MATH 205 HOMEWORK #5 OFFICIAL SOLUTION

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1 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION Problem 1: An inner product on a vector space V over F is a bilinear map, : V V F satisfying the extra conditions v, w = w, v, and v, v 0, with equality if and only if v = 0. (a) Show that the standard dot product on R n is an inner product. (b) Show that (f, g) f(x)g(x) dx is an inner product on C ([0, 1], R). (c) Suppose that F is ordered. Prove that for any v, w V, v, w 2 v, v w, w. When does equality hold? What standard inequality in trigonometry does this reflect when V = R n? (d) We say that two vectors v, w in V are orthogonal if v, w = 0. Suppose that T : V V is a linear transformation satisfying T v, w = v, T w for all v, w. Show that eigenvectors of T with different eigenvalues are orthogonal. Solution: Note that if we can show that a bilinear form is symmetric, then it is linear in the first variable if and only if it is linear in the second. Thus showing linearity in one variable is enough. (a) Let v = (a 1,..., a n ) and w = (b 1,..., b n ). The standard dot product is symmetric, as v, w = a 1 b a n b n = b 1 a b n a n = w, v. Let λ R and u = (c 1,..., c n ). Then v+λu, w = (a 1 +λc 1 )b 1 + +(a n +λc n )b n = a 1 b 1 + +a n b n +λ(c 1 b 1 + +c n b n ) = v, w +λ u, w. Thus, is bilinear. In addition, v, v = a a 2 n 0, with equality if and only if a i = 0 for all i. Thus the standard dot product is an inner products, as desired. (b) Note that f, g = f(x)g(x) dx = g(x)f(x) dx = g, f, so that, is symmetric. Let λ R, h C ([0, 1], R). Then (f(x) + λh(x))g(x) dx = f(x)g(x) dx + λ h(x)g(x) dx, so that, is bilinear. Lastly, suppose f(x) 0, so that there exists x 0 such that f(x 0 ) 0. Then there exists some δ such that f(x) > f(x 0 )/2 for x x 0 < δ. Then 1 0 f(x) 2 dx x0 +δ x 0 f(x 0 )/2 2 dx = f(x 0 )/2 2 δ > 0. Thus if f 0 then f, f > 0; clearly if f = 0 f, f = 0. Thus the second property of an inner product holds as well, and we are done. 1

2 2 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION (c) First note that if either v or u is 0 then the inequality is an equality and holds trivially; thus we now consider the case when the two are nonzero. Let w = u u, v / v, v v. Then we have Now consider w, v = u, v u, v = 0. 0 w, w = u, u 2 u, v 2 / v, v + u, v 2 / v, v = u, u u, v 2 / v, v. Rearranging this, we get u, u v, v u, v 2, as desired. Equality will hold exactly when w = 0, which means that u is a scalar multiple of v. Note that in the classical case, this reduces to the inequality cos 2 θ 1. (d) Suppose that u is an eigenvector of T with eigenvalue λ and v is an eigenvector of T with eigenvalue ρ. Then λ u, v = T u, v = u, T v = ρ u, v. Thus if λ ρ we must have u, v = 0. In other words, u and v are orthogonal. Problem 2: Show that (F ) = F. Conclude that it is not the case that V and V are always isomorphic. Solution: Suppose that f (F ). Then f is uniquely determined by its value on a basis of F. Taking the standard basis {e i } i=1, where e i has a 1 in the ith position and 0s elsewhere, we see that the data of f consists exactly of a scalar in F for each i I. These add and scale pointwise. Thus f corresponds exactly to a vector in F. Since F is never isomorphic to F we see that V is not always isomorphic to V. Problem 3: Suppose that F is a field containing F and V is an F -vector space. If we consider F to be an F -vector space, we can form the tensor product F V, which is naturally an F -vector space. Show that it is also an F -vector space. This is called the change of base of V. Solution: In order to show that F V is an F -vector space we need to define scalar multiplication. For a pure tensor α v we define scalar multiplication by λ F by λ(α v) = (λα) v. If this is well-defined then it is clearly associative, since multiplication in F is associative. Similarly, it is unital: if λ = 1 then it clearly returns the same vector. We also have and λ(α v + β w) = (λα) v + (λβ) w (λ + ρ)(α v) = ((λ + ρ)α) v = (λα + ρα) v = (λα) v + (ρα) v. Thus if scalar multiplication is well-defined then it satisfies the axioms of a vector space. We have defined scalar multiplication by its action on the generators of F V. To show that this is well-defined we need to check that it preserves the relations; in other words, that the scalar

3 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION 3 multiple of any element in A 0 stays in A 0. We therefore check: λ((α + β) v α v β v) = (λ(α + beta)) v (λα) v (λβ) v = ((λα) + (λβ)) v (λα) v (λβ) v A 0. λ(α (v + v ) α v α v ) = (λα) (v + v ) (λα) v (λα) v A 0. λ((aα) v a(α v)) = (λaα) v (λa)(α v) = (aλα) v (a(λ(α v))) = (aλα) v a((λα) v) A 0. λ(α (av) a(α v)) = (λα) (av) a(λ(α v)) = (λα) (av) a((λα) v) A 0. Thus multiplication by λ preserves A 0, and is thus well-defined on F V. Thus F V is an F -vector space. Problem 4: Prove the universal property for tensor products. In other words, show that for any vector spaces U, V, and W there is a bijection { } { } bilinear maps linear maps. U V W U V W Solution: Note that there exists a bilinear map ϕ : U V U V which maps (u, v) to u v. Thus we have a function { } { linear maps bilinear maps R: U V W U V W defined by mapping a linear map T : U V W to the bilinear map T ϕ. We need to show that this map is a bijection. FIrst we check that R is injective. Recall that S = {u v u U, v V } is a spanning set for U V ; thus any linear map T : U V W is uniquely defined by its values on S. if T, T : U V W satisfy T ϕ = T ϕ; then by definition T S = T S ; thus T = T. Thus R is injective. Now we need to check that R is surjective. Let T : U V W be a bilinear map. We need to show that there exists T : U V W such that T ϕ = T. We define T (u v) = T (u, v). If this is well-defined then it satisfies the desired condition; thus we need to check that it is zero on A 0. We check this on the generators of A 0 : T ((u + u ) v u v u v ) = T (u + u, v) T (u, v) T (u, v) = T (u + u, v) T (u + u, v) = 0 T (u (v + v ) u v u v ) = T (u, v + v ) T (u, v) T (u, v ) = T (u, v + v ) T (u, v + v ) = 0 T ((λu) v λ(u v)) = T (λu, v) λt (u, v) = λt (u, v) λt (u, v) = 0 T (u (λv) λ(u v)) = T (u, λv) λt (u, v) = λt (u, v) λt (u, v) = 0. Thus it is well-defined and R is surjective. Problem 5: Let {u i } m i=1 and {v j} n j=1 be bases of U and V, respectively. Show that a general element w = m n i=1 j=1 w iju i v j is the sum of r pure tensors if and only if the m n matrix (w ij ) has rank at most r. }

4 4 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION Solution: Note that if we replace the basis {u i } m i=1 with the basis (u 1,..., λu i + u j, u i+1,..., u m ) then the relationship between the matrix (w ij ) and the matrix for w written in terms of the new basis is the same, except that column j turns into C j λ 1 C i. Thus we can do column reductions on the matrix (w ij ) by replacing the basis {u i } m i=1. Similarly, we can do row reductions by replacing the basis {v j } n j=1. A matrix has rank r if and only if it can be reduced to a matrix with r 1 s, with no two 1 s in any row or column, using row and column reductions. If the matrix (w ij ) has only r entries then we can rewrite w as a sum of r pure tensors. Thus if the matrix has rank r then by doing changes of basis corresponding to row and column operations we can write w as a sum of r pure tensors. Note that if w = (a 1 u a m u m ) (b 1 v b n v n ) then w ij = a i b j, so that (w ij ) has rank 1, and every row is a scalar multiple of another row. Thus if w = x 1 y x r y r each row in (w ij ) is in the span of the vectors x 1,..., x r, so the row space of (w ij ) has dimension at most r. Thus the rank of the matrix (w ij ) is at most r, as desired. Problem 6: This next problem will investigate tensor products of modules. Note that the definition of linear and bilinear still work for rings instead of fields. Suppose that R is a commutative ring with unit, and M and N are R-modules. We define the tensor product M R N to be the R-module such that there is a bijection { } { } bilinear maps linear maps M N S M R N S for any R-module S. (a) Suppose that M = R m and N = R n. Show that M R N = R mn. (b) Suppose that R S is a homomorphism of rings. Show that S is an R-module. Show that S R M is an S-module. (c) Now suppose that R = Z. As we discussed before, Z-modules are just abelian groups. What is Z Z Z/nZ? What is Z/pZ Z Z/qZ for not necessarily distinct primes p and q? Find a general description of Z/mZ Z Z/nZ. Solution: Note that the construction of the tensor product that we did for vector spaces works equally well in the case of R-modules, since we did not use any properties of fields in the construction. Before we begin the problem we will show that for any R-modules M, M and N, (M M ) N = (M N) (M N). We define ϕ : (M M ) N (M N) (M N) by mappint (m, m ) n to (m n, m n). To check that this is a well-defined morphism of R-modules we need to check that it is 0 on the generators of A 0. We check: ϕ(((m, m ) + (m 0, m 0)) n (m, m ) n (m 0, m 0) n) = ((m + m 0 ) n, (m + m 0) n) (m n, m n) (m 0 n, m 0 n) ϕ((m, m ) (n + n ) = ((m + m 0 ) n m n m 0 n, (m + m 0) n m n m 0 n) = (0, 0) (M N) (M N). (m, m ) n (m, m ) n ) = (m (n + n ), m (n + n )) (m n, m n) (m n, m n ) = (m (n + n ) m n m n, m (n + n ) m n m n ) = (0, 0) (M N) (M N).

5 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION 5 The other two relations follow analogously. We define φ : (M N) (M N) (M M ) N by mapping (m n, m n ) to (m, 0) n + (0, m ) n. We check that this is well-defined: φ((m + m 0 ) n m n m 0 n, (m + m 0) n m n m 0 n ) = (m + m 0, 0) n (m, 0) n (m 0, 0) n +(m + m 0, 0) n (m, 0) n (m 0, 0) n = 0. The other relations follow analogously. Thus these are well-defined maps. We have and φ(ϕ((m, m ) n)) = φ(m n, m n) = (m, 0) n + (0, m ) n = (m, m ) n ϕ(φ(m n, m n )) = ϕ((m, 0) n + (0, m ) n ) = (m n, 0 n) + (0 n, m n ) = (m n + 0 n, 0 n + m n ) = (m n, m n ). Thus these are mutually inverse isomorphisms. We make one extra observation before starting. For any m M, n N, Indeed, 0 n = m 0 = 0 M R N. 0 n = (0 + 0) n = 0 n + 0 n, and subtracting 0 n from each side gives us the desired equality. The other one follows analogously. We are now ready to solve the problem. (a) Note that R n is the n-fold product of R with itself. Thus applying the above we know that R n R R m = (R R R m ) n = R mn. The last step follows because R R R m = R m. Indeed, we note that for any pure tensor r v R R R m we have r v = (r 1) v = r(1 v) = 1 rv. So we have a map R R R m R m which takes r v to rv. We also have an inverse map which takes v R m to 1 v. (b) The proof we wrote in problem 3 works verbatim here. (c) We define mutually inverse isomorphisms Z Z Z/n Z/n and Z/n Z Z Z/n. The first takes a [b] to [ab], and the second takes [b] to 1 [b]. Thus Z Z Z/n = Z/n. Now suppose that m and n are relatively prime numbers. Let a b Z/m Z Z/n. Since m and n are relatively prime, n has a multiplicative inverse n in Z/m. Then we have a b = (1a) b = (nn a) b = n(n a b) = (n a) (nb). But n = 0 in Z/n, so nb = 0. Thus a b = (n a) 0 = 0, so all pure tensors are equal to 0. Therefore Z/m Z/n = 0 if m and n are relatively prime. On the other hand, if m n then Z/m Z Z/n = Z/m. Indeed, note that a b = (a 1) b = a(1 b) = a(1 (b 1)) = ab(1 1). Thus Z/m Z Z/n is generated by the multiples of 1 1, and is thus isomorphic to Z/k for some k. Note that m(1 1) = (m 1) 1 = 0 1 = 0, so k m. On the other hand, we have a surjective homomorphism Z/m Z Z/n Z/m given by a b ab. Thus m k, and we see that k = m, as desired.

6 6 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION Write m = p k 1 1 pk l l and n = p m 1 1 p m l l, where some of the k i or m j can be 0. We can then write Z/mZ = Z/p k 1 1 Z Z/pk l l Z and Z/nZ = Z/p m 1 1 Z/p m l l. Then Z/mZ Z Z/nZ l r = Z/p k i i Z Z/pm j j Z. i=1 j=1 If i j then that summand is the trivial group. On the other hand, if i = j then the tensor product is just the smaller power of p i. Thus we have Z/mZ Z Z/nZ l = Z/p min(k i,m i ) i Z = Z/ gcd(m, n)z. i=1