Transcription of HOMEWORK 2 { solutions
1 Math 4377/6308 Advanced Linear Algebra IDr. Vaughn Climenhaga, PGH 651 AFall 2013 HOMEWORK 2 solutionsDue4pm Wednesday, September 4. You will be graded not onlyon the correctness of your answers but also on the clarity and com-pleteness of your communication. Write in complete whether or not each of the following is a subspace your answer.(a)X1={(x,y) R2|x+y= 0}Solution.[4 points]Yes,X1is a subspace. Given any (x,y),(x ,y ) X1andc R, we must check that (cx+x ,cy+y ) X1. Indeed,(cx+x ) + (cy+y ) =c(x+y) + (x +y ) =c 0 + 0 = 0.(b)X2={(x,y) R2|x 1 = 0}Solution.[4 points]No,X2is not a subspace. It does notcontain (0,0). (It also fails to be closed under addition or scalarmultiplication.)(c)X3={(x,y) R2|xy= 0}Solution.[4 points]No,X3is not a subspace.
2 It is not closedunder addition: (1,0) X3and (0,1) X3, but their sum (1,1)is not inX3.(d)X4={(1,0),(0,1)}Solution.[4 points]No,X4is not a subspace. It does not containthe zero vector. (It also fails to be closed under addition or scalarmultiplication.)(e)X5= span{(1,0),(0,1)}Math 4377/6308 HOMEWORK 2 solutionsPage 2 of 4 Solution.[4 points]Yes,X5is a subspace; the span of any setof vectors is always a that ifXandYare subspaces ofV, then so areX YandX+ [10 points]Given anyv1,v2 X Yand anyc K,we havev1,v2 Xandv1,v2 Y(by the definition of intersection).Thus the subspace property ofXandYimplies thatcv1+v2 Xandcv1+v2 Y, and in particularcv1+v2 X Y. ThusX Ysatisfies the subspace property, and by Proposition in the notes, itis a +Y, we observe that given anyv1,v2 X+Yandc K,there existx1,x2 Xandy1,y2 Ysuch thatv1=x1+y1andv2=x2+y2.
3 (This is from the definition of the sum of two subspaces.)Nowcv1+v2=c(x1+y1) + (x2+y2) = (cx1+x2) + (cy1+y2), and sinceXandYare subspaces, we deduce thatcx1+x2 Xandcy1+y2 Y,so their sumcv1+v2is inX+Y. ThusX+Yis a subspace. (It isnon-empty because bothXandYare.) that if0 L V, thenLis linearly [5 points]The set{0}is linearly dependent because 1 0=0. ThusLcontains a linearly dependent set, hence by the previousexercise it is linearly which of the following subsets ofR3are linearly indepen-dent.(a)S1={(1,1,0),(3,0,0)}Math 4377/6308 HOMEWORK 2 solutionsPage 3 of 4 Solution.[5 points]Ifa(1,1,0) +b(3,0,0) =0, thena+ 3b= 0(from the first coordinate) anda= 0 (from the second), so weconclude thata=b= 0, henceS1is linearly independent.(b)S2= span{(1,1,0),(3,0,0)}Solution.
4 [5 points]No,S2is not linearly independent. It con-tains the zero vector, so by the previous exercise it is linearlydependent.(c)S3={(2,0,1),(1,1,0) ,(0,0,1)}Solution.[5 points]Ifa(2,0,1) +b(1,1,0) +c(0,0,1) =0, then(abc)is in the null space of(2 1 00 1 01 0 1). Row reduction shows that thenull space is trivial, soS3is linearly a vector space, letS Vbe a spanning set, and letL Vbe linearly independent.(a)Show that ifS S V, thenS is [5 points]Given anyv V, there arex1,..,xn Sandc1,..cn Ksuch that nj=1cjxj=v. Becausexj S aswell, this demonstrates thatv span(S ), and soS is spanning.(b)Show that ifL L, thenL is linearly [5 points]Ifx1,..,xn L andc1,..,cn Karesuch that nj=1cjxj=0, then we also havex1,..,xn L, andso linear independence ofLimplies thatc1= =cn=0.
5 Thisshows thatL is linearly 4377/6308 HOMEWORK 2 solutionsPage 4 of that the set{sinx,cosx,sin(2x)} C1(R) is linearly : ifa,b,c Rare such thatasinx+bcosx+csin(2x) = 0foreveryx R, then in particular, the equation is true forx= 0, /4, that this implies thata=b=c= [10 points]Supposea,b,c Rare such thatasinx+bcosx+csin(2x) = 0 for everyx R. Then in particular, forx= 0,x= /4 andx= /2 we geta 0 +b 1 +c 0 = 0a 22+b 22+c 1 = 0a 1 +b 0 +c 0 = other words,(abc)is in the null space of the matrix 010 22 221100 Row reduction shows that the null space is trivial and soa=b=c= the set is linearly independent.
