MATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)

Transcription

1 MATH 205 HOMEWORK #2 OFFICIAL SOLUTION Problem 2: Do problems 7-9 o page 40 of Hoffma & Kuze. (7) We will prove this by cotradictio. Suppose that W 1 is ot cotaied i W 2 ad W 2 is ot cotaied i W 1. The there exist w 1 W 1 \W 2 ad w 2 W 2 \W 1. Cosider the vector w 1 + w 2 ; this is i W 1 W 2 because it is a vector space. However, w 1 + w 2 is ot i W 1, because if it were the w 2 = (w 1 + w 2 ) w 1 W 1, a cotradictio. Aalogously, w 1 + w 2 is ot i W 2. Thus w 1 + w 2 is ot i W 1 W 2, a cotradictio. Thus this caot happe, ad we must have either W 1 W 2 or W 2 W 1. (8) (a) Suppose that f, g V e. The ad (f + g)(x) = f(x) + g(x) = f( x) + g( x) = (f + g)( x) (cf)(x) = cf(x) = cf( x) = (cf)( x). So V e is a subspace of V. Aalogously, if f, g V o the ad (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x) (cf)(x) = cf(x) = cf( x) = (cf)( x). Thus V o is a subspace of V. (b) Let f be ay fuctio. The we defie f o (x) = f(x) f( x) f e (x) = f(x) + f( x). The f = (f o + f e )/2, so f V o + V e. Sice f was arbitrary, V = V o + V e. (c) Suppose that f V e V o. The f(x) = f( x) ad f(x) = f( x), so f( x) = f( x). Therefore 2f( x) = 0, ad (sice the characteristic of F is ot 2 i Hoffma & Kuze), f( x) = 0. Note that if the characteristic of f is equal to 2 the this does ot work: the fuctio f(x) = 1 is both odd ad eve. (9) Sice W 1 + W 2 = V we kow that for all v V there exist w 1 W 1 ad w 2 W 2 such that w 1 + w 2 = v; thus all that remais to show is that this represetatio is uique. Suppose that it is t, so that v = w 1 + w 2 = w 1 + w 2. The 0 = (w 1 w 1 ) + (w 2 w 2 ), ad thus w 1 w 1 = w 2 w 2. w 1 w 1 W 1 ad w 2 w 2 W 2, so we see that w 1 w 1 W 1 W 2. Sice W 1 W 2 = {0}, w 1 = w 1 ; aalogously, w 2 = w 2, ad we see that the represetatio was uique. Problem 3: Give a example of a oempty subset U of R 2 such that U is a abelia group uder additio but U is ot a subspace of R 2. Let U be the subgroup of poits with iteger coordiates. The it is a group uder additio, but it is ot ivariat uder scalar multiplicatio for ay scalar i R\Z. Thus it is ot a subspace. Problem 4: Fid bases for (a) The subspace { x x x = 0} of F. (b) The subspace {p F [x] p(3) = 0} of F [x]. 1

2 2 MATH 205 HOMEWORK #2 OFFICIAL SOLUTION (a) A basis is e 1 e, e 2 e,..., e 1 e. Note that these are clearly liearly idepedet, sice each has a 1 i a differet coordiate; i additio, they are i the subspace. To see that they re a basis ote that for ay x i the subspace, x 1 (e 1 e ) + x 2 (e 2 e ) + + x 1 (e 1 e ) = x. Alterately, we observe that the subspace is ot all of F, so it must have dimesio at most 1. O the other had, it cotais 1 liearly idepedet vectors, so it must have dimesio exactly 1 ad this must be a basis. (b) A basis is {(x 3) 1}. Clearly, each of these polyomials satisfies p(3) = 0; i additio, each of them is a differet degree so they are liearly idepedet. To see that they spa all such polyomials, cosider ay polyomial p with p(3) = 0. Cosider the Taylor series of p aroud 3. Its costat term is 0, ad all other terms are multiples of (x 3) m for various m s, so it must be i the spa of these. Problem 5: Let V be vector space over F, ad let α, β, γ be three liearly idepedet vectors. For which fields F will α + β, β + γ, γ + α also be liearly idepedet? We chaim that if char F 2 the α + β, β + γ, γ + α are liearly idepedet. Ideed, suppose that a(β + γ) + b(α + γ) + c(α + beta) = 0 for a, b, c F. Rewritig this we get that (b + c)α + (a + c)β + (a + b)γ = 0 so b + c = a + c = a + b = 0. Thus b = c ad a = b, so a = c. Pluggig this ito the secod term we get 2a = 0, so (as char F 2) a = 0. Thus b = c = 0 as well, as desired. However, if char F = 2 this is ot the case. I this case, (α + β) + (β + γ) = α + 2β + γ = α + γ, so two of the vectors sum to the last. Thus they are ot liearly idepedet. Problem 6: Suppose that W ad W are subspaces of a fiite dimesioal vector space V such that dim W + dim W > dim V. Show that W W {0}. Let {v 1,..., v k } be a basis for W 1 ad {v k+1,..., v } be a basis for W 2. We will show that if W 1 W 2 = {0} the {v 1,..., v } is a liearly idepedet set; this will mea i tur that dim V = dim W 1 + dim W 2, cotradictig the assumptio i the problem. Suppose that the set {v 1,..., v } is ot liearly idepedet. The we ca write a 1 v a v = 0 for some a i F, ot all zero. Rewritig this, we get a 1 v a k v k = a k+1 v k+1 a v. The left-had side is i W 1 ad the right-had side is i W 2, so a 1 v a k v k W 1 W 2. Sice this is {0}, we must have a 1 = = a k = 0, as the v s are a basis; aalogously, a k+1 = = a = 0. Thus the set {v 1,..., v } is liearly idepedet, as desired. Problem 7: Costruct a fuctio f : R R such that f(a + b) = f(a) + f(b) a, b R but f is ot a liear map. Show that this is impossible if f is cotiuous. Note that the set {1, π} is liearly idepedet over R. Complete it to a Q-basis of R, ad defie f(v) = 1 for all v i the basis; complete this to a Q-liear fuctio R R. The f(π) = 1 πf(1), so this fuctio is ot R-liear. However, sice it is Q-liear it must satisfy f(a + b) = f(a) + f(b) for all a ad b.

3 MATH 205 HOMEWORK #2 OFFICIAL SOLUTION 3 Ay two cotiuous fuctios that agree o a dese subset of the domai must be equal. Suppose that f : R R is ay fuctio such that f(a + b) = f(a) + f(b) for all a ad b. The f(a) = f(( 1)a) + f(a) = f(( 2)a) + 2f(a) = = f(a). Thus ( m ) ( ) 1 f a = mf a = mf(a); dividig both sides by we get f((m/)a) = (m/)f(a). Therefore ay such fuctio must be Q-liear; i particular, for all α Q we must hae f(α) = αf(1). Thus the fuctio f ad the fuctio g(x) = xg(1) agree o Q, which is a dese subset of R. Sice both are cotiuous they agree everywhere ad f is liear. Problem 8: A module M over a rig R is a algebraic structure which satisfies all of the same axioms that a vector space over a field does; however, the scalars are assumed to be a rig ad ot a field. (a) Show that a module over Z is a abelia group, ad that a liear trasformatio is a homomorphism of abelia groups. (b) Go over the proof that every vector space has a basis ad poit out oe place where it fails if the scalars are a rig ad ot a field. (c) Give a example of a module over Z which has o basis. (a) Ay module is a abelia group by defiitio, so it remais to show that ay abelia group ca be thought of as a Z-module. Let G be a abelia group, ad for ay Z ad g G defie g = g + g. times We eed to check that this is a well-defie module actio; i particular, we eed to check that (g + g ) = g + g. We have (g + g ) = g + g + + g + g = g + + g + g + + g = g + g, as desired. (b) Oe place where the proof fails is i the assumptio that a sigleto set is liearly idepedet. Note that for istace whe M = Z/pZ we have p 1 = 0 eve though p 0 i Z; so it s possible that a ozero multiple of a vector is zero. Thus a sigleto set ca be liearly depedet. I particular, ote that whe M = Z/pZ (or i ay case where the order of every elemet is fiite) there are o oempty liearly idepedet subsets of M, so a basis caot possibly exist. Aother place where the proof fails is i the proof that if somethig is ot i the spa of the maximal liearly idepedet subset the addig a vector v ot i the spa keeps the set liearly idepedet. The couterexamples are very similar to the oes above; the key poit is that we may ot be able to divide by the coefficiet of v. (c) As discussed before Z/pZ is such a module. Problem 9: We defie F to be the vector space of sequeces (x 1, x 2,...) with all x i F such that all but fiitely may of the x i s are 0. We defie F to be the vector space of all sequeces (x 1, x 2,...) with x i V. (a) Show that F has a coutable basis.

4 4 MATH 205 HOMEWORK #2 OFFICIAL SOLUTION (b) Show that if F is coutable the F does ot have a coutable basis. (Hit: what is the cardiality of F?) (c) Coclude that if F is coutable the F ad F are ot isomorphic. (a) Let e i be the vector that has a 1 i the i-th coordiate ad zeroes elsewhere. The e i is a basis of F, sice each sequece has oly fiitely may ozero terms. This basis is coutable. (b) If F is coutable we claim that ay vector space with a coutable basis is coutable. Recall that the product of two coutable sets is coutable; thus by iductio F = F 1 F is coutable as well. Suppose that V has a coutable basis {v 1, v 2,...}. The V = 1 spa(v 1,..., v ). Each of the terms of the uio is isomorphic as vector spaces to F ad thus is coutable. A coutable uio of coutable sets is coutable, ad therefore V must be coutable as well. However, F is ot coutable. Ideed, suppose that it were; let {v 1, v 2,...} be a eumeratio of it. Cosider the vector w costructed by lettig the i-th coordiate be 1 (v i ) i, where (v i ) i is the i-th coordiate of v i. We claim that w v for all, cotradictig the assumptio that we had eumerated everythig i F. If w = v m the all coordiates must be the same; however, w m (v m ) m, so this ca t be the case. Sice m was arbitrary, w caot be i the list. Thus F is ucoutable, ad therefore ca t have a coutable basis. (c) Suppose that F ad F were isomorphic. The i particular we would have a bijectio of sets betwee them. However, this caot be, as they are differet cardialities. Therefore they caot be isomorphic. Problem 10: Let L : F F be the left-shift operator which takes (x 1, x 2,...) to (x 2, x 3,...). For ay x F, defie ρ x = (1, x, x 2, x 3,...) F. (a) Show that L is a liear trasformatio. What is its kerel? What is its image? (b) Show that the set {ρ x x F } is liearly idepedet. (Hit: For all x F, Lρ x = xρ x.) (c) Use the previous part to show that if F is ucoutable the it does t have a coutable spaig set. Coclude that F ad F are ever isomorphic. (a) Note that L(x 1 +y 1, x 2 +y 2...) = (x 2 +y 2, x 3 +y 3,...) = (x 2, x 3,...)+(y 2, y 3,...) = L(x 1, x 2,...)+L(y 1, y 2,...). Similarly, L(cx 1, cx 2,...) = (cx 2, cx 3,...) = c(x 2, x 3,...) = cl(x 1, x 2,...). Thus L is a liear trasformatio. Its kerel cosists of all sequeces such that x i = 0 for i > 1. Its image if all of F, as (x 1, x 2,...) = L(0, x 1, x 2,...). (b) Suppose that the set is ot liearly idepedet; the there exists a miimal liearly depedet subset {ρ x1,..., ρ x }. Thus there exist coefficiets a 1,..., a such that a 1 ρ x1 + + a ρ x = 0.

5 MATH 205 HOMEWORK #2 OFFICIAL SOLUTION 5 Sice was miimal, all a i 0. Apply L to this equatio. The we get a 1 x 1 ρ x1 + + a x ρ x = 0. If x 1 = 0 the all other x i 0 ad we get that {ρ x2,..., ρ x } is liearly depedet, cotradictig the miimality of. Thus x 1 0. But the we ca divide both sides by x 1 to get x 2 x a 1 ρ x1 + a 2 ρ x2 + + a ρ x = 0. x 1 x 1 Subtractig this from the origial equatio gives us a 2 (1 x 2 )ρ x1 + + a (1 x )ρ x = 0. x 1 x 1 Sice all x i s were distict, all coefficiets of this equatio are ozero, ad we agai coclude that {ρ x2,..., ρ x } is liearly depedet, cotradictig the miimality of. thus we see that such a miimal could ot exist, ad the set must be liearly idepedet. (c) The set {ρ x x F } is liearly idepedet, ad thus it ca be completed to a basis. Sice F is ucoutable the cardiality of this basis is at least ucoutable. Sice every spaig set cotais a basis, we see that F ca t have a coutable spaig set. However, F does have a coutable spaig set, so it ca t be isomorphic to F.