Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Homework 4 1.9.1. Claim: For any x, y, and n ∈ N, n (x + y) = n X n k=0 k xk y n−k . Proof: We will prove the above equation by induction on n. Suppose first that n = 1. Then (x + y)1 = 10 x + 11 y, so the equation holds. Now suppose the equation holds for n = l. For n = l + 1, we know that (x + y)l+1 = (x + y)(x + y)l . So by assumption, l+1 (x + y) l X l k l−k = (x + y) x y . k k=0 Using the distributive property, we have (x + y) l+1 l l X l k+1 l−k X l k l−k+1 x y . x y + = k k k=0 k=0 If we now group together the monomials of the same bidegree, we have l+1 (x + y) =x l+1 +y l+1 l X l l ( + )xk y l+1−k . + k k−1 k=1 Then by Lemma 1.9.3, this is l+1 (x + y) = l+1 X l+1 k k=0 xk y l+1−k . 1.10.2. Claim: The set C4 = {1, i, −1, −i} is a group under complex multiplication. Proof: First, we check that the set C4 is closed under multiplication. We must check that for every a, b, ∈ C4 , ab ∈ C4 . We see 12 = (−1)2 = 1 ∈ C4 and i2 = (−i)2 = −1 ∈ C4 . Also, 1b = b and −1b = −b ∈ C4 for each b ∈ C4 . Finally, i(−i) = 1 ∈ C4 . Hence C4 is closed under complex multiplication. Complex multiplication is always associative, so we must check for an identity element and inverses to prove that C4 is a group. 1 ∈ C4 is the multiplicative identity, since 1a = a 1 for all a ∈ C4 . The inverse of −1 is −1 since (−1)2 = 1. The inverse of i is −i, and the inverse of −i is i, since i(−i) = 1. This shows that C4 is a group under complex multiplication. 1.10.4. Claim: The group Z4 is isomorphic to C4 . Proof: Consider the bijection of sets [0] ↔ 1, [1] ↔ i, [2] ↔ −1, [3] ↔ −i. We check the multiplication tables for C4 and Z4 and see that this bijection is an isomorphism of the groups C4 and Z4 . Z4 [0] [1] [2] [3] C4 1 i −1 −i [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] 1 1 i −1 −i i i −1 −i 1 −1 −1 −i 1 i [3] [3] [0] [1] [2] −i −i 1 i −1 1.10.9. Claim: The set of affine transformations of Rn is a group under composition of maps. Proof: Consider the set of affine maps T (~x) = S(~x) + ~b, where T is an invertible linear transformation from Rn to itself, and ~b is a vector. First, we show this set is closed under composition. If T1 = S1 +~b1 and T2 = S2 +~b2 , then T1 ◦T2 (~x) = T1 (S2 (~x)+ b~2 ) = S1 (S2 (~x)+ ~b2 ) +~b1 . Using the properties of linear transformations, this is S1 S2 (~x) + (S1~b2 +~b1 ). Since S1 and S2 are invertible, S1 S2 is invertible, and T1 ◦ T2 is an affine linear transformation. Function composition is always associative, so we must check that there is an identity and inverse functions for each affine map. First, if we take S = 1n the n × n identity matrix, and ~b = 0, then we claim this is the identity under composition. To check, for any other affine transformation T , 1n T (~x) = 1n S(~x) + 1n~b = S(~x) = T (~x), and T 1n (~x) = T (1n ~x) + ~b = T (~x). Hence 1n is the identity affine transformation. Now we show that T (~x) = S(~x) + ~b has an inverse transformation. Let T −1 (~x) = S −1 (~x) − S −1~b. Then T ◦ T −1 (~x) = T (S −1 (~x) − S −1~b) = S(S −1 (~x) − S −1~b) +~b = ~x. This is the same as 1n (~x). Now we check T −1 ◦ T (~x) = T −1 (S(~x) +~b) = S −1 (S(~x) +~b) − S −1~b = ~x. This is again the same as 1n (~x). Therefore T ◦ T −1 = 1n and T −1 ◦ T = 1n , so T has 2 an inverse under composition. This completes the proof that affine transformations are a group under function composition. 1.11.2. Claim: Let K be any field. Then K[x] is a commutative ring under polynomial addition and multiplication. Further, 1 is the multiplicative identity and the only units are constant polynomials. Proof: First, we show that K[x] is a group under addition. Let f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0 and g(x) = bm xm + am−1 xm−1 + · · · + b1 x + b0 be polynomials in K[x]. Say n ≥ m. Then f (x) + g(x) = an xn + · · · am+1 xm+1 + (am + bm )xm + · · · + (a1 + b1 )x + (a0 + b0 ). This is another polynomial in K[x], since K is closed under addition. Further, polynomial addition is associative since addition in K is associative. The identity polynomial is 0, and 0 + f (x) = f (x) = f (x) + 0. Each polynomial f (x) has an additive inverse −f (x) = −an xn + · · · + (−a1 )x + a0 . Now that we see K[x] is a group under addition, we check that multiplication is associative in K[x]. But this follows from the facts that multiplication and addition in K are associative. So now we check the distributive property. Let h(x) = cl xl + · · · c1 x + c0 . Say n ≥ m ≥ l. Then f (x)(g(x) + h(x)) = f (x) = m X ! (bi + ci )xi i=0 n i XX aj (bi−j + ci−j )xi i=0 j=0 = n X i X aj bi−j xi + aj ci−j xi i=0 j=0 = f (x)g(x) + f (x)h(x). (Note that in this formula we pretend n = m = l, or you can take bi = 0 and cj = 0 for i > m or j > l.) This shows that K[x] is a ring. It is commutative since multiplication in K is commutative. The element 1 ∈ K is the identity in K[x] as well. Suppose now that f (x) ∈ K[x] is a unit, and g(x) is its inverse. Then g(x)f (x) = 1. This means that the degree of g(x) added to the degree of f (x) is 0. Hence f (x) and g(x) are both constant polynomials in K. 3