Homework 5 M 373K Mark Lindberg and Travis Schedler 1. Artin, Chapter 3, Exercise.1. Prove that the numbers of the form a + b, where a and b are rational numbers, form a subfield of C. Let F be the numbers of the form a + b, where a and b are rational numbers. Then Let a + b and c + d be in F. Then a + c + (b + d) F because the rationals are closed under addition. Let a + b F. Then a b F, and a + b + ( a b ) = 0. Let a+b and c+d be in F. Then (a+b )(c+d ) = ac+bd+(ad+bc) F, because the rationals are closed under addition and multiplication. Let a+b F, a+b 0. Then 1 a+b = a b = a a b a b a b F, because the rationals are closed under multiplication, subtraction, and division.a b 0, because if it were, b = ±a, which is not rational. a = 1, b = 0 1 F. Thus, since all necessary properties hold, F is a subfield of C.. Artin, Chapter 3, Exercise.3. Compute the product polynomial (x 3 + 3x + 3x + 1)(x 4 + 4x 3 + 6x + 4x + 1) when the coefficients are regarded as elements of the field F 7. Explain your answer. (x 3 + 3x + 3x + 1)(x 4 + 4x 3 + 6x + 4x + 1) = (x + 1) 3 (x + 1) 4 = (x + 1) 7 = x 7 + 7x 6 + 1x 5 + 35x 4 + 35x 3 + 1x + 7x + 1 = x 7 + 1, because 0 = 7 = 1 = 35 in F 7. Alternative answer: This equals (x + 1) 7, and in general by the binomial theorem, for p any prime, (a + b) = a p + b p + p 1 ( p ) ( i=1 i a i b p i, with the last term vanishing in F p because p ) i is a multiple of p if 1 i p 1. Therefore the Freshman s Dream holds for p-th powers in F p, that is, (a + b) p = a p + b p. Thus in our case, we get x 7 + 1 7 = x 7 + 1 in F 7. 3. Artin, Chapter 3, Exercise.5. Determine the primes p such that the matrix 1 0 A = 0 3 1 0 is invertible, when its entries are considered to be in F p. det(a) = 1 3 1 0 0 1 = 6 ( ) = 6 + 4 = 10 = 5. = 0 only in F and F 5, for p prime, so for all primes p, such that p and p 5, det(a) 0, and so A will be invertible. Conversely, in F or F 5, the determinant is zero, so the matrix is not invertible. 4. Artin, Chapter[ 3, Exercise [.10. [ Interpreting [ matrix entries in the field F, prove that the 0 0 1 0 1 1 0 1 four matrices,,, form a field. Note: This gives one way to 0 0 0 1 1 0 1 1 define the (unique up to isomorphism) field of order four! b 1
We already know that Mat (F ) is a ring, so we only have to show that these elements give a subring (they are closed under addition and multiplication and include 0 and 1), that this subring is commutative, that all nonzero elements in the subring have multiplicative inverses, and that in this ring 1 0, i.e., the ring is not the zero ring. Let 0 = [ 0 0 0 0, I = [ 1 0 0 1, C = [ 1 1 1 0, D = [ 0 1 1 1. Then 0 is the additive identity and I the multiplicative identity. Closure under addition: Note that 0 + X = X and X + X = 0 for all X. So we only have to check I + C = D, I + D = C, C + D = I, and we are closed under addition. Closure under multiplication: Since 0X = X0 = 0 for all X and IX = X = XI for all X, we only have to consider CD, DC, C, and D. But C = D and CD = I, hence also C 3 = I, and we get DC = C C = I and D = C 4 = C. Commutativity: we conclude also from the previous point that, since D = C, C and D commute, and everything always commutes with I and 0. Inverses: Again, we saw C and D were inverses, and I is its own iverse. The identity I 0 is clear. Thus, all of the properties are fulfilled, and we have formed a field of order 4. 5. Artin, Chapter 3, Exercise 3.. Which of the following subsets is a subspace of the vector space F n n of n n matrices with coefficients in F? (a) Symmetric matrices (A = A t ). Let A, B be such that A = A t and B = B t. Then (A + B) t = A t + B t = A + B, and so the symmetric matrices are closed under addition. Let A be a symmetric matrix. Then (ca) t = ca t = ca, and so the symmetric matrices are closed under scalar multiplication. 0 t = 0, so the zero matrix is symmetric. Therefore, the symmetric matrices are a subspace of the vector space F n n of n n. (b) Invertible matrices By a property shown earlier, a matrix A is invertible if and only if det(a) 0. But det(0) = 0, and so 0 is not invertible, and the invertible matrices are not a subspace. (c) Upper triangular matrices. Upper triangular matrices are closed under addition, by observation. Since c0 = 0 for any c R, again by observation, the upper triangular matrices are closed under scalar multiplication. 0 is an upper triangular matrix by definition. Therefore, the upper triangular matrices are a subspace of the vector space F n n of n n. 6. Artin, Chapter 3, Exercise 4.. Let W[ R 4 be the space of solutions of the system of 1 3 linear equations AX = 0, where A =. Find a basis for W. 1 1 3 0
[ 1 0 1 3 Row reduction on A yields B =. By an earlier result, the set of solutions 0 1 4 3 to AX = 0 and BX = 0 must be the same, so we can now consider this new case. Let x 1 X = x x 3 x 4 Then x 1 x 3 + 3x 4 = 0 and x + 4x 3 3x 4 = 0, or x 1 = x 3 3x 4 and x = 4x 3 + 3x 4. Thus, our solutions must be of the form x 3 3x 4 X = x 3 + 3x 4 x 4 Letting x 3 = 0, x 4 = 1 and x 3 = 1, x 4 = 1 by turns, we get the set 1 3 1, 3 0. 0 1 We can see that this obviously spans the set of solutions, as 1 3 x 3 3x 4 x 3 1 = x 3 4 0 = x 3 + 3x 4 0 1 x 4 Then we show that they are linearly independent. Let x 3, x 4 be such that 1 3 x 3 3x 4 x 3 1 = x 3 4 0 = x 3 + 3x 4 = 0. 0 1 x 4 Then x 3 = x 4 = 0 by the last two entries, and so by definition, they must be linearly independent. Then, by a property from the book, the set is a basis for W. 7. Artin, Chapter 3, Exercise 4.3. Prove that the three functions x, cos(x), and e x are linearly independent. Let a, b, c R such that ax + b cos(x) + ce x = 0 x R. Then it also holds after we differentiate. So differentiating twice, b cos(x) + ce x = 0 as well. Adding, we get ax + ce x = 0. Differentiating twice, we get ce x = 0. Thus c = 0. Then a = 0 from ax + ce x = 0 and b = 0 from b cos(x) + ce x = 0. So, by the definition of linear indepenencen, x, cos(x), and e x are linearly independent. 3
8. Artin, Chapter 3, Exercise 4.5. Let V = F n be the space of column vectors. Prove that every subspace W of V is the space of solutions of some system of homogeneous linear equations AX = 0. Let C be an n m matrix of whose columns span a subspace W of V. Note that rk(c) = dim W (as we can see by taking the columns to actually be a basis, so that row reduction on C t gives no zero rows hence rk(c t ) is the size of the basis, i.e., dim W, and rk(c) = rk(c t )). Then consider the homogeneous system C t y = 0, where y is a column vector of length n since C t is an m n matrix. Let U be the vector space of solutions to this system. By reducing C to reduced row-echelon form, we see that U is spanned by n rk(c) vectors, and in fact these are linearly independent. Thus dim U = n rk(c) = n dim W. Let B be an n k matrix whose columns span U. Again rk(b) = dim U. Let W be the space of solutions to the system B t x = 0. Again, we see that dim(w ) = n rk(b) = n dim U = n (n dim W ) = dim W. On the other hand it is clear that W W, and since the dimensions are equal, we conclude W = W. Therefore W is actually the vector space of solutions to the system B t x = 0. 9. Compute the size of GL n (F p ). Observe from an earlier property (chapter 1, and linear algebra courses) of linear algebra that an n n matrix is invertible if and only if all columns are linearly independent. Then we consider the column of GL n (F p ). The first column must be non-zero, but there are p possible different value for the first entry, p for the second entry, and so on, up to p for the nth entry. Since these are independent, by a property of counting, we multiply them, to obtain a total of p n possible column vectors, and p n 1 non-zero column vectors. The first column can be any of these. Then consider the second column. There are p multiples of the first column, and still p n total possible column vectors. Since the zero vector is included in the p multiples of the first column, there are p n p non-zero column vectors which are not multiples of, and thus are linearly independent from, the first column. This process continues for every column, so let us consider the kth column in the general sense. There are p n possible columns, and we have p k 1 previous columns which are all linearly independent. This means that there are p multiples of each of these columns, and so there are a total of p k 1 possible combinations of these columns essentially, the size of their span. Then we have p n p k 1 possible values for column k. Since each column is chosen separately, and we have column 1 and column, and so on, there are a total n of (p n p k 1 ) possible matrices in which every column is linearly independent, which k=1 means that there are the same number of matrices in GL n (F p ). Bonus: Do the same for GL n (F p k). (It is not any more work!) What about SL n (F p k)? All of the above statements still hold for elements of GL n (F p k), and so we can apply the same count to say that GL n (F p k) = n i=1 (p kn p k(i 1) ). Then SL n (F p k) is the kernel of the determinant map det : GL n (F p k) F p k, which is obviously surjective (e.g., by taking diagonal matrices with all entries but the first equal to one); thus SL n (F p k) = GL n (F p k) /(p k 1). 10. Prove that GL (F ) = S 3 as follows: Observe that GL (F ) acts on the set X = F \ {0} of nonzero vectors of F, which has size three, so we get a homomorphism GL (F ) 4
Perm(X) = S 3. Show that this is bijective (hence an isomorphism). We know from basic matrix {[ [ properties [ } that I GL (F ), and for any column vector 1 0 1 x X = F \ {0} =,,, I 0 1 1 x = x. We also have that for any g 1, g GL (F ), x X, by properties of matrix multiplication, g 1 (g x) = (g 1 g )x. Thus, G acts on the set X. Then by problem 1 from homework 4, we have that f : G Perm(X), f(g) g X is a homomorphism. Then we observe that the only matrix g GL (F ) such that g x i = x i is g = I, and therefore the kernel of f is the identity, which means that, by an earlier problem, f is injective. We have by the previous problem that there are ( 1)( ) = 3 = 6 elements in GL (F ), and we know that since X has 3 elements, there are 3! = 6 elements in Perm(X). Then, by Problem 1, part c, from homework 3, since f is an invective homomorphism between two finite groups of the same size, it must be bijective. Thus, by definition, f is an isomorphism, and so GL (F ) = Perm(X) = S 3, and so by the transitivity of isomorphism, GL (F ) = S 3. 5