6 The Smith Canonical Form

Transcription

1 6 The Smith Canonical Form 61 Equivalence of Polynomial Matrices DEFINITION 61 A matrix P M n n (F [x]) is called a unit in M n n (F [x]) if Q M n n (F [x]) such that P Q = I n Clearly if P and Q are units, so is P Q THEOREM 61 A matrix P M n n (F [x]) is a unit in M n n (F [x]) if and only if det P = c, where c F and c proof only if Suppose P is a unit Then P Q = I n and det P Q = det P det Q = det I n = 1 However det P and det Q belong to F [x], so both are in fact non zero elements of F if Suppose P M n n (F [x]) satisfies det P = c, where c F and c Then P adj P = (det P )I n = ci n Hence P Q = I n, where Q = c 1 adj P M n n (F [x]) Hence P is a unit in M n n (F [x]) EXAMPLE [ x x P = x 1 x ] M 2 2 (F [x]) is a unit, as det P = 1 THEOREM 62 Elementary row matrices in M n n (F [x]) are units: (i) E ij : interchange rows i and j of I n ; (ii) E i (t): multiply row i of I n by t F, t ; (iii) E ij (f): add f times row j of I n to row i, f F [x] In fact det E ij = 1; det E i (t) = t; det E ij (f) = 1 Similarly for elementary column matrices in M n n (F [x]): F ij, F i (t), F ij (f) 12

2 Remark: It follows that a product of elementary matrices in M n n (F [x]) is a unit Later we will be able to prove that the converse is also true DEFINITION 62 Let A, B M m n (F [x]) Then A is equivalent to B over F [x] if units P M m m (F [x]) and Q M n n (F [x]) exist such that P AQ = B THEOREM 63 Equivalence of matrices over F [x] defines an equivalence relation on M m n (F [x]) 611 Determinantal Divisors DEFINITIONS 61 Let A M m n (F [x]) Then for 1 k min (m, n), let d k (A) denote the gcd of all k k minors of A d k (A) is sometimes called the k th determinantal divisor of A Note: gcd (f 1,, f n ) at least one of f 1,, f n is non zero ρ(a), the determinantal rank of A, is defined to be the largest integer r for which there exists a non zero r r minor of A THEOREM 64 For 1 k ρ(a), we have d k (A) Also d k (A) divides d k+1 (A) for 1 k ρ(a) 1 proof Let r = ρ(a) Then there exists an r r non zero minor and hence d r (A) Then because each r r minor is a linear combination over F [x] of (r 1) (r 1) minors of A, it follows that some (r 1) (r 1) minor of A is also non zero and hence d r 1 (A) ; also d r 1 (A) divides each minor of size r 1 and consequently divides each minor of size r; hence d r 1 (A) divides d r (A), the gcd of all minors of size r This argument can be repeated with r replaced by r 1 and so on THEOREM 65 Let A, B M m n (F [x]) Then if A is equivalent to B over F [x], we have (i) ρ(a) = ρ(b) = r; 121

3 (ii) d k (A) = d k (B) for 1 k r proof Suppose P AQ = B, where P and Q are units First consider P A The rows of P A are linear combinations over F [x] of the rows of A, so it follows that each k k minor of P A is a linear combination of the k k minors of A Similarly each column of (P A)Q is a linear combinations over F [x] of the columns of P A, so it follows that each k k minor of B = (P A)Q is a linear combination over F [x] of the k k minors of P A and consequently of the k k minors of A It follows that all minors of B with size k > ρ(a) must be zero and hence ρ(b) ρ(a) However B is equivalent to A, so we deduce that ρ(a) ρ(b) and hence ρ(a) = ρ(b) Also d k (B) is a linear combination over F [x] of all k k minors of B and hence of all k k minors of A Hence d k (A) d k (B) and by symmetry, d k (B) d k (A) Hence d k (A) = d k (B) if 1 k r 62 Smith Canonical Form THEOREM 66 (Smith canonical form) Every non zero matrix A M m n (F [x]) with r = ρ(a) is equivalent to a matrix of the form D = f 1 f 2 f r = P AQ where f 1,, f r F [x] are monic, f k f k+1 for 1 k r 1, P is a product of elementary row matrices, and Q is a product of elementary column matrices DEFINITION 63 The matrix D is said to be in Smith canonical form proof This is presented in the form of an algorithm which is in fact used by Cmat to find unit matrices P and Q such that P AQ is in Smith canonical form 122

4 Our account is based on that in the book Rings, Modules and Linear Algebra, by B Hartley and TO Hawkes We describe a sequence of elementary row and column operations over F [x], which when applied to a matrix A with a 11 either yields a matrix C of the form f 1 C = C where f 1 is monic and divides every element of C, or else yields a matrix B in which b 11 and deg b 11 < deg a 11 (28) Assuming this, we start with our non zero matrix A By performing suitable row and column interchanges, we can assume that a 11 Now repeatedly perform the algorithm mentioned above Eventually we must reach a matrix of type C, otherwise we would produce an infinite strictly decreasing sequence of non negative integers by virtue of inequalities of type (28) On reaching a matrix of type C, we stop if C = Otherwise we perform the above argument on C and so on, leaving a trail of diagonal elements as we go Two points must be made: (i) Any elementary row or column operation on C corresponds to an elementary operation on C, which does not affect the first row or column of C (ii) Any elementary operation on C gives a new C whose new entries are linear combinations over F [x] of the old ones; consequently these new entries will still be divisible by f 1 Hence in due course we will reach a matrix D which is in Smith canonical form We now detail the sequence of elementary operations mentioned above Case 1 a 1j in row 1 with a 11 not dividing a 1j Then a 1j = a 11 q + b, by Euclid s division theorem, where b and deg b < deg a 11 Subtract q times column 1 from column j and then interchange columns 1 and j This yields a matrix of type B mentioned above 123

5 Case 2 a i1 in column 1 with a 11 not dividing a i1 Proceed as in Case 1, operating on rows rather than columns, again reaching a matrix of type B Case 3 Here a 11 divides every element in the first row and first column Then by subtracting suitable multiples of column 1 from the other columns, we can replace all the entries in the first row other than a 11 by Similarly for the first column We then have a matrix of the form E = e 11 If e 11 divides every element of E, we have reached a matrix of type C Otherwise e ij not divisible by e 11 We then add row i to row 1, thereby reaching Case 1 E EXAMPLE 62 (of the Smith Canonical Form) [ 1 + x 2 x A = x 1 + x We want D = P AQ in Smith canonical form So we construct the augmented matrix work on rows work on columns x 2 x 1 1 x 1 + x 1 R 1 R 1 xr 2 1 x 1 x x 1 + x 1 C 2 C 2 + x 2 C 1 1 x 1 1 x 2 1 x 1 + x + x 3 1 R 2 R 2 xr 1 1 x 1 1 x 2 x 1 + x x + x 3 1 P D Q Invariants are f 1 = 1, f 2 = 1 + x + x 3 Note also ] f 1 = d 1 (A), f 2 = d 2(A) d 1 (A) 124

6 621 Uniqueness of the Smith Canonical Form THEOREM 67 Every matrix A M m n (F [x]) is equivalent to precisely one matrix is Smith canonical form proof Suppose A is equivalent to a matrix B in Smith canonical form That is, f 1 B = f r and f 1 f 2 f r Then r = ρ(a), the determinantal rank of A But if 1 k r, d k (A) = d k (B) = f 1 f 2 f k and so the f i are uniquely determined by f 1 = d 1 (A) f 2 = d 2(A) d 1 (A) f r = d r(a) d r 1 (A) 63 Invariant factors of a polynomial matrix DEFINITION 64 The polynomials f 1,, f r in the Smith canonical form of A are called the invariant factors of A 3 Note: Cmat calls the invariant factors of xi B, where B M n n (F ), the similarity invariants of B We next find these similarity invariants They are 1, 1,, 1, d }{{} 1,, d s n s where d 1,, d s are what earlier called the invariant factors of T B 3 NB This is a slightly different, though similar, form of invariant factor to that we met a short while ago 125

7 LEMMA 61 The Smith canonical form of xi n C(d) where d is a monic polynomial of degree n is diag (1,, 1, d) }{{} n 1 proof Let d = x n + a n 1 x n a F [x], so xi n C(d) = x a 1 x a 1 1 a 2 x a n 2 1 x + a n 1 Now use the row operation R 1 R 1 + xr 2 + x 2 R x n 1 R n to obtain d 1 x a 1 1 a 2 x a n 2 1 x + a n 1 (think about it!) and then column operations C 2 C 2 + xc 1,, C n 1 C n 1 + xc n 2 and then C n C n + a 1 C 1 + a 2 C a n 2 C n 2 + (x + a n 1 )C n 1 yielding d

8 Trivially, elementary operations now form the matrix diag (1,, 1, d) }{{} n 1 THEOREM 68 Let B M n n (F ) Then if the invariant factors of B are d 1,, d s, then the invariant factors of xi n B are Then 1,, 1, d }{{} 1, d 2,, d s n s proof There exists non-singular P M n n (F ) such that P 1 BP = P 1 (xi n B)P = xi n = s C(d k ) k=1 s C(d k ) k=1 s (xi mk C(d k )) where m k = deg d k k=1 But by the lemma, each xi mk C(d k ) is equivalent over F [x] to diag (1,, 1, d k ) and hence xi n B is equivalent to 1 s diag (1,, 1, d k ) 1 k=1 d 1 EXAMPLE 63 Find the invariant factors of B = M 4 4(Q) d s

9 by finding the Smith canonical form of xi 4 B Solution: xi 4 B = We start off with the row operations x 2 1 x 1 1 x x 2 R 1 R 1 (x 2)R 2 R 1 R 2 R 4 R 4 + R 1 and get (column ops) 1 x 1 (x 1)(x 2) 1 x 1 x 2 1 x 2 1 (x 1)(x 2) 1 x 1 x 2 1 x x 1 (x 1)(x 2) x 2 1 x x 1 x(x 1)(x 2) (x 1)(x 2) 1 x(x 2) {= (x 1) 2 } 1 1 x(x 1)(x 2) (x 1)(x 2) (x 1) 2 Now, for brevity, we work just on the 2 2 block in the bottom right corner: [ ] (x 1)(x 2) x(x 1)(x 2) (x 1) 2 128

10 C 2 C 2 xc 1 R 1 R 1 + R 2 C 2 C 2 C 1 C 1 C 2 C 2 C 2 (x 2)C 1 R 2 R 2 + (x 1)R 1 [ ] (x 1)(x 2) (x 1) 2 [ ] (x 1)(x 2) (x 1) 2 (x 1) 2 [ ] (x 1)(x 2) x 1 (x 1) 2 [ ] x 1 (x 1)(x 2) (x 1) 2 [ ] x 1 (x 1) 2 (x 2)(x 1) 2 [ ] x 1 (x 2)(x 1) 2 and here we stop, as we have a matrix in Smith canonical form Thus xi 4 B 1 1 x 1 (x 1) 2 (x 2) so the invariant factors of B are the non-trivial ones of xi 4 B, ie (x 1) and (x 1) 2 (x 2) Also, the elementary divisors of B are so the Jordan canonical form of B is (x 1), (x 1) 2 and (x 2) J 2 (1) J 1 (1) J 1 (2) THEOREM 69 Let A, B M n n (F ) Then A is similar to B xi n A is equivalent to xi n B xi n A and xi n B have the same Smith canonical form proof 129

11 Obvious If P 1 AP = B, P M n n (F ) then P 1 (xi n A)P = xi n P 1 AP = xi n B If xi n A and xi n B are equivalent over F [x], then they have the same invariant factors and so have the same non-trivial invariant factors That is, A and B have the same invariant factors and hence are similar Note: It is possible to start from xi n A and find P M n n (F ) such that P 1 AP = s C(d k ) k=1 where P 1 (xi n B)Q 1 = diag (1,, 1, d 1,, d s ) (See Perlis, Theory of matrices, p 144, Corollary 8 1 and p 137, Theorem 7 9) THEOREM 61 Every unit in M n n (F [x]) is a product of elementary row and column matrices Proof: Problem sheet 7, Question 12 13