SSM:Linear AlgebraSection 4.1Chapter 44.11. Not a subspace since it does not contain the neutral element, that is, the functionf(t) = 0,for allt.3. This subsetVis a subspace ofP2:•The neutral elementf(t) = 0 (for allt) is inVsincef0(1) =f(2) = 0.•Iffandgare inV(so thatf0(1) =f(2) andg0(1) =g(2)), then(f+g)0(1) = (f0+g0)(1) =f0(1) +g0(1) =f(2) +g(2) = (f+g)(2), so thatf+gis inV.•Iffis inV(so thatf0(1) =f(2)) andkis any constant, then (kf)0(1) = (kf0)(1) =kf0(1) =kf(2) = (kf)(2), so thatkfis inV.Iff(t) =a+bt+ct2thenf0(t) =b+ 2ct, andfis inViff0(1) =b+ 2c=a+ 2b+ 4c=f(2), ora+b+ 2c= 0. The general element ofVis of theformf(t) = (-b-2c) +bt+ct2=b(t-1) +c(t2-2), so thatt-1, t2-2 is a basis ofV.5. Ifp(t) =a+bt+ct2thenp(-t) =a-bt+ct2and-p(-t) =-a+bt-ct2.Comparing coefficients we see thatp(t) =-p(-t) for alltif (and only if)a=c= 0.The general element of the subset is of the formp(t) =bt.These polynomials form a subspace ofP2, with basist.7. The setVof diagonal 3×3 matrices is a subspace ofR3×3:a. The zero matrix000000000is inV,b. IfA=a000b000candB=p000q000rare inV, then so is their sumA+B=a+p000b+q000c+r.91
Chapter 4SSM:Linear Algebrac. IfA=a000b000cis inV, then so iskA=ka000kb000kc, for all constantsk.9. Not a subspace; consider multiplication with a negative scalar.I3belongs to the set, but-I3doesn’t.11. Not a subspace:I3is in rref, but the scalar multiple 2I3isn’t.13. Not a subspace: (1,2,4,8, . . .) and (1,1,1,1, . . .) are both geometric sequences, but theirsum (2,3,5,9, . . .) is not, since the ratios of consecutive terms fail to be equal, for example,326=53.15. The setWof all square-summable sequences is a subspace ofV:•The sequence (0,0,0, . . .) is inW.•Suppose (xn) and (yn) are inW. Note that the inequality (xn+yn)2≤2x2n+ 2y2nholdsfor alln, since 2x2n+ 2y2n-(xn+yn)2=x2n+y2n-2xnyn= (xn-yn)2≥0.Thus∑∞n=1(xn+yn)2≤2∑∞n=1x2n+ 2∑∞n=1y2nconverges, so that the sequence (xn+yn) is inWas well.•If (xn) is inW(so that∑∞n=1x2nconverges), then (kxn) is inWas well, for any constantk, since∑∞n=1(kxn)2=k2∑∞n=1x2nwill converge.17. LetEijbe then×mmatrix with a 1 as itsijth entry, and zeros everywhere else. AnyAinRn×mcan be written as the sum of allaijEij, and theEijare linearly independent,so that they form a basis ofRn×m. Thus dim(Rn×m) =nm.19.a+bic+di=a10+bi0+c01+d0iThe vectors10,i0,01,0iform a basis ofC2as areallinear space, so that dim(C2) =4.21. Use Summary 4.1.6. The general element of the subspace isa00d=a1000+d0001.Thus1000,0001is a basis of the subspace; thedimension is 2.92
SSM:Linear AlgebraSection 4.123. Proceeding as in Exercise 21, we find the basis1000,0010,0001; the dimensionis 3.