim ker A 0 = dim U j 1 + n r

29 Approximating the commutative rank Bläser, Jindal & Pandey (2017) r = max. rk in A = A 1,..., A k goal: find A A: rk A (1 ɛ)r) l = min(k, 1/ɛ ) Iteration if rk A 0 (1 ɛr) then: length of Wong seq. for A 0 and x 1 A x k A k rk A 0 /(r rk A 0 ) (1 ɛ)/ɛ = 1/ɛ 1 try A 0 = A 0 + ω 1 A i ω l A il replace A 0 with A 0 if better terminate if no improvement Cost: (kn) 1/ɛ poly

30 Thin Wong sequences dim U j+1 = dim U j + 1 (j = 0,..., l 1) basic case n = rk A b 11 b 12 b 17 B 18 b 19 b 19 b 11 b 12 b 17 B 18 b 21 b 22 b 27 B 28 b 21 b 22 b 27 B b 76 b 77 B 78 b 76 b 77 B 78 B 87 B 88 b 97 B 98 b 97 B 98 B 87 B 88 (7 = l + 1) find A A with no zero diag entry in the big upper left part find A + λa 0 with invertible lower right diag block (B 88 )

31 Thin Wong sequences (2) hardness of rank of diagonal A i over F of constant size q 3: reduction from coloring with q colors: vertices: v 1,..., v k, edges e 1,..., e n +1 if e t = {v i, v j }, j > i (A i ) tt = 1 if e t = {v i, v j }, j < i (i = 1,..., k) 0 otherwise. special instances: pencils: A = A 0, A 1 A 1,..., A k of rank one: find smallest l, A il... A i1 ker A 0 im A 0 A A 0, A i1,..., A il poly method for A spanned by A 0 and unknown rank one matrices

32 Wong sequences - remarks & problems Triangularizable spaces of full rank:... A GL GL.... Would be length 1 for rk A n rank of the diagonal part triangularization by conjugation Dual Wong sequence could recover part of triang structure Shortening length of Wong with A 0 + λa i? Example: A 0 triangular, A 1,..., A k diagonal Nicer classes? Nice classes for length 2?

33 Wong sequences - remarks & problems (2) Length and blowup size length 2 current blowup size (sufficient to increase rk A 0 ) thinness at a single step block triang current blowup size rk A 0 /4 relation with final blowup size? commutative rank for bounded blowup size? Rank of generators rank one: blowup size 1 rank 2: current blowup size 2 (looks so) final blowup size???? in special cases, e.g., (skew) symmetric? rank c: bound on current blowup size?

34 Modules modules for the free algebra B = F X 1,..., X t n-dimensional (left) B-module: V = F n, : B V V bilinear commutes with of B: (a b) v = a (b v) Notation: av = a v input data: linear maps L 1,..., L t : V V (n n matrices) action of X 1,..., X t multiplication tables in groups could take smaller (finite dim.) B Isomorphisms V, V, given by L 1,..., L t M n (F ), L 1,..., L t M n (F ) φ : V V bijective linear X i φ(v) = φ(x i v) L i φ = φ L i

35 Module morphisms (2) Homomorphisms V, V, given by L 1,..., L t M n (F ), L 1,..., L t M n (F ) Hom(V, V ) := {φ : V V lin. : φ L i = L i φ} subspace of Lin n n(f ) solutions of the lin. constraints φ L i = L i φ Isomorphism: n = n, invertible transf. from Hom(V, V ) Isomorphism: a full rank matrix Hom(V, V ) (n = n) In P: Chistov, I & Karpinski (1997) over many fields; Brooksbank & Luks (08); I, Karpinski & Saxena (010) all fields Length 1 Wong sequence in a special case...

36 Module morphisms (3) submodule Common invariant subspaces for L i Block triangular form of L i simple (irreducible) modules: no proper submodules direct sum ( ) external : space V V Li ; action L i internal : V = V 1 V 2 (of subspaces), V 1, V 2 submodule block diagonal form for L i indecomposable modules: no such decomp. Krull-Schmidt: uniqueness of decomposition into indecomposables: isomorphism types and multiplicities factorization of numbers

37 Module isomorphism - the decision version An new ncrank-based method Key observation: Hom(V, V ) M d (F ) = Hom(V d, V d ) Consequence: (assume dim V = dim V = n): Proof. V = V ncrk (Hom(V, V ) = n V = V rk Hom(V, V ) = n ncrk Hom(V, V ) = n V d = V d for some d V = V by (Krull-Schmidt)

38 Hardness of injectivity Are spaces Hom(V, V ) special? NO: every matrix space is essentially Hom(V, V ) Construction: I, Karpinski & Saxena (2010) A 1,..., A k arbitrary n n V, V modules for F x 1,..., x n, x n+1 dim V = n + 1, dim V = n + k ( L j = ( L j = ) e j n + 1 n + 1 e j = 1 j (j n) ) Â j, n + k n + k ( ) Â j = A (j) 1 A (j) : jth columns of A k 1,..., A k (j n) Another slicing of the 3-tensor A: (Âj ) il = (A l ) ij ( ( ) In n L n+1 = L 0), n+1 = In n 0k k

39 Hardness of injectivity (2) ( B Hom(V, V ): n + k n + 1-matrices C = α 1 B M n(f ), α =. F n s.t.: α k CL j = L jc (j = 1,..., n) α ) Proof. B = A α, where A α = α 1 A α k A k CL j = Hom(V, V ) C α = ( B (j) ) L j C = ( α1 A (j) α k A (j) k )) = ( Aα ( ) (j) A α ) injective A α α nonsingular

40 Module isomorphism - the semisimple case V = V semisimple (the indecomposable components are simple) important property: every submodule is a direct summand Assume φ : Hom(V, V ) not invertible. Let V 0 ker φ simple, let V = V 0 W 0. Let V = V 1 V t, V i irreducible By Krull-Schmidt, i s.t. V i im φ and V i = V 0. ψ 0 : V 0 V i isomorphism extend to V V i V : ψ(v + w) := ψ 0 (v) (v V 0, w W ) ψ ker φ = V 0 im φ length 1 Wong

41 Semisimple module algorithm - remarks Actually, finds max rank morphisms between semisimple module An application to decrease dimension of representation of simple algebras (Babai & Rónyai 1990, revised presentation): B = M n (F ), (unknown isomorphism) Every (unital) module is a direct sum of copies of F n V B-module of dimension nr, r = g.c.d(n, r), sr = tn + r, U = B by left mult. find injective φ Hom(U t, V s ), W = im φ V s /W is a B-module of dim. nr. Example: n prime, z any zero divisor in B V = Bz left ideal as module of dimension nr r = 1, construct n-dimensional module isomorphism with M n(f ).

42 Module isomorphism - the general case Reduction to finding minimum size sets of module generators ( surjective morphisms from free modules) H = Hom(V, V ) closed under multiplication: a matrix algebra Hom(V, V ) left H-module if V = V : isomorphism H-mod. generator of Hom(V, V ) the length 1 property of the semisimple case can be exploited to finding min. size sets of generators (I, Karpinski & Saxena (2010)) Surjectivity from free modules Remark: free can be weakened to projective

43 Hidden rank one generators I, Karpinski, Qiao, Santha & Saxena (2014) A = A 0, A 1,..., A k rk A i = 1, but A i unknown (i = 1,..., k) Assume A 0 = ( Ir ), l: smallest s.t. A l ker A 0 im A 0 i 1,..., i l : A il... A i1 ker A 0 im A 0 A l s A ij A s 1 ker A 0 im A 0 when s j For s < j: A ij A s 1 ker A 0 = (0), (otherwise A l j s ker A 0 A il... A ij A s 1 ker A 0 = im A il im A 0 ) For j > s: A l s im A j im A 0, similarly A j : space of solutions for A l s X A s 1 ker A 0 im A 0 (s {1,..., l} \ {j}) system of lin eq.

44 Hidden rank one generators (2) Compute bases for A 1,..., A l A l A 1 ker A 0 A l A 1 ker A 0 im A 0 key property: A il... A i1 im A 0 if i j j for some j Find B i A i : B l B 1 ker A 0 im A 0 Pick a basis element B 1 for A 1 s.t. A l A 2 B 1 ker A 0 im A 0 ; Then B 2 from basis for A 2 s.t. A l A 3 B 2 B 1 ker A 0 im A 0 ; etc.... (B B l ) l ker A 0 = B l... B 1 ker A 0 modulo im A 0 Find λ: λa 0 + B B l has rank rk A 0. rationality issues: rank one generators do not need to be rational example: field extension known rank one generators: over F, even if small

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