MATH 110 LINEAR ALGEBRA HOMEWORK 1 CHU WEE LIM 1 Let us suppose x y z F such that x z y z There exists an additive inverse of z i e we can nd z F such that z z z z 0F Then x z y z x z z y z z x z z y z z x 0F y 0F x y 2 Since Z2 has only 2 elements to verify the axioms we can check all possible cases E g to check distributivity 0 0 0 0 0 0 0 since they are both 0 0 0 1 0 0 0 1 since they are both 0 0 1 0 0 1 0 0 since they are both 0 0 1 1 0 1 0 1 since they are both 0 1 0 0 1 0 1 0 since they are both 0 1 0 1 1 0 1 1 since they are both 1 1 1 0 1 1 1 0 since they are both 1 1 1 1 1 1 1 1 since they are both 0 3 To prove that Zp forms a eld we rst have to make sense of what Zp actually is and what the addition product operations are Consider the set of integers Z For i 0 1 p 1 let Ai be the subset of Z consisting of all j such that j i mod p i e p j i For example A0 comprises of all multiples of p while A1 1 p 1 2p 1 Then the disjoint union of A0 Ap 1 is Z In other words if i j then Ai Aj while A0 A1 Ap 1 Z Now Zp is the set A0 A1 Ap 1 that s right it s a set of sets To add Ai and Aj we pick any elements a Ai and b Aj Now Ai Aj is simply the unique Ak which contains the integer a b Likewise Ai Aj is the unique Al which contains the integer a b There is a slight caveat here what if we pick a di erent a Ai and b Aj It turns out that since a b ab a b b b a a we still have ab a b mod p Date September 8 1 2 CHU WEE LIM To make sense of the above abstract de nition let us take p 7 Then we have A0 A1 A6 A0 0 7 14 A1 1 8 15 etc Suppose we want to compute A3 A4 Let us pick elements from A3 and A4 say 10 A3 and 18 A4 Then 18 10 180 A5 so we have A3 A4 A5 Now that we re done with the de nition of Zp associativity and commutativity becomes clear These follow immediately from the fact that addition and multiplication on integers are associative and commutative For example to show that Ai Aj Aj Ai let us pick a Ai and b Aj Then Ai Aj resp Aj Ai is the unique Ak which contains a b resp b a Since a and b are integers we have a b b a The tricky part is to verify that A1 Ap 1 have inverses Let a Ai where i 0 There are two ways we can proceed We can use the fact that if p q are coprime integers then there exist integers c d such that pc qd 1 Hence since a is not a multiple of p and p is prime a and p must be relatively prime Thus we can nd c d Z such that ac pd 1 This means ac 1 mod p and so the unique Ak which contains c must be the multiplicative inverse of Ai Or if we re forced to use the hint provided in the problem consider a 2a p 1 a Since p is prime and a is not a multiple of p none of the above numbers is a multiple of p So each of them must belong to some Ak Now if none of them belongs to A1 then we re left with A2 A3 Ap 1 p 2 sets By pigeonhole principle two of the numbers say ma and na must belong to the same set whence ma na m n a is a multiple of p which contradicts the fact that m n are distinct elements of 1 2 p 1 Hence one of the na s must belong to A1 which gives na 1 mod p 1 2 Vector Spaces Problem 1 a b c d e f g h i True This is one of the axioms False Corollary 1 to Theorem 1 1 False E g a b 1 and x 0V is the zero vector False E g a 0 and x y can be any two distinct vectors True We may regard a vector as a column vector False It should have m rows and n columns False E g x2 and x 3 can be added to give x2 x 3 False E g x2 x and x2 of degree 2 can be added to give x of degree 1 True The leading coe cient of xn is still nonzero after mutiplying with a nonzero scalar j True since c 0 can be written as c x0 and x0 1 k True That s the de nition of F S F on page 9 example 3 MATH 110 HOMEWORK 1 3 Problem 7 The function f F S R takes 0 1 1 3 while the function g takes 0 1 1 3 Hence f g takes 0 1 1 2 and 1 3 3 6 Since h takes 0 2 and 1 6 as well we see that f g h Problem 9 To prove corollary 1 let s suppose 0 and 0 are both additive identities of V i e 0 x 0 x x for all x V Now if we let x 0 we get 0 0 0 And if we let x 0 we get 0 0 0 0 0 Hence this shows that 0 0 For corollary 2 suppose y and y are both additive inverses of x i e x y x y 0 Then by cancellation law y y For theorem 1 2c we have a0 a0 a 0 0 a0 a0 0 By cancellation law a0 0 Problem 12 An easy shortcut is to cheat and apply the results in 1 3 here The set W of odd functions is a subset of V F R R Let us prove that W is in fact a subspace of V First the 0 function is clearly odd since it takes t to 0 0 Now to show that W is closed under addition and scalar multiplication let f g W and c R be a scalar f g t f t g t f t g t f t g t f g t c f t c f t c f t c f t c f t Hence f g and c f are odd and hence in W …