2 Polynomials over a field


 Lester Benson
 1 years ago
 Views:
Transcription
1 2 Polynomials over a field A polynomial over a field F is a sequence (a 0, a 1, a 2,, a n, ) where a i F i with a i = 0 from some point on a i is called the i th coefficient of f We define three special polynomials 0 = (0, 0, 0, ) 1 = (1, 0, 0, ) x = (0, 1, 0, ) The polynomial (a 0, ) is called a constant and is written simply as a 0 Let F [x] denote the set of all polynomials in x If f 0, then the degree of f, written deg f, is the greatest n such that a n 0 Note that the polynomial 0 has no degree a n is called the leading coefficient of f F [x] forms a vector space over F if we define λ(a 0, a 1, ) = (λa 0, λa 1, ), λ F DEFINITION 21 (Multiplication of polynomials) Let f = (a 0, a 1, ) and g = (b 0, b 1, ) Then fg = (c 0, c 1, ) where EXAMPLE 21 c n = a 0 b n + a 1 b n a n b 0 n = a i b n i = i=0 0 i,0 j i+j=n a i b j x 2 = (0, 0, 1, 0, ), x 3 = (0, 0, 0, 1, 0, ) More generally, an induction shows that x n = (a 0, ), where a n = 1 and all other a i are zero If deg f = n, we have f = a a 1 x + + a n x n 20
2 THEOREM 21 (Associative Law) f(gh) = (fg)h PROOF Take f, g as above and h = (c 0, c 1, ) Then f(gh) = (d 0, d 1, ), where d n = (fg) i h j i+j=n = = i+j=n u+v+j=n Likewise (fg)h = (e 0, e 1, ), where e n = ( u+v+j=n u+v=i f u g v h j f u g v ) h j f u g v h j Some properties of polynomial arithmetic: fg = gf 0f = 0 1f = f f(g + h) = fg + fh f 0 and g 0 fg 0 The last statement is equivalent to The we deduce that and deg(fg) = deg f + deg g fg = 0 f = 0 or g = 0 fh = fg and f 0 h = g 21 Lagrange Interpolation Polynomials Let P n [F ] denote the set of polynomials a 0 + a 1 x + + a n x n, where a 0,, a n F Then a 0 + a 1 x + + a n x n = 0 implies that a 0 = 0,, a n = 0 P n [F ] is a subspace of F [x] and 1, x, x 2,, x n form the standard basis for P n [F ] 21
3 If f P n [F ] and c F, we write f(c) = a 0 + a 1 c + + a n c n This is the value of f at c This symbol has the following properties: (f + g)(c) = f(c) + g(c) (λf)(c) = λ(f(c)) (f g)(c) = f(c)g(c) DEFINITION 22 Let c 1,, c n+1 be distinct members of F Then the Lagrange interpolation polynomials p 1,, p n+1 are polynomials of degree n defined by EXAMPLE 22 p i = n+1 j=1 j i ( x cj c i c j ), 1 i n + 1 p 1 = p 2 = ( ) x c1 c 2 c 1 etc ( ) x c2 c 1 c 2 ( ) x c3 ( c 1 c 3 ) x c3 c 2 c 3 ( ) x cn+1 ( c 1 c n+1 ) x cn+1 c 2 c n+1 We now show that the Lagrange polynomials also form a basis for P n [F ] PROOF Noting that there are n + 1 elements in the standard basis, above, we see that dim P n [F ] = n + 1 and so it suffices to show that p 1,, p n+1 are LI We use the following property of the polynomials p i : { 1 if i = j p i (c j ) = δ ij = 0 if i j Assume that a 1 p a n+1 p n+1 = 0 where a i F, 1 i n + 1 Evaluating both sides at c 1,, c n+1 gives a 1 p 1 (c 1 ) + + a n+1 p n+1 (c 1 ) = 0 a 1 p 1 (c n+1 ) + + a n+1 p n+1 (c n+1 ) = 0 22
4 Hence a i = 0 i as required COROLLARY 21 If f P n [F ] then Proof: We know that a a a n+1 0 = 0 a a a n+1 0 = 0 a a a n+1 1 = 0 f = f(c 1 )p f(c n+1 )p n+1 f = λ 1 p λ n+1 p n+1 for some λ i F Evaluating both sides at c 1,, c n+1 then, gives as required f(c 1 ) = λ 1, f(c n+1 ) = λ n+1 COROLLARY 22 If f P n [F ] and f(c 1 ) = 0,, f(c n+1 ) = 0 where c 1,, c n+1 are distinct, then f = 0 (Ie a nonzero polynomial of degree n can have at most n roots) COROLLARY 23 If b 1,, b n+1 are any scalars in F, and c 1,, c n+1 are again distinct, then there exists a unique polynomial f P n [F ] such that f(c 1 ) = b 1,, f(c n+1 ) = b n+1 ; namely f = b 1 p b n+1 p n+1 23
5 EXAMPLE 23 Find the quadratic polynomial f = a 0 + a 1 x + a 2 x 2 P 2 [R] such that f(1) = 8, f(2) = 5, f(3) = 4 Solution: f = 8p 1 + 5p 2 + 4p 3 where p 1 = p 2 = p 3 = (x 2)(x 3) (1 2)(1 3) (x 1)(x 3) (2 1)(2 3) (x 1)(x 2) (3 1)(3 2) 22 Division of polynomials DEFINITION 23 If f, g F [x], we say f divides g if h F [x] such that g = fh For this we write f g, and f g denotes the negation f does not divide g Some properties: f g and g 0 deg f deg g and thus of course f 1 deg f = Euclid s Division Theorem Let f, g F [x] and g 0 Then q, r F [x] such that f = qg + r, (3) where r = 0 or deg r < deg g Moreover q and r are unique Outline of Proof: 24
6 If f = 0 or deg f < deg g, (3) is trivially true (taking q = 0 and r = f) So assume deg f deg g, where f = a m x m + a m 1 x m 1 + a 0, g = b n x n + + b 0 and we have a long division process, viz: a m b 1 n x m n + b n x n + + b 0 a m x m + a m 1 x m a 0 a m x m etc (See S Perlis, Theory of Matrices, p111) 222 Euclid s Division Algorithm f = q 1 g + r 1 with deg r 1 < deg g g = q 2 r 1 + r 2 with deg r 2 < deg r 1 r 1 = q 3 r 2 + r 3 with deg r 3 < deg r 2 r n 2 = q n r n 1 + r n with deg r n < deg r n 1 r n 1 = q n+1 r n Then r n = gcd(f, g), the greatest common divisor of f and g ie r n is a polynomial d with the property that 1 d f and d g, and 2 e F [x], e f and e g e d (This defines gcd(f, g) uniquely up to a constant multiple) We select the monic (ie leading coefficient = 1) gcd as the gcd Also, u, v F [x] such that r n = gcd(f, g) = uf + vg find u and v by forward substitution in Euclid s algorithm; viz r 1 = f + ( q 1 )g r 2 = g + ( q 2 )r 1 25
7 = g + ( q 2 )(f + ( q 1 )g) = g + ( q 2 )f + (q 1 q 2 )g = ( q 2 )f + (1 + q 1 q 2 )g r n = ( ) f + ( ) g }{{} }{{} u v In general, r k = s k f + t k g for 1 k n, where r 1 = f, r 0 = g, s 1 = 1, s 0 = 0, t 1 = 0, t 0 = 1 and s k = q k s k 1 + s k 2, t k = q k t k 1 + t k 2 for 1 k n (Proof by induction) The special case gcd(f, g) = 1 (ie f and g are relatively prime) is of great importance: here u, v F [x] such that uf + vg = 1 EXERCISE 21 Find gcd(3x 2 + 2x + 4, 2x 4 + 5x + 1) in Q[x] and express it as uf + vg for two polynomials u and v 23 Irreducible Polynomials DEFINITION 24 Let f be a nonconstant polynomial Then, if g f or g is a constant g = constant f we call f an irreducible polynomial Note: (Remainder theorem) f = (x a)q + f(a) where a F So f(a) = 0 iff (x a) f EXAMPLE 24 f(x) = x 2 + x + 1 Z 2 [x] is irreducible, for f(0) = f(1) = 1 0, and hence there are no polynomials of degree 1 which divide f 26
8 THEOREM 22 Let f be irreducible Then if f g, gcd(f, g) = 1 and u, v F [x] such that uf + vg = 1 PROOF Suppose f is irreducible and f g Let d = gcd(f, g) so d f and d g Then either d = cf for some constant c, or d = 1 But if d = cf then So d = 1 as required f d and d g f g a contradiction COROLLARY 24 If f is irreducible and f gh, then f g or f h Proof: Suppose f is irreducible and f gh, f g We show that f h By the above theorem, u, v such that uf + vg = 1 ufh + vgh = h f h THEOREM 23 Any nonconstant polynomial is expressible as a product of irreducible polynomials where representation is unique up to the order of the irreducible factors Some examples: PROOF (x + 1) 2 = x 2 + 2x + 1 = x inz 2 [x] (x 2 + x + 1) 2 = x 4 + x in Z 2 [x] (2x 2 + x + 1)(2x + 1) = x 3 + x inz 3 [x] = (x 2 + 2x + 2)(x + 2) inz 3 [x] 27
9 Existence of factorization: If f F [x] is not a constant polynomial, then f being irreducible implies the result Otherwise, f = f 1 F 1, with 0 < deg f 1, deg F 1 < deg f If f 1 and F 1 are irreducible, stop Otherwise, keep going Eventually we end with a decomposition of f into irreducible polynomials Uniqueness: Let cf 1 f 2 f m = dg 1 g 2 g n be two decompositions into products of constants (c and d) and monic irreducibles (f i, g j ) Now f 1 f 1 f 2 f m = f 1 g 1 g 2 g n and since f i, g i are irreducible we can cancel f 1 and some g j Repeating this for f 2,, f m, we eventually obtain m = n and c = d in other words, each expression is simply a rearrangement of the factors of the other, as required THEOREM 24 Let F q be a field with q elements Then if n N, there exists an irreducible polynomial of degree n in F [x] PROOF First we introduce the idea of the Riemann zeta function: 1 ζ(s) = n s = 1 n=1 p prime 1 p 1 s To see the equality of the latter expressions note that and so RHS = = 1 1 x = x i = 1 + x + x 2 + i=0 ( ) 1 p p prime is i=0 ( s s + ) ( = s s s s + 1 ) 3 2s + 28
10 note for the last step that terms will be of form ( 1 p a 1 1 pa R R ) s up to some prime p R, with a i 0 i = 1,, R and as R, the prime factorizations p a 1 1 pa R R map onto the natural numbers, N We let N m denote the number of monic irreducibles of degree m in F q [x] For example, N 1 = q since x + a, a F q are the irreducible polynomials of degree 1 Now let f = q deg f, and 0 = 0 Then we have fg = f g since deg fg = deg f + deg g and, because of the uniqueness of factorization theorem, f monic Now the left hand side is 1 f s = f monic and irreducible f s = = = and RHS = n=0 n=0 f monic and deg f = n q n q ns 1 f s (there are q n monic polynomials of degree n) 1 q n(s 1) n= n=1 q s 1 1 ( 1 q 1 ) Nn ns 29
11 Equating the two, we have q s 1 = 1 ( 1 q 1 ) Nn (4) ns n=1 We now take logs of both sides, and then use the fact that ( ) 1 x n log = if x < 1; 1 x n so (4) becomes log so 1 1 q (s 1) = k=1 n=1 1 ( 1 q 1 ) Nn ns N n log n=1 1 kq (s 1)k = k=1 Putting x = q s, we have k=1 = q k kq sk = q k x k k = = k=1 n=1 N n n=1 m=1 N n n=1 m=1 k=1 mn=k ( 1 1 q ns 1 mq mns nn n kq ks mn=k n mnq mns x k nn n, and since both sides are power series, we may equate coefficients of x k to obtain q k = nn n (5) mn=k nn n = n k We can deduce from this that N n > 0 as n (see Berlekamp s Algebraic Coding Theory ) Now note that N 1 = q, so if k is a prime say k = p, (5) gives q p = N 1 + pn p = q + pn p N p = qp q > 0 as q > 1 and p 2 p 30 )
12 This proves the theorem for n = p, a prime But what if k is not prime? Equation (5) also tells us that Now let k 2 Then q k kn k q k = kn k + n k n k kn k + n k n k k/2 kn k + n=1 nn n q n (as nn n q n ) q n k/2 < kn k + q n (adding 1) n=0 = kn k + q k/ q 1 (sum of geometric series) so But q t+1 1 q 1 < q t+1 if q 2, q k < kn k + q k/2 +1 N k > qk q k/2 +1 k 0 if q k q k/2 +1 Since q > 1 (we cannot have a field with a single element, since the additive and multiplicative identities cannot be equal by one of the axioms), the latter condition is equivalent to k k/2 + 1 which is true and the theorem is proven 31
13 24 Minimum Polynomial of a (Square) Matrix Let A M n n (F ), and g = ch A Then g(a) = 0 by the Cayley Hamilton theorem DEFINITION 25 Any non zero polynomial g of minimum degree and satisfying g(a) = 0 is called a minimum polynomial of A Note: If f is a minimum polynomial of A, then f cannot be a constant polynomial For if f = c, a constant, then 0 = f(a) = ci n implies c = 0 THEOREM 25 If f is a minimum polynomial of A and g(a) = 0, then f g (In particular, f ch A ) PROOF Let g(a) = 0 and f be a minimum polynomial Then where r = 0 or deg r < deg f Hence g = qf + r, g(a) = q(a) 0 + r(a) 0 = r(a) So if r 0, the inequality deg r < deg f would give a contradict the definition of f Consequently r = 0 and f g Note: It follows that if f and g are minimum polynomials of A, then f g and g f and consequently f = cg, where c is a scalar Hence there is a unique monic minimum polynomial and we denote it by m A EXAMPLES (of minimum polynomials): 1 A = 0 m A = x 2 A = I n m A = x 1 3 A = ci n m A = x c 4 A 2 = A and A 0 and A I n m A = x 2 x EXAMPLE 25 F = Q and A =
14 Now A c 0 I 3, c 0 Q, so m A x c 0, A 2 = 3A 2I 3 m A = x 2 3x + 2 This is an special case of a general algorithm: (Minimum polynomial algorithm) Let A M n n (F ) Then we find the least positive integer r such that A r is expressible as a linear combination of the matrices I n, A,, A r 1, say A r = c 0 + c 1 A + + c r 1 A r 1 (Such an integer must exist as I n, A,, A n2 form a linearly dependent family in the vector space M n n (F ) and this latter space has dimension equal to n 2 ) Then m A = x r c r 1 x r 1 c 1 x c 0 THEOREM 26 If f = x n + a n 1 x n a 1 x + a 0 F [x], then m C(f) = f, where C(f) = a a a a n 1 PROOF For brevity denote C(f) by A Then postmultiplying A by the respective unit column vectors E 1,, E n gives AE 1 = E 2 AE 2 = E 3 A 2 E 1 = E 3 AE n 1 = E n A n 1 E 1 = E n AE n = a 0 E 1 a 2 E 2 a n 1 E n = a 0 E 1 a 2 AE 1 a n 1 A n 1 E 1 = A n E 1, 33
15 so f(a)e 1 = 0 first column of f(a) zero Now although matrix multiplication is not commutative, multiplication of two matrices, each of which is a polynomial in a given square matrix A, is commutative Hence f(a)g(a) = g(a)f(a) if f, g F [x] Taking g = x gives f(a)a = Af(A) Thus f(a)e 2 = f(a)ae 1 = Af(A)E 1 = 0 and so the second column of A is zero Repeating this for E 3,, E n, we see that f(a) = 0 and thus m A f To show m A = f, we assume deg m A = t < n; say Now m A = x t + b t 1 x t b 0 m A (A) = 0 A t + b t 1 A t b 0 I n = 0 (A t + b t 1 A t b 0 I n )E 1 = 0, and recalling that AE 1 = E 2 etc, and t < n, we have E t+1 + b t 1 E t + + b 1 E 2 + b 0 E 1 = 0 which is a contradiction since the E i are independent, the coefficient of E t+1 cannot be 1 Hence m A = f Note: It follows that ch A = f Because both ch A and m A have degree n and moreover m A divides ch A EXERCISE 22 If A = J n (a) for a F, an elementary Jordan matrix of size n, show 34
16 that m A = (x a) n where A = J n (a) = a a a a (ie A is an n n matrix with a s on the diagonal and 1 s on the subdiagonal) Note: Again, the minimum polynomial happens to equal the characteristic polynomial here DEFINITION 26 (Direct Sum of Matrices) Let A 1,, A t be matrices over F Then the direct sum of these matrices is defined as follows: A A 2 A 1 A 2 A t = 0 A t Properties: 1 (A 1 A t ) + (B 1 B t ) = (A 1 + B 1 ) (A t + B t ) 2 If λ F, λ(a 1 A t ) = (λa 1 ) (λa t ) 3 (A 1 A t )(B 1 B t ) = (A 1 B 1 ) (A t B t ) 4 If f F [x] and A 1,, A t are square, f(a 1 A t ) = f(a 1 ) f(a t ) DEFINITION 27 If f 1,, f t F [x], we call f F [x] a least common multiple ( lcm ) of f 1,, f t if 35
17 1 f 1 f, f t f, and 2 f 1 e, f t e f e This uniquely defines the lcm up to a constant multiple and so we set the lcm to be the monic lcm EXAMPLES 21 If fg 0, lcm (f, g) fg (Recursive property) THEOREM 27 Also lcm (f 1,, f t+1 ) = lcm ( lcm (f 1,, f t ), f t+1 ) m A1 A t = lcm (m A1,, m At ), ch A1 A t = t ch Ai i=1 PROOF Let f = LHS and g = RHS Then Conversely, Thus f = g f(a 1 A t ) = 0 f(a 1 ) f(a t ) = 0 0 f(a 1 ) = 0,, f(a t ) = 0 m A1 f,, m At f g f m A1 g,, m At g g(a 1 ) = 0,, g(a t ) = 0 g(a 1 ) g(a t ) = 0 0 g(a 1 A t ) = 0 f = m A1 A t g EXAMPLE 26 Let A = C(f) and B = C(g) Then m A B = lcm (f, g) 36
18 Note: If f = cp a 1 1 pat t g = dp b 1 1 p bt t where c, d 0 are in F and p 1,, p t are distinct monic irreducibles, then gcd(f, g) = p min(a 1,b 1 ) 1 p min(at,bt) t, lcm (f, g) = p max(a 1,b 1 ) 1 p max(at,bt) t Note so min(a i, b i ) + max(a i, b i ) = a i + b i gcd(f, g) lcm (f, g) = fg EXAMPLE 27 If A = diag (λ 1,, λ n ), then m A = (x c 1 ) (x c t ), where c 1,, c t are the distinct members of the sequence λ 1,, λ n PROOF For A is the direct sum of the 1 1 matrices λ 1,, λ n having minimum polynomials x λ 1,, λ n Hence m A = lcm (x λ 1,, x λ n ) = (x c 1 ) (x c t ) We know that m A ch A Hence if ch A = p a 1 1 pat t where a 1 > 0,, a t > 0, and p 1,, p t are distinct monic irreducibles, then m A = p b 1 1 p bt t where 0 b i a i, i = 1,, t We soon show that each b i > 0, ie if p ch A p m A and p is irreducible then 37
19 25 Construction of a field of p n elements (where p is prime and n N) Let f be a monic irreducible polynomial of degree n in Z p [x] that is, F q = Z p here For instance, n = 2, p = 2 x 2 + x + 1 = f n = 3, p = 2 x 3 + x + 1 = f or x 3 + x = f Let A = C(f), the companion matrix of f Then we know f(a) = 0 We assert that the set of all matrices of the form g(a), where g Z p [x], forms a field consisting of precisely p n elements The typical element is b 0 I n + b 1 A + + b t A t where b 0,, b t Z p We need only show existence of a multiplicative inverse for each element except 0 (the additive identity), as the remaining axioms clearly hold So let g Z p [x] such that g(a) 0 We have to find h Z p [x] satisfying Note that g(a) 0 f g, since g(a)h(a) = I n f g g = ff 1 and hence g(a) = f(a)f 1 (A) = 0f 1 (A) = 0 Then since f is irreducible and f g, there exist u, v Z p [x] such that uf + vg = 1 Hence u(a)f(a) + v(a)g(a) = I n and v(a)g(a) = I n, as required We now show that our new field is a Z p vector space with basis consisting of the matrices I n, A,, A n 1 Firstly the spanning property: By Euclid s division theorem, g = fq + r 38
20 where q, r Z p [x] and deg r < deg g So let where r 0,, r n 1 Z p Then r = r 0 + r 1 x + + r n 1 x n 1 g(a) = f(a)q(a) + r(a) = 0q(A) + r(a) = r(a) = r 0 I n + r 1 A + + r n 1 A n 1 Secondly, linear independence over Z p : Suppose that r 0 I n + r 1 A + + r n 1 A n 1 = 0, where r 0, r 1,, r n 1 Z p Then r(a) = 0, where r = r 0 + r 1 x + + r n 1 x n 1 Hence m A = f divides r Consequently r = 0, as deg f = n whereas deg r < n if r 0 Consequently, there are p n such matrices g(a) in the field we have constructed Numerical Examples EXAMPLE 28 Let p = 2, n = 2, f = x 2 + x + 1 Z 2 [x], and A = C(f) Then [ ] [ ] A = =, and F 4 = { a 0 I 2 + a 1 A a 0, a 1 Z 2 } = { 0, I 2, A, I 2 + A } We construct addition and multiplication tables for this field, with B = I 2 + A (as an exercise, check these): 0 I 2 A B 0 0 I 2 A B I 2 I 2 0 B A A A B 0 I 2 B B A I I 2 A B I 2 0 I 2 A B A 0 A B I 2 B 0 B I 2 A 39
21 EXAMPLE 29 Let p = 2, n = 3, f = x 3 + x + 1 Z 2 [x] Then A = C(f) = = and our eightmember field F 8 (usually denoted by GF (8) [ GF corresponds to Galois Field, in honour of Galois]) is F 8 = { a 0 I 3 + a 1 A + a 2 A 2 a 0, a 1, a 2 Z 2 } = { 0, I 3, A, A 2, I 3 + A, I 3 + A 2, A + A 2, I 3 + A + A 2 } Now find (A 2 + A) 1 Solution: use Euclid s algorithm Hence x 3 + x + 1 = (x + 1)(x 2 + x) + 1 x 3 + x (x + 1)(x 2 + x) = 1 A 3 + A + I 3 + (A + I 3 )(A 2 + A) = I 3 (A + I 3 )(A 2 + A) = I 3 Hence (A 2 + A) 1 = A + I 3 THEOREM 28 Every finite field has precisely p n elements for some prime p the least positive integer with the property that 1 } {{ } = 0 p p is then called the characteristic of the field Also, if x F, a field of q elements, then it can be shown that if x 0, then x q 1 = 1 In the special case F = Z p, this reduces to Fermat s Little Theorem:, x p 1 1 (mod p), if p is prime not dividing x 40
22 26 Characteristic and Minimum Polynomial of a Transformation DEFINITION 28 (Characteristic polynomial of T : V V ) Let β be a basis for V and A = [T ] β β Then we define ch T = ch A This polynomial is independent of the basis β: PROOF ( ch T is independent of the basis) If γ is another basis for V and B = [T ] γ γ, then we know A = P 1 BP where P is the change of basis matrix [I V ] γ β Then ch A = ch P 1 BP = det(xi n P 1 BP ) where n = dim V = det(p 1 (xi n )P P 1 BP ) = det(p 1 (xi n B)P ) = det P 1 ch B det P = ch B DEFINITION 29 If f = a a t x t, where a 0,, a t F, we define Then the usual properties hold: f(t ) = a 0 I V + + a t T t f, g F [x] (f+g)(t ) = f(t )+g(t ) and (fg)(t ) = f(t )g(t ) = g(t )f(t ) LEMMA 21 f F [x] [f(t )] β β = f ([T ] β β ) Note: The CayleyHamilton theorem for matrices says that ch A (A) = 0 Then if A = [T ] β β, we have by the lemma so ch T (T ) = 0 V [ ch T (T )] β β = ch T (A) = ch A (A) = 0, 41
23 DEFINITION 210 Let T : V V be a linear transformation over F Then any polynomial of least positive degree such that f(t ) = 0 V is called a minimum polynomial of T We have corresponding results for polynomials in a transformation T to those for polynomials in a square matrix A: g = qf + r g(t ) = q(t )f(t ) + r(t ) Again, there is a unique monic minimum polynomial of T is denoted by m T and called the minimum polynomial of T Also note that because of the lemma, For (with A = [T ] β β ) m T = m [T ] β β (a) m A (A) = 0, so m A (T ) = 0 V Hence m T m A (b) m T (T ) = 0 V, so [m T (T )] β β = 0 Hence m T (A) = 0 and so m A m T EXAMPLES 22 T = 0 V m T = x T = I V m T = x 1 T = ci V m T = x c T 2 = T and T 0 V and T I V m T = x 2 x 261 M n n (F [x]) Ring of Polynomial Matrices Example: [ x x 5 ] + 5x + 1 M x (Q[x]) ] [ ] [ + x x = x 5 [ we see that any element of M n n (F [x]) is expressible as x m A m + x m 1 A m A 0 ] [ where A i M n n (F ) We write the coefficient of x i after x i, to distinguish these entities from corresponding objects of the following ring 42 ]
24 262 M n n (F )[y] Ring of Matrix Polynomials This consists of all polynomials in y with coefficients in M n n (F ) Example: [ ] [ ] [ ] [ ] 0 1 y y y + M (F )[y] THEOREM 29 The mapping given by Φ : M n n (F )[y] M n n (F [x]) Φ(A 0 + A 1 y + + A m y m ) = A 0 + xa x m A m where A i M n n (F ), is a 1 1 correspondence and has the following properties: Φ(X + Y ) = Φ(X) + Φ(Y ) Φ(XY ) = Φ(X)Φ(Y ) Φ(tX) = tφ(x) t F Also Φ(I n y A) = xi n A A M n n (F ) THEOREM 210 ((Left) Remainder theorem for matrix polynomials) where Let B m y m + + B 0 M n n (F )[y] and A M n n (F ) Then B m y m + + B 0 = (I n y A)Q + R R = A m B m + + AB 1 + B 0 and Q = C m 1 y m C 0 where C m 1,, C 0 are computed recursively: B m = C m 1 B m 1 = AC m 1 + C m 2 B 1 = AC 1 + C 0 43
25 PROOF First we verify that B 0 = AC 0 + R: R = A m B m = A m C m 1 +A m 1 B m 1 A m C m 1 + A m 1 C m AB 1 A 2 C 1 + AC 0 +B 0 B 0 = B 0 + AC 0 Then (I n y A)Q + R = (I n y)(c m 1 y m C 0 ) A(C m 1 y m C 0 ) + A m B m + + B 0 = C m 1 y m + (C m 2 AC m 1 )y m (C 0 AC 1 )y + AC 0 + R = B m y m + B m 1 y m B 1 y + B 0 Remark There is a similar right remainder theorem THEOREM 211 If p is an irreducible polynomial dividing ch A, then p m A PROOF (From Burton Jones, Linear Algebra ) Let m A = x t + a t 1 x t a 0 and consider the matrix polynomial in y Φ 1 (m A I n ) = I n y t + (a t 1 I n )y t (a 0 I n ) = (I n y A)Q + A t I n + A t 1 (a t 1 I n ) + + a 0 I n = (I n y A)Q + m T (A) = (I n y A)Q Now take Φ of both sides to give m A I n = (xi n A)Φ(Q) and taking determinants of both sides yields {m A } n = ch A det Φ(Q) 44
26 So letting p be an irreducible polynomial dividing ch A, we have p {m A } n and hence p m A Alternative simpler proof (MacDuffee): m A (x) m A (y) = (x y)k(x, y), where k(x, y) F [x, y] Hence m A (x)i n = m A (xi n ) m A (A) = (xi n A)k(xI n, A) Now take determinants to get m A (x) n = ch A (x) det k(xi n, A) Exercise: If (x) is the gcd of the elements of adj(xi n A), use the equation (xi n a)adj(xi n A) = ch A (x)i n and an above equation to deduce that m A (x) = ch A (x)/ (x) EXAMPLES 23 With A = 0 M n n (F ), we have ch A = x n and m A = x A = diag (1, 1, 2, 2, 2) M 5 5 (Q) Here ch A = (x 1) 2 (x 2) 3 and m A = (x 1)(x 2) DEFINITION 211 A matrix A M n n (F ) is called diagonable over F if there exists a non singular matrix P M n n (F ) such that where λ 1,, λ n belong to F P 1 AP = diag (λ 1,, λ n ), THEOREM 212 If A is diagonable, then m A is a product of distinct linear factors PROOF If P 1 AP = diag (λ 1,, λ n ) (with λ 1,, λ n F ) then m A = m P 1 AP = m diag (λ 1,, λ n ) = (x c 1 )(x c 2 ) (x c t ) where c 1,, c t are the distinct members of the sequence λ 1,, λ n The converse is also true, and will (fairly) soon be proved 45
27 EXAMPLE 210 A = J n (a) We saw earlier that m A = (x a) n so if n 2 we see that A is not diagonable DEFINITION 212 (Diagonable LTs) T : V V is called diagonable over F if there exists a basis β for V such that [T ] β β is diagonal THEOREM 213 A is diagonable T A is diagonable PROOF (Sketch) Suppose P 1 AP = diag (λ 1,, λ n ) letting P = [P 1 P n ] we see that Now premultiplying by P and T A (P 1 ) = AP 1 = λ 1 P 1 T A (P n ) = AP n = λ n P n and we let β be the basis P 1,, P n over V n (F ) Then λ 1 [T A ] β β = λ 2 λn Reverse the argument and use Theorem 117 THEOREM 214 Let A M n n (F ) Then if λ is an eigenvalue of A with multiplicity m, (that is (x λ) m is the exact power of x λ which divides ch A ), we have nullity (A λi n ) m 46
28 REMARKS (1) If m = 1, we deduce that nullity (A λi n ) = 1 For the inequality 1 nullity (A λi n ) always holds (2) The integer nullity (A λi n ) is called the geometric multiplicity of the eigenvalue λ, while m is referred to as the algebraic multiplicity of λ PROOF Let v 1,, v r be a basis for N(A λi n ), where λ is an eigenvalue of A having multiplicity m Extend this linearly independent family to a basis v 1,, v r, v r+1,, v n of V n (F ) Then the following equations hold: Av 1 = λv 1 Av r = λv r Av r+1 = b 11 v b n1 v n Av n = b 1n r v b nn r v n These equations can be combined into a single matrix equation: A[v 1 v r v r+1 v n ] = [Av 1 Av r Av r+1 Av n ] = [λv 1 λv r b 11 v b n1 v n b 1n r v b nn r v n ] [ ] λir B = [v 1 v n ] 1 0 B 2 Hence if P = [v 1 v n ], we have [ P 1 λir B AP = 1 0 B 2 ] Then ch A = ch P 1 AP = ch λir ch B2 = (x λ) r ch B2 and because (x λ) m is the exact power of x λ dividing ch A, it follows that nullity (A λi n ) = r m THEOREM 215 Suppose that ch T = (x c 1 ) a1 (x c t ) at Then T is diagonable if nullity (T c i I v ) = a i for 1 i t 47
29 PROOF We first prove that the subspaces Ker (T c i I V ) are independent (Subspaces V 1,, V t are called independent if v v t = 0, v i V i, i = 1, t, v 1 = 0,, v t = 0 Then dim (V V t ) = dim (V 1 ) + + dim V t )) Assume that v v t = 0, where v i Ker (T c i I v ) for 1 i t Then Similarly we deduce that T (v v t ) = T (0) c 1 v c t v t = 0 c 2 1v c 2 t v t = 0 c t 1 1 v c t 1 t v t = 0 We can combine these t equations into a single matrix equation 1 1 c 1 c t v 1 o = c t 1 1 c t 1 v t 0 t However the coefficient matrix is the Vandermonde matrix, which is non singular as c i c j if i j, so we deduce that v 1 = 0,, v t = 0 Hence with V i = Ker (T c i I V ), we have Hence dim (V V t ) = t dim V i = i=1 V = V V t t a i = dim V Then if β i is a basis for V i for i i t and β = β 1 β t, it follows that β is a basis for V Moreover t [T ] β β = (c i I ai ) i=1 48 i=1
30 and T is diagonable EXAMPLE Let A = (a) We find that ch A = (x 3) 2 (x 9) Next we find bases for each of the eigenspaces N(A 9I 3 ) and N(A 3I 3 ): First we solve (A 3I 3 )X = 0 We have A 3I 3 = Hence the eigenspace consists of vectors X = [x, y, z] t satisfying x = y+z, with y and z arbitrary Hence y + z 1 1 X = y = y 1 + z 0, z 0 1 so X 11 = [ 1, 1, 0] t and X 12 = [1, 0, 1] t form a basis for the eigenspace corresponding to the eigenvalue 3 Next we solve (A 9I 3 )X = 0 We have A 9I 3 = Hence the eigenspace consists of vectors X = [x, y, z] t satisfying x = z and y = z, with z arbitrary Hence z 1 X = z z = z 1 1 and we can take X 21 = [ 1, 1, 1] t as a basis for the eigenspace corresponding to the eigenvalue 9 Then P = [X 11 X 12 X 21 ] is non singular and P 1 AP =
31 THEOREM 216 If m T = (x c 1 ) (x c t ) for c 1,, c t distinct in F, then T is diagonable and conversely Moreover there exist unique linear transformations T 1,, T t satisfying I V = T T t, T = c 1 T c t T t, T i T j = 0 V if i j, T 2 i = T i, 1 i t Also rank T i = a i, where ch T = (x c 1 ) a1 (x c t ) at Remarks 1 T 1,, T t are called the principal idempotents of T 2 If g F [x], then g(t ) = g(c 1 )T g(c t )T t For example T m = c m 1 T c m t T t 3 If c 1,, c t are non zero (that is the eigenvalues of T are non zero), the T 1 is given by T 1 = c 1 1 T c 1 t T t Formulae 2 and 3 are useful in the corresponding matrix formulation PROOF Suppose m T = (x c 1 ) (x c t ), where c 1,, c t are distinct Then ch T = (x c 1 ) a1 (x c t ) at To prove T is diagonable, we have to prove that nullity (T c i I V ) = a i, 1 i t Let p 1,, p t be the Lagrange interpolation polynomials based on c 1,, c t, ie t ( ) x cj p i =, 1 i t Then In particular, j=1 j i c i c j g F [x] g = g(c 1 )p g(c t )p t g = 1 1 = p p t 50
32 and g = x x = c 1 p c t p t Hence with T i = p i (T ), I V = T T t T = c 1 T c t T t Next m T = (x c 1 ) (x c t ) p i p j (p i p j )(T ) = 0 V if i j if i j p i (T )p j (T ) = 0 V or T i T j = 0 V if i j Then T 2 i = T i (T T t ) = T i I V = T i Next 0 V = m T (T ) = (T c 1 I V ) (T c t I V ) Hence dim V = nullity 0 V t nullity (T c i I V ) i=1 t a i = dim V i=1 Consequently nullity (T c i I V ) = a i, 1 i t and T is therefore diagonable Next we prove that rank T i = a i From the definition of p i, we have nullity p i (T ) t nullity (T c j I V ) = j=1 j i t a j = dim V a i j=1 j i Also p i (T )(T c i I V ) = 0, so Im (T c i I V ) Ker p i (T ) Hence dim V a i nullity p i (T ) and consequently nullity p i (T ) = dim (V ) a i, so rank p i (T ) = a i We next prove the uniqueness of T 1,, T t Suppose that S 1,, S t also satisfy the same conditions as T 1,, T t Then T i T = T T i = c i T i S j T = T S j = c j S j T i (T S j ) = T i (c j S j ) = c j T i S j = (T i T )S j = c i T i S j 51
33 so (c j c i )T i S j = 0 V and T i S j = 0 V if i j Hence T i = T i I V = T i ( S i = I V S i = ( t S j ) = T i S i j=1 t T j )S i = T i S i Hence T i = S i Conversely, suppose that T is diagonable and let β be a basis of V such that A = [T ] β β = diag (λ 1,, λ n ) Then m T = m A = (x c 1 ) (x c t ), where c 1,, c t are the distinct members of the sequence λ 1,, λ n COROLLARY 25 If j=1 ch T = (x c 1 ) (x c t ) with c i distinct members of F, then T is diagonable Proof: Here m T = ch T and we use theorem 33 EXAMPLE 211 Let A = [ 0 a b 0 ] a, b F, ab 0, Then A is diagonable if and only if ab = y 2 for some y F For ch A = x 2 ab, so if ab = y 2, ch A = x 2 y 2 = (x + y)(x y) which is a product of distinct linear factors, as y y here Conversely suppose that A is diagonable Then as A is not a scalar matrix, it follows that m A is not linear and hence m A = (x c 1 )(x c 2 ), where c 1 c 2 Also ch A = m A, so ch A (c 1 ) = 0 Hence c 2 1 ab = 0, or ab = c 2 1 For example, take F = Z 7 and let a = 1 and b = 3 consequently A is not diagonable Then ab y 2 and 52
Similarity and Diagonalization. Similar Matrices
MATH022 Linear Algebra Brief lecture notes 48 Similarity and Diagonalization Similar Matrices Let A and B be n n matrices. We say that A is similar to B if there is an invertible n n matrix P such that
More informationChapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm.
Chapter 4, Arithmetic in F [x] Polynomial arithmetic and the division algorithm. We begin by defining the ring of polynomials with coefficients in a ring R. After some preliminary results, we specialize
More informationQuotient Rings and Field Extensions
Chapter 5 Quotient Rings and Field Extensions In this chapter we describe a method for producing field extension of a given field. If F is a field, then a field extension is a field K that contains F.
More informationLINEAR ALGEBRA W W L CHEN
LINEAR ALGEBRA W W L CHEN c W W L Chen, 1997, 2008 This chapter is available free to all individuals, on understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied,
More information1 Sets and Set Notation.
LINEAR ALGEBRA MATH 27.6 SPRING 23 (COHEN) LECTURE NOTES Sets and Set Notation. Definition (Naive Definition of a Set). A set is any collection of objects, called the elements of that set. We will most
More informationFactoring Polynomials
Factoring Polynomials Sue Geller June 19, 2006 Factoring polynomials over the rational numbers, real numbers, and complex numbers has long been a standard topic of high school algebra. With the advent
More informationUnique Factorization
Unique Factorization Waffle Mathcamp 2010 Throughout these notes, all rings will be assumed to be commutative. 1 Factorization in domains: definitions and examples In this class, we will study the phenomenon
More informationThe Division Algorithm for Polynomials Handout Monday March 5, 2012
The Division Algorithm for Polynomials Handout Monday March 5, 0 Let F be a field (such as R, Q, C, or F p for some prime p. This will allow us to divide by any nonzero scalar. (For some of the following,
More informationsome algebra prelim solutions
some algebra prelim solutions David Morawski August 19, 2012 Problem (Spring 2008, #5). Show that f(x) = x p x + a is irreducible over F p whenever a F p is not zero. Proof. First, note that f(x) has no
More informationFactorization Algorithms for Polynomials over Finite Fields
Degree Project Factorization Algorithms for Polynomials over Finite Fields Sajid Hanif, Muhammad Imran 20110503 Subject: Mathematics Level: Master Course code: 4MA11E Abstract Integer factorization is
More informationH/wk 13, Solutions to selected problems
H/wk 13, Solutions to selected problems Ch. 4.1, Problem 5 (a) Find the number of roots of x x in Z 4, Z Z, any integral domain, Z 6. (b) Find a commutative ring in which x x has infinitely many roots.
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationa 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2.
Chapter 1 LINEAR EQUATIONS 1.1 Introduction to linear equations A linear equation in n unknowns x 1, x,, x n is an equation of the form a 1 x 1 + a x + + a n x n = b, where a 1, a,..., a n, b are given
More informationit is easy to see that α = a
21. Polynomial rings Let us now turn out attention to determining the prime elements of a polynomial ring, where the coefficient ring is a field. We already know that such a polynomial ring is a UF. Therefore
More informationNOTES ON LINEAR TRANSFORMATIONS
NOTES ON LINEAR TRANSFORMATIONS Definition 1. Let V and W be vector spaces. A function T : V W is a linear transformation from V to W if the following two properties hold. i T v + v = T v + T v for all
More informationMathematics Course 111: Algebra I Part IV: Vector Spaces
Mathematics Course 111: Algebra I Part IV: Vector Spaces D. R. Wilkins Academic Year 19967 9 Vector Spaces A vector space over some field K is an algebraic structure consisting of a set V on which are
More informationRESULTANT AND DISCRIMINANT OF POLYNOMIALS
RESULTANT AND DISCRIMINANT OF POLYNOMIALS SVANTE JANSON Abstract. This is a collection of classical results about resultants and discriminants for polynomials, compiled mainly for my own use. All results
More informationI. GROUPS: BASIC DEFINITIONS AND EXAMPLES
I GROUPS: BASIC DEFINITIONS AND EXAMPLES Definition 1: An operation on a set G is a function : G G G Definition 2: A group is a set G which is equipped with an operation and a special element e G, called
More informationCHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY
January 10, 2010 CHAPTER SIX IRREDUCIBILITY AND FACTORIZATION 1. BASIC DIVISIBILITY THEORY The set of polynomials over a field F is a ring, whose structure shares with the ring of integers many characteristics.
More informationIntroduction to Algebraic Geometry. Bézout s Theorem and Inflection Points
Introduction to Algebraic Geometry Bézout s Theorem and Inflection Points 1. The resultant. Let K be a field. Then the polynomial ring K[x] is a unique factorisation domain (UFD). Another example of a
More informationPOLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS
POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS RUSS WOODROOFE 1. Unique Factorization Domains Throughout the following, we think of R as sitting inside R[x] as the constant polynomials (of degree 0).
More informationPROBLEM SET 6: POLYNOMIALS
PROBLEM SET 6: POLYNOMIALS 1. introduction In this problem set we will consider polynomials with coefficients in K, where K is the real numbers R, the complex numbers C, the rational numbers Q or any other
More informationThe Characteristic Polynomial
Physics 116A Winter 2011 The Characteristic Polynomial 1 Coefficients of the characteristic polynomial Consider the eigenvalue problem for an n n matrix A, A v = λ v, v 0 (1) The solution to this problem
More informationOrthogonal Diagonalization of Symmetric Matrices
MATH10212 Linear Algebra Brief lecture notes 57 Gram Schmidt Process enables us to find an orthogonal basis of a subspace. Let u 1,..., u k be a basis of a subspace V of R n. We begin the process of finding
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationApplied Linear Algebra I Review page 1
Applied Linear Algebra Review 1 I. Determinants A. Definition of a determinant 1. Using sum a. Permutations i. Sign of a permutation ii. Cycle 2. Uniqueness of the determinant function in terms of properties
More informationLecture 3: Finding integer solutions to systems of linear equations
Lecture 3: Finding integer solutions to systems of linear equations Algorithmic Number Theory (Fall 2014) Rutgers University Swastik Kopparty Scribe: Abhishek Bhrushundi 1 Overview The goal of this lecture
More informationby the matrix A results in a vector which is a reflection of the given
Eigenvalues & Eigenvectors Example Suppose Then So, geometrically, multiplying a vector in by the matrix A results in a vector which is a reflection of the given vector about the yaxis We observe that
More informationU.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009. Notes on Algebra
U.C. Berkeley CS276: Cryptography Handout 0.1 Luca Trevisan January, 2009 Notes on Algebra These notes contain as little theory as possible, and most results are stated without proof. Any introductory
More informationAu = = = 3u. Aw = = = 2w. so the action of A on u and w is very easy to picture: it simply amounts to a stretching by 3 and 2, respectively.
Chapter 7 Eigenvalues and Eigenvectors In this last chapter of our exploration of Linear Algebra we will revisit eigenvalues and eigenvectors of matrices, concepts that were already introduced in Geometry
More informationUniversity of Lille I PC first year list of exercises n 7. Review
University of Lille I PC first year list of exercises n 7 Review Exercise Solve the following systems in 4 different ways (by substitution, by the Gauss method, by inverting the matrix of coefficients
More informationModern Algebra Lecture Notes: Rings and fields set 4 (Revision 2)
Modern Algebra Lecture Notes: Rings and fields set 4 (Revision 2) Kevin Broughan University of Waikato, Hamilton, New Zealand May 13, 2010 Remainder and Factor Theorem 15 Definition of factor If f (x)
More informationT ( a i x i ) = a i T (x i ).
Chapter 2 Defn 1. (p. 65) Let V and W be vector spaces (over F ). We call a function T : V W a linear transformation form V to W if, for all x, y V and c F, we have (a) T (x + y) = T (x) + T (y) and (b)
More information7. Some irreducible polynomials
7. Some irreducible polynomials 7.1 Irreducibles over a finite field 7.2 Worked examples Linear factors x α of a polynomial P (x) with coefficients in a field k correspond precisely to roots α k [1] of
More informationMATH 423 Linear Algebra II Lecture 38: Generalized eigenvectors. Jordan canonical form (continued).
MATH 423 Linear Algebra II Lecture 38: Generalized eigenvectors Jordan canonical form (continued) Jordan canonical form A Jordan block is a square matrix of the form λ 1 0 0 0 0 λ 1 0 0 0 0 λ 0 0 J = 0
More informationminimal polyonomial Example
Minimal Polynomials Definition Let α be an element in GF(p e ). We call the monic polynomial of smallest degree which has coefficients in GF(p) and α as a root, the minimal polyonomial of α. Example: We
More informationElementary Number Theory We begin with a bit of elementary number theory, which is concerned
CONSTRUCTION OF THE FINITE FIELDS Z p S. R. DOTY Elementary Number Theory We begin with a bit of elementary number theory, which is concerned solely with questions about the set of integers Z = {0, ±1,
More information1 VECTOR SPACES AND SUBSPACES
1 VECTOR SPACES AND SUBSPACES What is a vector? Many are familiar with the concept of a vector as: Something which has magnitude and direction. an ordered pair or triple. a description for quantities such
More informationThe cyclotomic polynomials
The cyclotomic polynomials Notes by G.J.O. Jameson 1. The definition and general results We use the notation e(t) = e 2πit. Note that e(n) = 1 for integers n, e(s + t) = e(s)e(t) for all s, t. e( 1 ) =
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More information4: EIGENVALUES, EIGENVECTORS, DIAGONALIZATION
4: EIGENVALUES, EIGENVECTORS, DIAGONALIZATION STEVEN HEILMAN Contents 1. Review 1 2. Diagonal Matrices 1 3. Eigenvectors and Eigenvalues 2 4. Characteristic Polynomial 4 5. Diagonalizability 6 6. Appendix:
More informationCONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12
CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.
More informationCONTROLLABILITY. Chapter 2. 2.1 Reachable Set and Controllability. Suppose we have a linear system described by the state equation
Chapter 2 CONTROLLABILITY 2 Reachable Set and Controllability Suppose we have a linear system described by the state equation ẋ Ax + Bu (2) x() x Consider the following problem For a given vector x in
More informationIntroduction to finite fields
Introduction to finite fields Topics in Finite Fields (Fall 2013) Rutgers University Swastik Kopparty Last modified: Monday 16 th September, 2013 Welcome to the course on finite fields! This is aimed at
More informationGREATEST COMMON DIVISOR
DEFINITION: GREATEST COMMON DIVISOR The greatest common divisor (gcd) of a and b, denoted by (a, b), is the largest common divisor of integers a and b. THEOREM: If a and b are nonzero integers, then their
More informationDiagonalisation. Chapter 3. Introduction. Eigenvalues and eigenvectors. Reading. Definitions
Chapter 3 Diagonalisation Eigenvalues and eigenvectors, diagonalisation of a matrix, orthogonal diagonalisation fo symmetric matrices Reading As in the previous chapter, there is no specific essential
More information3 Factorisation into irreducibles
3 Factorisation into irreducibles Consider the factorisation of a nonzero, noninvertible integer n as a product of primes: n = p 1 p t. If you insist that primes should be positive then, since n could
More informationTheory of Matrices. Chapter 5
Chapter 5 Theory of Matrices As before, F is a field We use F[x] to represent the set of all polynomials of x with coefficients in F We use M m,n (F) and M m,n (F[x]) to denoted the set of m by n matrices
More information1 = (a 0 + b 0 α) 2 + + (a m 1 + b m 1 α) 2. for certain elements a 0,..., a m 1, b 0,..., b m 1 of F. Multiplying out, we obtain
Notes on realclosed fields These notes develop the algebraic background needed to understand the model theory of realclosed fields. To understand these notes, a standard graduate course in algebra is
More informationIRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL. 1. Introduction
IRREDUCIBLE OPERATOR SEMIGROUPS SUCH THAT AB AND BA ARE PROPORTIONAL R. DRNOVŠEK, T. KOŠIR Dedicated to Prof. Heydar Radjavi on the occasion of his seventieth birthday. Abstract. Let S be an irreducible
More informationMATH10040 Chapter 2: Prime and relatively prime numbers
MATH10040 Chapter 2: Prime and relatively prime numbers Recall the basic definition: 1. Prime numbers Definition 1.1. Recall that a positive integer is said to be prime if it has precisely two positive
More informationA note on companion matrices
Linear Algebra and its Applications 372 (2003) 325 33 www.elsevier.com/locate/laa A note on companion matrices Miroslav Fiedler Academy of Sciences of the Czech Republic Institute of Computer Science Pod
More informationPROOFS BY DESCENT KEITH CONRAD
PROOFS BY DESCENT KEITH CONRAD As ordinary methods, such as are found in the books, are inadequate to proving such difficult propositions, I discovered at last a most singular method... that I called the
More informationChapter 13: Basic ring theory
Chapter 3: Basic ring theory Matthew Macauley Department of Mathematical Sciences Clemson University http://www.math.clemson.edu/~macaule/ Math 42, Spring 24 M. Macauley (Clemson) Chapter 3: Basic ring
More informationEigenvalues, Eigenvectors, Matrix Factoring, and Principal Components
Eigenvalues, Eigenvectors, Matrix Factoring, and Principal Components The eigenvalues and eigenvectors of a square matrix play a key role in some important operations in statistics. In particular, they
More informationFactoring of Prime Ideals in Extensions
Chapter 4 Factoring of Prime Ideals in Extensions 4. Lifting of Prime Ideals Recall the basic AKLB setup: A is a Dedekind domain with fraction field K, L is a finite, separable extension of K of degree
More informationLINEAR ALGEBRA. September 23, 2010
LINEAR ALGEBRA September 3, 00 Contents 0. LUdecomposition.................................... 0. Inverses and Transposes................................. 0.3 Column Spaces and NullSpaces.............................
More information12 Greatest Common Divisors. The Euclidean Algorithm
Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 12 Greatest Common Divisors. The Euclidean Algorithm As mentioned at the end of the previous section, we would like to
More informationModélisation et résolutions numérique et symbolique
Modélisation et résolutions numérique et symbolique via les logiciels Maple et Matlab Jeremy Berthomieu Mohab Safey El Din Stef Graillat Mohab.Safey@lip6.fr Outline Previous course: partial review of what
More informationInner Product Spaces and Orthogonality
Inner Product Spaces and Orthogonality week 34 Fall 2006 Dot product of R n The inner product or dot product of R n is a function, defined by u, v a b + a 2 b 2 + + a n b n for u a, a 2,, a n T, v b,
More information4.6 Null Space, Column Space, Row Space
NULL SPACE, COLUMN SPACE, ROW SPACE Null Space, Column Space, Row Space In applications of linear algebra, subspaces of R n typically arise in one of two situations: ) as the set of solutions of a linear
More informationFactoring Algorithms
Factoring Algorithms The p 1 Method and Quadratic Sieve November 17, 2008 () Factoring Algorithms November 17, 2008 1 / 12 Fermat s factoring method Fermat made the observation that if n has two factors
More informationa 1 x + a 0 =0. (3) ax 2 + bx + c =0. (4)
ROOTS OF POLYNOMIAL EQUATIONS In this unit we discuss polynomial equations. A polynomial in x of degree n, where n 0 is an integer, is an expression of the form P n (x) =a n x n + a n 1 x n 1 + + a 1 x
More information9. POLYNOMIALS. Example 1: The expression a(x) = x 3 4x 2 + 7x 11 is a polynomial in x. The coefficients of a(x) are the numbers 1, 4, 7, 11.
9. POLYNOMIALS 9.1. Definition of a Polynomial A polynomial is an expression of the form: a(x) = a n x n + a n1 x n1 +... + a 1 x + a 0. The symbol x is called an indeterminate and simply plays the role
More informationRecall that two vectors in are perpendicular or orthogonal provided that their dot
Orthogonal Complements and Projections Recall that two vectors in are perpendicular or orthogonal provided that their dot product vanishes That is, if and only if Example 1 The vectors in are orthogonal
More informationSec 4.1 Vector Spaces and Subspaces
Sec 4. Vector Spaces and Subspaces Motivation Let S be the set of all solutions to the differential equation y + y =. Let T be the set of all 2 3 matrices with real entries. These two sets share many common
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More information4. Matrix inverses. left and right inverse. linear independence. nonsingular matrices. matrices with linearly independent columns
L. Vandenberghe EE133A (Spring 2016) 4. Matrix inverses left and right inverse linear independence nonsingular matrices matrices with linearly independent columns matrices with linearly independent rows
More informationPYTHAGOREAN TRIPLES KEITH CONRAD
PYTHAGOREAN TRIPLES KEITH CONRAD 1. Introduction A Pythagorean triple is a triple of positive integers (a, b, c) where a + b = c. Examples include (3, 4, 5), (5, 1, 13), and (8, 15, 17). Below is an ancient
More informationIntroduction to Finite Fields (cont.)
Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number
More information2.1: MATRIX OPERATIONS
.: MATRIX OPERATIONS What are diagonal entries and the main diagonal of a matrix? What is a diagonal matrix? When are matrices equal? Scalar Multiplication 45 Matrix Addition Theorem (pg 0) Let A, B, and
More information5.1 Commutative rings; Integral Domains
5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following
More informationCHAPTER 5. Number Theory. 1. Integers and Division. Discussion
CHAPTER 5 Number Theory 1. Integers and Division 1.1. Divisibility. Definition 1.1.1. Given two integers a and b we say a divides b if there is an integer c such that b = ac. If a divides b, we write a
More informationDecember 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B. KITCHENS
December 4, 2013 MATH 171 BASIC LINEAR ALGEBRA B KITCHENS The equation 1 Lines in twodimensional space (1) 2x y = 3 describes a line in twodimensional space The coefficients of x and y in the equation
More informationMATH 304 Linear Algebra Lecture 18: Rank and nullity of a matrix.
MATH 304 Linear Algebra Lecture 18: Rank and nullity of a matrix. Nullspace Let A = (a ij ) be an m n matrix. Definition. The nullspace of the matrix A, denoted N(A), is the set of all ndimensional column
More informationAlgebra 3: algorithms in algebra
Algebra 3: algorithms in algebra Hans Sterk 20032004 ii Contents 1 Polynomials, Gröbner bases and Buchberger s algorithm 1 1.1 Introduction............................ 1 1.2 Polynomial rings and systems
More informationMath 115A HW4 Solutions University of California, Los Angeles. 5 2i 6 + 4i. (5 2i)7i (6 + 4i)( 3 + i) = 35i + 14 ( 22 6i) = 36 + 41i.
Math 5A HW4 Solutions September 5, 202 University of California, Los Angeles Problem 4..3b Calculate the determinant, 5 2i 6 + 4i 3 + i 7i Solution: The textbook s instructions give us, (5 2i)7i (6 + 4i)(
More information3. Let A and B be two n n orthogonal matrices. Then prove that AB and BA are both orthogonal matrices. Prove a similar result for unitary matrices.
Exercise 1 1. Let A be an n n orthogonal matrix. Then prove that (a) the rows of A form an orthonormal basis of R n. (b) the columns of A form an orthonormal basis of R n. (c) for any two vectors x,y R
More informationr + s = i + j (q + t)n; 2 rs = ij (qj + ti)n + qtn.
Chapter 7 Introduction to finite fields This chapter provides an introduction to several kinds of abstract algebraic structures, particularly groups, fields, and polynomials. Our primary interest is in
More informationMATH1231 Algebra, 2015 Chapter 7: Linear maps
MATH1231 Algebra, 2015 Chapter 7: Linear maps A/Prof. Daniel Chan School of Mathematics and Statistics University of New South Wales danielc@unsw.edu.au Daniel Chan (UNSW) MATH1231 Algebra 1 / 43 Chapter
More informationLectures notes on orthogonal matrices (with exercises) 92.222  Linear Algebra II  Spring 2004 by D. Klain
Lectures notes on orthogonal matrices (with exercises) 92.222  Linear Algebra II  Spring 2004 by D. Klain 1. Orthogonal matrices and orthonormal sets An n n realvalued matrix A is said to be an orthogonal
More informationThe Dirichlet Unit Theorem
Chapter 6 The Dirichlet Unit Theorem As usual, we will be working in the ring B of algebraic integers of a number field L. Two factorizations of an element of B are regarded as essentially the same if
More informationQUADRATIC RECIPROCITY IN CHARACTERISTIC 2
QUADRATIC RECIPROCITY IN CHARACTERISTIC 2 KEITH CONRAD 1. Introduction Let F be a finite field. When F has odd characteristic, the quadratic reciprocity law in F[T ] (see [4, Section 3.2.2] or [5]) lets
More informationCS 103X: Discrete Structures Homework Assignment 3 Solutions
CS 103X: Discrete Structures Homework Assignment 3 s Exercise 1 (20 points). On wellordering and induction: (a) Prove the induction principle from the wellordering principle. (b) Prove the wellordering
More informationTHREE DIMENSIONAL GEOMETRY
Chapter 8 THREE DIMENSIONAL GEOMETRY 8.1 Introduction In this chapter we present a vector algebra approach to three dimensional geometry. The aim is to present standard properties of lines and planes,
More informationModule MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013
Module MA3411: Abstract Algebra Galois Theory Appendix Michaelmas Term 2013 D. R. Wilkins Copyright c David R. Wilkins 1997 2013 Contents A Cyclotomic Polynomials 79 A.1 Minimum Polynomials of Roots of
More informationMath 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field
Math 34560 Abstract Algebra I Questions for Section 23: Factoring Polynomials over a Field 1. Throughout this section, F is a field and F [x] is the ring of polynomials with coefficients in F. We will
More information26. Determinants I. 1. Prehistory
26. Determinants I 26.1 Prehistory 26.2 Definitions 26.3 Uniqueness and other properties 26.4 Existence Both as a careful review of a more pedestrian viewpoint, and as a transition to a coordinateindependent
More informationMATH10212 Linear Algebra. Systems of Linear Equations. Definition. An ndimensional vector is a row or a column of n numbers (or letters): a 1.
MATH10212 Linear Algebra Textbook: D. Poole, Linear Algebra: A Modern Introduction. Thompson, 2006. ISBN 0534405967. Systems of Linear Equations Definition. An ndimensional vector is a row or a column
More informationMatrix Representations of Linear Transformations and Changes of Coordinates
Matrix Representations of Linear Transformations and Changes of Coordinates 01 Subspaces and Bases 011 Definitions A subspace V of R n is a subset of R n that contains the zero element and is closed under
More informationA number field is a field of finite degree over Q. By the Primitive Element Theorem, any number
Number Fields Introduction A number field is a field of finite degree over Q. By the Primitive Element Theorem, any number field K = Q(α) for some α K. The minimal polynomial Let K be a number field and
More informationWinter Camp 2011 Polynomials Alexander Remorov. Polynomials. Alexander Remorov alexanderrem@gmail.com
Polynomials Alexander Remorov alexanderrem@gmail.com Warmup Problem 1: Let f(x) be a quadratic polynomial. Prove that there exist quadratic polynomials g(x) and h(x) such that f(x)f(x + 1) = g(h(x)).
More informationGröbner Bases and their Applications
Gröbner Bases and their Applications Kaitlyn Moran July 30, 2008 1 Introduction We know from the Hilbert Basis Theorem that any ideal in a polynomial ring over a field is finitely generated [3]. However,
More informationKevin James. MTHSC 412 Section 2.4 Prime Factors and Greatest Comm
MTHSC 412 Section 2.4 Prime Factors and Greatest Common Divisor Greatest Common Divisor Definition Suppose that a, b Z. Then we say that d Z is a greatest common divisor (gcd) of a and b if the following
More informationChapter 17. Orthogonal Matrices and Symmetries of Space
Chapter 17. Orthogonal Matrices and Symmetries of Space Take a random matrix, say 1 3 A = 4 5 6, 7 8 9 and compare the lengths of e 1 and Ae 1. The vector e 1 has length 1, while Ae 1 = (1, 4, 7) has length
More informationAlgebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the 201213 school year.
This document is designed to help North Carolina educators teach the Common Core (Standard Course of Study). NCDPI staff are continually updating and improving these tools to better serve teachers. Algebra
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings These notes are a summary of some of the important points on divisibility in polynomial rings from 17 and 18 of Gallian s Contemporary Abstract Algebra. Most of the important
More informationSection 2.1. Section 2.2. Exercise 6: We have to compute the product AB in two ways, where , B =. 2 1 3 5 A =
Section 2.1 Exercise 6: We have to compute the product AB in two ways, where 4 2 A = 3 0 1 3, B =. 2 1 3 5 Solution 1. Let b 1 = (1, 2) and b 2 = (3, 1) be the columns of B. Then Ab 1 = (0, 3, 13) and
More informationDirect Methods for Solving Linear Systems. Matrix Factorization
Direct Methods for Solving Linear Systems Matrix Factorization Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011
More informationNotes on Determinant
ENGG2012B Advanced Engineering Mathematics Notes on Determinant Lecturer: Kenneth Shum Lecture 918/02/2013 The determinant of a system of linear equations determines whether the solution is unique, without
More information