Exercise 2.A.11 Proof. - Stanford University

1 Math 113 Homework 2 SolutionsSolutions by Guanyang Wang, with edits by Tom from the , ..,vmis linearly independent inVandw thatv1, .., vm, wis linearly independent if and only ifw / span(v1, .., vm) supposev1, .., vm, wis linearly independent. Then ifw span(v1, .., vm),we can writewas the linear combination ofv1, .., vm, that isw=a1v1+..+ both sides of the equation by w, we havea1v1+..+amvm+ ( w) = 0 Therefore we can write 0 asa1v1+..+amvm+( w), so there existsa1, a2, .., am, 1,not all 0, such thata1v1+..+amvm+ ( w) = 0. by the definition of linear de-pendence, we havev1, ..vm, wis linearly dependent, which contradicts our initialassumption. Thus we havew / span(v1, .., vm).Conversely, supposew / span(v1, .., vm). Ifv1, .., vm, wis linearly dependent,then by the linear dependence lemma(Lemma ), we havevj span(v1.)

2 , vj 1)for somejorw span(v1, .., vm). But sincev1, .., vmis linearly independent,there is noj {1, .., m}such thatvj span(v1, .., vj 1). Meanwhile we havew / span(v1, .., vm) by our assumption. Thereforev1, .., vm, wis linearly independent. Exercise or disprove: there exists a basisp0, p1, p2, p3ofP3(F)such that none of the polynomialsp0, p1, p2, p3has degree will show thatp0= 1p1=xp2=x3+x2p3=x3is a basis forP3(F). Note that none of these polynomials has degree in the book states that ifVis a finite dimensional vector space,and we have a spanning list of vectors of length dimV, then that list is a basis. Itis shown in the book thatP3(F) has dimension 4. Since this list has 4 vectors, weonly need to show that it spansP3(F).Supposep(x) =a0+a1x+a2x2+a3x3 P3(F).

3 We need to findb0, .. , (x) =b0p0+ +b3p3. Note thatp2 p3=x2. So letb0=a0, b1=a1,b2=a2andb3=a3 a2. Then,b0p0+b1p1+b2p2+b3p3=a0+a1x+a2(x2+x3 ) + (a3 a2)x3=a0+a1x+a2x2+a2x3+a3x3 a2x3=a0+a1x+a2x2+a3x3=p(x)12So we can can writep(x) as a linear combination ofp0, p1, p2andp3. Thusp0, p1, p2andp3spanP3(F). Thus, they form a basis forP3(F). Therefore, there exists abasis ofP3(F) with no polynomial of degree 2. Exercise or give a counterexample: Ifv1, v2, v3, v4is a basis ofVandUis a subspace ofVsuch thatv1, v2 Uandv3/ Uandv4/ U, thenv1, v2is a basis of statement above is false. TakeV=R4, letv1= (1,0,0,0),v2=(0,1,0,0),v3= (0,0,1,0),v4= (0,0,0,1), it is the standard basis ofR4(seeexample (a)). LetU={(a, b, c, c) :a, b, c R}We havev1 U, v2 U, v3/ U, v4/ U. Now we will provev1, v2does notspan U.

4 For anyw span(v1, v2),w=a1v1+a2v2=a1(1,0,0,0) +a2(0,1,0,0) =(a1, a2,0,0). Letu= (0,0,1,1), we haveu Ubutu / span(v1, v2).By definition of basis, we havev1, v2is not a basis of U. Exercise thatVis finite dimensional andUis a subspace ofVsuch that dimU= dimV. Prove thatU= dimU= dimV=n. Then we can find a basisu1, .. , , .. , unis a basis ofU, it is a linearly independent set. Proposition that ifVis finite dimensional, then every linearly independent list of vectorsinVof length dimVis a basis forV. The listu1, .. , unis a list ofnlinearlyindependent vectors inV(because it forms a basis forU, and becauseU V.)Since dimV=n,u1, .. , unis a basis means thatu1, .. , unspansV. Thus, we can express anyv Vas a linearcombination ofu1, .. , un. But eachuiis an element ofU.

5 SinceUis a vectorspace, any linear combination of elements ofUis also inU. Thus anyv Vis alsoan element ofU. ThereforeV haveU VsinceUis a subspace ofV, and we have just shown thatV ,U=V. Exercise (a) LetU={p P4(F) :p(2) =p(5) =p(6)}. Find a basis ofU.(b) Extend the basis in part (a) to a basis ofP4(F).(c) Find a subspace W ofP4(F) such thatP4(F) =U (a) A basis ofUis1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6)Each polynomial in the list above is inU. To verify that the list above is indeed abasis ofU, first note that the list above is linearly independent. Supposea, b, c Randa+b(x 2)(x 5)(x 6) +c(x 2)2(x 5)(x 6) = 0for everyx R. Without explicitly expanding the left side of the equation above,we can see that the left side has acx4term. Because the right side has nox4term,this implies thatc= 0.

6 Becausec= 0, we see that the left side has abx3term,which implies thatb= 0. Becauseb=c= 0, the equation becomesa= the equation above impliesa=b=c= 0. Hence the list 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6) is linearly independent inU. Now we are going toprove dimU= 3, then Proposition implies that 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6) is a basis ofU. Since we already know 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6) is linearly independent inU, we have dimU 3, thus we justneed to prove dimU {p P4(F) :p(2) =p(5)}. We know thatVis a proper subspace ofP4(F), since (x) =xis a polynomial inP4(F) that is not inV(sincef(2) = 2whilef(5) = 5). We already know dim(P4(F)) = 5 from Example Usingthe result in Exercise , we know that dimV <dimP4(F) sinceVis a propersubspace ofP4(F), so dimV 4.

7 Similarly, we knowUis a proper subspace ofV, because (x) = (x 2)(x 5) is a polynomial that is inVbut not inU(sinceq(5) = 0 whileq(6) = 4). Applying Exercise again, we conclude thatdimU <dimV, so dimU conclude that dimU= 3. By Prop. , we can conclude that 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6) is a basis ofU.(b)The list1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2is a basis ofP4(F).First we prove that 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2is , b, c, d, e Randa+b(x 2)(x 5)(x 6) +c(x 2)2(x 5)(x 6) +dx+ex2= 0 Without explicitly expanding the left side of the equation above, we can see thatthe left side has acx4term. Because the right side has nox4term, this impliesthatc= 0. Becausec= 0, we see that the left side has abx3term, which impliesthatb= 0. Becauseb=c= 0, the left side has aex2term which implies thatb= 0.

8 Becauseb=c=e= 0, the left side has adxterm which implies thatd= 0, the equation above becomesa= the equation above impliesa=b=c=d=e= 0. Hence the list1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2is linearly independent inP4(F).Notice that this linearly independent list has length 5, meanwhile dimP4(F) = 5(see Example ). Using Proposition , we can conclude that 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2is a basis ofP4(F).(c) Denote the subspacespan(x, x2) byW. Since1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2forms a basis ofP4(F), we know thatx, x2is linearly independent, thusx, x2isa basis ofW(see Definition ) and we have dimW= 2 (see Definition ).From (a) we know that 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6) is a basis ofU, and dimU= 3. Now we want to proveP4(F) =U we prove thatUandWis a direct sum.

9 Supposef U W, then we canwritefasf=a1+a2(x 2)(x 5)(x 6) +a3(x 2)2(x 5)(x 6)( sincef U)andf=a4x+a5x2( sincef W)4 Combining the two equalities together we havea1+a2(x 2)(x 5)(x 6) +a3(x 2)2(x 5)(x 6) =a4x+a5x2 Adding both sides of the equality by (a4x+a5x2) and using the property ofadditive inverse, we havea1+a2(x 2)(x 5)(x 6) +a3(x 2)2(x 5)(x 6) + ( (a4x+a5x2))=a4x+a5x2+ ( (a4x+a5x2))=0So we havea1+a2(x 2)(x 5)(x 6) +a3(x 2)2(x 5)(x 6) + ( a4)x+ ( a5)x2= 0 Since 1,(x 2)(x 5)(x 6),(x 2)2(x 5)(x 6), x, x2is linearly independent,using Definition we have:a1=a2=a3= a4= a5= 0 Which is equivalent toa1=a2=a3=a4=a5= 0So we havef= 0x2+ 0x= 0, thusUandWis a direct sum (see Proposition ).Then we proveU W=P4(F). Using Theorem , we havedim(U+W) = dimU+ dimW dim(U W)Consider the right side of the equation.

10 Since dimU= 3, dimW= 2, dim(U W) = 0. We have dim(U+W) = 5. The vector spaceU+Wis a subspace ofP4(F), so using the result in Exercise , we haveU+W=P4(F). We haveproved thatUandWis a direct sum, therefore we haveU W=P4(F). Exercise thatUandWare subspaces ofR8such that dimU=3, dimW= 5, andU+W=R8. Prove thatR8=U know from Theorem thatdim(U+W) = dimU+ dimW dim(U W)First consider the left hand side of the equation. Here we haveU+W=R8, sodim(U+W) = dim(R8) = consider the right hand side of the equation. Since dimU= 3, dimW= 5, the right hand of the equation equals to 8 dim(U W).Therefore we have 8 dim(U W) = 8, so dim(U W) = 0, which impliesU W={0}, so we haveR8=U W(see Proposition from our textbook). Exercise both five-dimensional subspaces thatU W6={0}.