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Home , Bilinear form, Bilinear map, Dot product, Linear map

MATH 205 HOMEWORK #5 OFFICIAL SOLUTION

Problem 1: An inner on a V over F is a bilinear h·, ·i : V × V → F satisfying the extra conditions •h v, wi = hw, vi, and •h v, vi ≥ 0, with equality if and only if v = 0. n (a) Show that the standard on R is an inner product. R ∞ (b) Show that (f, g) 7→ 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 , hv, wi2 ≤ hv, vihw, wi. When does equality hold? What standard in trigonometry does this reflect when n V = R ? (d) We say that two vectors v, w in V are orthogonal if hv, wi = 0. Suppose that T : V → V is a linear transformation satisfying hT v, wi = hv, T wi for all v, w. Show that eigenvectors of T with different eigenvalues are orthogonal.

Solution: Note that if we can show that a is symmetric, then it is linear in the first variable if and only if it is linear in the second. Thus showing in one variable is enough.

(a) Let v = (a1, . . . , an) and w = (b1, . . . , bn). The standard dot product is symmetric, as

hv, wi = a1b1 + ··· + anbn = b1a1 + ··· + bnan = hw, vi.

Let λ ∈ R and u = (c1, . . . , cn). Then hv+λu, wi = (a1+λc1)b1+···+(an+λcn)bn = a1b1+···+anbn+λ(c1b1+···+cnbn) = hv, wi+λhu, wi. Thus h·, ·i is bilinear. In , 2 2 hv, vi = a1 + ··· + an ≥ 0,

with equality if and only if ai = 0 for all i. Thus the standard dot product is an inner products, as desired. (b) Note that Z Z hf, gi = f(x)g(x) dx = g(x)f(x) dx = hg, fi,

∞ so that h·, ·i is symmetric. Let λ ∈ R, h ∈ C ([0, 1], R). Then Z Z Z (f(x) + λh(x))g(x) dx = f(x)g(x) dx + λ h(x)g(x) dx,

so that h·, ·i is bilinear. Lastly, suppose f(x) 6= 0, so that there exists x0 such that f(x0) 6= 0. Then there exists some δ such that |f(x)| > |f(x0)/2| for |x − x0| < δ. Then

Z 1 Z x0+δ 2 2 2 f(x) dx ≥ |f(x0)/2| dx = |f(x0)/2| δ > 0. 0 x0 Thus if f 6= 0 then hf, fi > 0; clearly if f = 0 hf, fi = 0. Thus the second property of an inner product holds as well, and we are done. 1 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 − hu, vi/hv, viv. Then we have hw, vi = hu, vi − hu, vi = 0. Now consider

0 ≤ hw, wi = hu, ui − 2hu, vi2/hv, vi + hu, vi2/hv, vi = hu, ui − hu, vi2/hv, vi.

Rearranging this, we get hu, uihv, vi ≥ hu, vi2, as desired. Equality will hold exactly when w = 0, which means that u is a multiple of v. Note that in the classical case, this reduces to the inequality cos2 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

λhu, vi = hT u, vi = hu, T vi = ρhu, vi.

Thus if λ 6= ρ we must have hu, vi = 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 of ⊕∞ ∞ F . Taking the {ei}i=1, where ei 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 . 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 0 is a field containing F and V is an F -vector space. If we consider F 0 to be an F -vector space, we can form the product F 0 ⊗ V , which is naturally an F -vector space. Show that it is also an F 0-vector space. This is called the change of base of V .

Solution: In order to show that F 0⊗V is an F 0-vector space we need to define scalar . For a pure tensor α ⊗ v we define by λ ∈ F 0 by

λ(α ⊗ v) = (λα) ⊗ v.

If this is well-defined then it is clearly associative, since multiplication in F 0 is associative. Similarly, it is unital: if λ = 1 then it clearly returns the same vector. We also have

λ(α ⊗ v + β ⊗ w) = (λα) ⊗ v + (λβ) ⊗ w and (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 0 ⊗ V . To show that this is well-defined we need to check that it preserves the relations; in other words, that the scalar MATH 205 HOMEWORK #5 OFFICIAL SOLUTION 3 multiple of any element in A0 stays in A0. 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 ∈ A0. 0 0 0 0 λ(α ⊗ (v + v ) − α ⊗ v − α ⊗ v ) = (λα) ⊗ (v + v ) − (λα) ⊗ v − (λα) ⊗ v ∈ A0. λ((aα) ⊗ v − a(α ⊗ v)) = (λaα) ⊗ v − (λa)(α ⊗ v) = (aλα) ⊗ v − (a(λ(α ⊗ v)))

= (aλα) ⊗ v − a((λα) ⊗ v) ∈ A0. λ(α ⊗ (av) − a(α ⊗ v)) = (λα) ⊗ (av) − a(λ(α ⊗ v))

= (λα) ⊗ (av) − a((λα) ⊗ v) ∈ A0. 0 0 Thus multiplication by λ preserves A0, and is thus well-defined on F ⊗ V . Thus F ⊗ V is an F 0-vector space.

Problem 4: Prove the for tensor products. In other words, show that for any vector spaces U, V , and W there is a  bilinear maps   linear maps  ←→ . U × V → W U ⊗ V → W

Solution: Note that there exists a ϕ : U × V → U ⊗ V which maps (u, v) to u ⊗ v. Thus we have a  linear maps   bilinear maps  R: −→ U ⊗ V → W U × V → W defined by mapping a 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 for U ⊗ V ; thus any linear map T : U ⊗ V → W is uniquely defined by its values on S. if 0 0 0 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 0 : U ⊗ V → W such that T 0 ◦ ϕ = T . We define T 0(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 A0. We check this on the generators of A0: T 0((u + u0) ⊗ v − u ⊗ v − u0 ⊗ v0) = T (u + u0, v) − T (u, v) − T (u0, v) = T (u + u0, v) − T (u + u0, v) = 0 T 0(u ⊗ (v + v0) − u ⊗ v − u ⊗ v0) = T (u, v + v0) − T (u, v) − T (u, v0) = T (u, v + v0) − T (u, v + v0) = 0 T 0((λu) ⊗ v − λ(u ⊗ v)) = T (λu, v) − λT (u, v) = λT (u, v) − λT (u, v) = 0 T 0(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.

m n Problem 5: Let {ui}i=1 and {vj}j=1 be bases of U and V , respectively. Show that a general Pm Pn element w = i=1 j=1 wijui ⊗ vj is the sum of r pure if and only if the m × n (wij) has at most r. 4 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION

m Solution: Note that if we replace the basis {ui}i=1 with the basis

(u1, . . . , λui + uj, ui+1, . . . , um) then the relationship between the matrix (wij) and the matrix for w written in terms of the new −1 basis is the same, except that column j turns into Cj − λ Ci. Thus we can do column reductions m on the matrix (wij) by replacing the basis {ui}i=1. Similarly, we can do row reductions by replacing n the basis {vj}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 (wij) 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 = (a1u1 + ··· + amum) ⊗ (b1v1 + ··· + bnvn) then wij = aibj, so that (wij) has rank 1, and every row is a scalar multiple of another row. Thus if w = x1 ⊗ y1 + ··· + xr ⊗ yr each row in (wij) is in the span of the vectors x1, . . . , xr, so the row space of (wij) has at most r. Thus the rank of the matrix (wij) 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 for rings instead of fields. Suppose that R is a commutative with , and M and N are R-modules. We define the M ⊗R N to be the R- such that there is a bijection  bilinear maps   linear maps  ←→ M × N → S M ⊗R N → S for any R-module S. m n mn (a) Suppose that M = R and N = R . Show that M ⊗R N = R . (b) Suppose that R → S is a 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 construc- tion. Before we begin the problem we will show that for any R-modules M, M 0 and N, (M ⊕ M 0) ⊗ N =∼ (M ⊗ N) ⊕ (M 0 ⊗ N). We define ϕ :(M ⊕ M 0) ⊗ N → (M ⊗ N) ⊕ (M 0 ⊗ N) by mappint (m, m0) ⊗ n to (m ⊗ n, m0 ⊗ n). To check that this is a well-defined of R-modules we need to check that it is 0 on the generators of A0. We check: 0 0 ϕ(((m, m ) + (m0, m0)) ⊗ n 0 0 0 0 0 0 −(m, m ) ⊗ n − (m0, m0) ⊗ n) = ((m + m0) ⊗ n, (m + m0) ⊗ n) − (m ⊗ n, m ⊗ n) − (m0 ⊗ n, m0 ⊗ n) 0 0 0 0 = ((m + m0) ⊗ n − m ⊗ n − m0 ⊗ n, (m + m0) ⊗ n − m ⊗ n − m0 ⊗ n) = (0, 0) ∈ (M ⊗ N) × (M 0 ⊗ N). ϕ((m, m0) ⊗ (n + n0) −(m, m0) ⊗ n − (m, m0) ⊗ n0) = (m ⊗ (n + n0), m0 ⊗ (n + n0)) − (m ⊗ n, m0 ⊗ n) − (m ⊗ n0, m0 ⊗ n0) = (m ⊗ (n + n0) − m ⊗ n − m ⊗ n0, m0 ⊗ (n + n0) − m0 ⊗ n − m0 ⊗ n0) = (0, 0) ∈ (M ⊗ N) × (M 0 ⊗ N). MATH 205 HOMEWORK #5 OFFICIAL SOLUTION 5

The other two relations follow analogously. We define φ :(M ⊗ N) × (M 0 ⊗ N) → (M × M 0) ⊗ N by mapping (m ⊗ n, m0 ⊗ n0) to (m, 0) ⊗ n + (0, m0) ⊗ n. We check that this is well-defined:

φ((m + m0) ⊗ n − m ⊗ n − m0 ⊗ n, 0 0 0 0 0 0 0 (m + m0) ⊗ n − m ⊗ n − m0 ⊗ n ) = (m + m0, 0) ⊗ n − (m, 0) ⊗ n − (m0, 0) ⊗ n 0 0 0 0 0 0 0 +(m + m0, 0) ⊗ n − (m , 0) ⊗ n − (m0, 0) ⊗ n = 0. The other relations follow analogously. Thus these are well-defined maps. We have φ(ϕ((m, m0) ⊗ n)) = φ(m ⊗ n, m0 ⊗ n) = (m, 0) ⊗ n + (0, m0) ⊗ n = (m, m0) ⊗ n and ϕ(φ(m ⊗ n, m0 ⊗ n0)) = ϕ((m, 0) ⊗ n + (0, m0) ⊗ n0) = (m ⊗ n, 0 ⊗ n) + (0 ⊗ n, m0 ⊗ n0) = (m ⊗ n + 0 ⊗ n, 0 ⊗ n0 + m0 ⊗ n0) = (m ⊗ n, m0 ⊗ n0). Thus these are mutually inverse . We make one extra observation before starting. For any m ∈ M, n ∈ N,

0 ⊗ n = m ⊗ 0 = 0 ∈ M ⊗R N. Indeed, 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 Rn is the n-fold product of R with itself. Thus applying the above we know that n m ∼ m n ∼ mn R ⊗R R = (R ⊗R R ) = R . m ∼ m The last step follows because R ⊗R R = R . Indeed, we note that for any pure tensor m r ⊗ v ∈ R ⊗R R we have r ⊗ v = (r · 1) ⊗ v = r(1 ⊗ v) = 1 ⊗ rv. m m So we have a map R ⊗R R → R which takes r ⊗ v to rv. We also have an inverse map which takes v ∈ Rm 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 . Let a ⊗ b ∈ Z/m ⊗Z Z/n. Since 0 m and n are relatively prime, n has a n in Z/m. Then we have a ⊗ b = (1a) ⊗ b = (nn0a) ⊗ b = n(n0a ⊗ b) = (n0a) ⊗ (nb). 0 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 7→ ab. Thus m|k, and we see that k = m, as desired. 6 MATH 205 HOMEWORK #5 OFFICIAL SOLUTION

k1 k` m1 m` Write m = p1 ··· p` and n = p1 ··· p` , where some of the ki or mj can be 0. We can then write ∼ k1 k` Z/mZ = Z/p1 Z × · · · × Z/p` Z and ∼ m1 m` Z/nZ = Z/p1 × · · · × Z/p` . Then ` r ∼ Y Y ki mj Z/mZ ⊗Z Z/nZ = Z/pi Z ⊗ Z/pj Z. i=1 j=1 If i 6= j then that summand is the trivial . On the other hand, if i = j then the tensor product is just the smaller of pi. Thus we have ` ∼ Y min(ki,mi) ∼ Z/mZ ⊗Z Z/nZ = Z/pi Z = Z/ gcd(m, n)Z. i=1