MATH 110: LINEAR ALGEBRAHOMEWORK #1CHU-WEE LIM(1) Let us suppose x, y, z ∈ F , such that x + z = y + z. There exists an additive inverseof z, i.e. we can find z∈ F such that z + z= z+ z =0F.Thenx + z = y + z ⇒ (x + z)+z=(y + z)+z⇒ x +(z + z)=y +(z + z)⇒x +0F= y +0F⇒ x = y.(2) Since Z2has 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 Zpformsafield,wefirsthavetomakesenseofwhatZpactually is,and what the addition/product operations are.Consider the set of integers Z.Fori =0, 1,...,p − 1, let Aibe the subset ofZ consisting of all j such that j ≡ i (mod p) (i.e. p|(j − i)). For example, A0comprises of all multiples of p, while A1= {...,1,p +1, 2p +1,...}. Then thedisjoint union of A0,...,Ap−1is Z. In other words, if i = j,thenAi∩ Aj= ∅; whileA0∪ A1∪···∪Ap−1= Z.Now, Zpis the set {A0,A1,...,Ap−1} (that’s right, it’s a set of sets). To add Aiand Aj, we pick any elements a ∈ Aiand b ∈ Aj. Now Ai+ Ajis simply the uniqueAkwhich contains the integer a + b. Likewise, Ai·Ajis the unique Alwhich containsthe integer a ·b. There is a slight caveat here: what if we pick a different a∈ Aiandb∈ Aj? It turns out that sinceab−ab = a(b−b)+b(a−a),we still have ab ≡ ab(mod p).Date: September 8.12 CHU-WEE LIMTo make sense of the above abstract definition, let us take p =7. ThenwehaveA0,A1,...,A6:A0= {...,0, 7, 14,...},A1= {...,1, 8, 15,...},etc.Suppose we want to compute A3· A4. Let us pick elements from A3and A4,say10 ∈ A3and 18 ∈ A4.Then18· 10 = 180 ∈ A5,sowehaveA3· A4= A5.Now that we’re done with the definition of Zp, associativity and commutativitybecomes clear. These follow immediately from the fact that addition and multi-plication on integers are associative and commutative. For example, to show thatAi+ Aj= Aj+ Ai, let us pick a ∈ Aiand b ∈ Aj.ThenAi+ Aj(resp. Aj+ Ai)is the unique Akwhich contains a + b (resp. b + a). Since a and b are integers, wehave a + b = b + a.ThetrickypartistoverifythatA1,...,Ap−1have inverses. Let a ∈ Ai,wherei = 0. There are two ways we can proceed.• Wecanusethefactthatifp, q are coprime integers, then there exist integers c, dsuch that pc + qd = 1. Hence since a is not a multiple of p,andp is prime, a andp must be relatively prime. Thus, we can find c, d ∈ Z, such that ac + pd =1.This means ac ≡ 1(modp) and so the unique Akwhich contains c must be themultiplicative inverse of Ai.• Or, if we’re forced to use the hint provided in the problem, considera, 2a,...,(p − 1)a.Since p is prime and a is not a multiple of p, none of the above numbers is amultiple of p. SoeachofthemmustbelongtosomeAk.Now,ifnoneofthembelongs to A1, then we’re left with A2,A3,...,Ap−1(p − 2 sets). By pigeonholeprinciple, two of the numbers (say ma and na)mustbelongtothesameset;whence ma − na =(m − n)a is a multiple of p which contradicts the fact thatm, n are distinct elements of {1, 2,...,p−1}. Hence, one of the na’s must belongto A1,whichgivesna ≡ 1(modp).§1.2: Vector SpacesProblem 1.(a) True. This is one of the axioms.(b) False. Corollary 1 to Theorem 1.1.(c) False. E.g. a = b =1,andx =0Vis the zero vector.(d) False. E.g. a =0,andx, y can be any two distinct vectors.(e) True. We may regard a vector as a column vector.(f) False. It should have m rows and n columns.(g) False. E.g. x2and x +3canbeaddedtogivex2+ x +3.(h) False. E.g. x2+ x and −x2(of degree 2) can be added to give x (of degree 1).(i) True. The leading coefficient (of xn) is still nonzero, after mutiplying with a nonzeroscalar.(j) True, since c = 0 can be written as c · x0and x0=1.(k) True. That’s the definition of F(S, F ) on page 9, example 3.MATH 110: HOMEWORK #1 3Problem 7. The function f ∈F(S, R)takes0→ 1, 1 → 3; while the function g takes0 → 1, 1 → 3. Hence f + g takes 0 → 1+1=2 and 1→ 3 + 3 = 6. Since h takes 0 → 2and 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,ifweletx =0,weget0+0=0. And if we letx =0,weget0=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 Wof odd functions is a subset of V = F(R, R). Let us prove that W is in fact a subspace ofV . First, the 0 function is clearly odd since it takes −t to 0 = −0. Now to show that W isclosed 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 as well.Problem 15. No, it is not a vector space over R. For instance, if (0, 1) ∈ V and i =√−1 ∈C,theni(0, 1) = (0,i) ∈ V .Problem 16. Yes, it is a vector space over Q, because multiplying a real number by arational number gives us a real number. Hence, multiplying an element of V by a rationalnumber gives us an element of V . The rest of the axioms are easy to verify.Problem 19. No, distributivity fails. E.g. we have 1(1, 1) = (1, 1) but 2(1, 1) = (2,12).Hence, this gives 1(1, 1) + 1(1, 1) = (1 + 1)(1, 1).§1.3: SubspacesProblem 1.(a) False. This is rather pedantic though. What happens if W is a subset of V that is avector space, under some other operations?(b) False. That’s why we need the axiom 0 ∈ W .(c) True. We can always take the zero subspace {0}  V .(d) False. It would be true for any two subspaces though.(e) True, since there are only n diagonal entries and all other entries are 0.(f) False. It’s the sum.(g) False. It’s not equal to R2per se, although it certainly is isomorphic (as