MATH2061_Linear Mathematics and Vector Calculus_201_tut02s

T HE U NIVERSITY OF S YDNEYP URE M ATHEMATICSLinear Mathematics2013Tutorial2(week3)—Solutions1.a)Explain why the set A= x x2 |x∈R is not a subspace of R2.b)Explain why the set B= 100 +t 110 |t∈R is not a subspace of R3.Solutiona)The set A is not closed under addition.For example, 11 ∈A and 24 ∈A,but11 + 24 = 35 ∈A.b) 000 ∈B.2.Vectors v1,v2,v3,v4,v5,v6are the vectors:= 000 v2= 052 v3= −440 v4= 111 v5= 201 v6= 021 v1Write each of the vectors v1,v2and v3as a linear combination of the three vectors{v4,v5,v6}, if possible.Solution To express a vector v∈R3as a linear combination of{v4,v5,v6},we have tofind scalars a,b,c∈R such that a v4+b v5+c v6=v.Writing v= x y z ,this is equivalent to solving the following system of linear equationsa+2b=x,a+2c=y,a+b+c=z.When v=v1= 000 no calculation is necessary because0=0v4+0v5+0v6.(The zero vector can always be written as a linear combination of any set of vectors.)We now solve the two cases where v=v2and v=v3simultaneously by computing with a “doubly augmented”matrix:1200−4,1025411120where the last two columns are the vectors v2and v3,ing Gaussian eliminationwefind that:1200−4 1025411120 R2:=R2−R1−−−−−−−→ 1200−40−225811120R3:=R3−R1−−−−−−−→ 1200−40−22580−1124R2←→R3R2:=R2×−1−−−−−−−→ 1200−401−1−2−40−2258R3:=R3+2R2−−−−−−−→ 1200−401−1−2−400010Thefirst column of the augmented matrix(or column4of the matrix),corresponds to v=v2.This system of equations is inconsistent,so there is no solution.That is,v2cannot be written as a linear combination of v4,v5and v6.The second column of the augmented matrix(or column5of the matrix),corresponds to v=v3.This system of equations has a solution;in fact,c is a free variable and there are an infinite number of solutions.Set c=t,a parameter.By back substitution,b=t−4and a=−2b−4=8−2t−4=4−2t.Hence,v3=(4−2t)v4+(t−4)v5+t v6,for any t∈R.In particular,taking t=0,we see that v3=4(v4−v5).3.Prove that the set S= x y ∈R2|y=4x is a subspace of R2. Solution S is clearly non-empty,since 00 satisfies the equation y=4x and is therefore in S. Now suppose that u= a4a and v= b4b both belong to S.Then u+v= a+b4a+4b = a+b4(a+b) ,and so u+v∈S,and S is closed under addition. If k∈R,then ku=k a4a = ka4(ka) ,and so k u∈S,and S is closed under scalar multiplication.4.Recall that F is the vector space of functions from R to R,with the usual operations of additionand scalar multiplication of functions.For each of the following subsets of F,write down two functions that belong to the subset,and determine whether or not the subset is a vector subspace of F.a)The set of polynomials of degree equal to3.b)The set{f:R−→R|f(1)=1}.c)The set{f:R−→R|f(−x)=−f(x)for all x∈R}.Solutiona)Both a and b,where a(x)=x3and b(x)=−x3,belong to this set.However,a(x)+b(x)=0,so a+b does not belong to the set.Hence this set is not closed under addition and so is not a vector subspace of F.b)Both a and b,where a(x)=1and b(x)=x,belong to this set.However,a(x)+b(x)=x+1,so(a+b)(1)=2,showing that a+b does not belong to this subset of F.Hence,the set is not closed under addition and so not a vector subspace of F.c )This is the set of all odd functions.Let F −be the set of odd functions.Then F −is clearly non-empty.(The zero function is an odd function,as is f (x )=x ,for example.)If f andg are two functions in F −then(f +g )(−x )=f (−x )+g (−x )=−f (x )−g (x )=−(f +g )(x ).Hence,F −is closed under addition.Also,if r ∈R then(rf )(−x )=r ·f (−x )=−r ·f (x )=−(rf )(x ),so r ·f ∈F −.Therefore,F −is a vector subspace of F .5.Let X =11−2, 235.Span(X )can be interpreted as a plane in 3dimensional space.Findthe equation of this plane.Solution The vector x y z∈Span(X )if and only if a 11−2+b 235= x y z,for some a,b ∈R .Applying Gaussian elimination we get12x 13y −25zR 2:=R 2−R 1R 3:=R 3+2R 1−−−−−−−→12x 01y −x 092x +z R 3:=R 3−9R2−−−−−−−→12x 01y −x 0011x −9y +zHence, x y z∈Span(X )if and only if 11x −9y +z =0.Therefore,Span(X )can be interpreted geometrically as the plane 11x −9y +z =0.6.For each of the sets of vectors X ⊂R 3below,explicitly describe all of the vectors in the subspace Span(X )of R 3.(Use set notation to describe Span(X ),and/or give a geometric interpretation.)a )X ={0}.b )X =111 .c )X =111, 222.d )X = 111 , 100.e )X = 111 , 100 , 110.Solution a )Span00 = 000.b )Span 111= r r r r ∈R .We can interpret this set geometrically as the straight line in 3-dimensional space passing through the origin and the point (1,1,1).c )As 222=2 111,Span 111 , 222 =Span 111 = r r r r ∈R .d )Span111, 100 = r 111+s 100 r,s ∈R = r +s r rr,s ∈R= x y z∈R 3 y =z .We can also describe Span(X )geometrically as the plane in R 3with equation y =z .e)Span 111 , 100 , 110 = r+s+t r+t r r,s,t∈R .We can analyze Span(X)alge-braically.A vector x y z ∈Span(X)if and only ifx y z =a 111 +b 100 +c 110 ,for some a,b,c∈ing Gaussian elimination we can simplify this to111x 101y100z R2:=R2−R1R3:=R3−R1−−−−−−−→ 111x0−10y−x0−1−1z−xR2:=R2×−1R3:=R3+R2−−−−−−−→ 111x010x−y00−1z−yHence,c=y−z,b=x−y and a=x−(x−y)−(y−z)=z.In particular,for any vector x y z we can alwaysfind suitable a,b,c∈R,so Span(X)=R3.Note that we now have given two quite different“explicit”descriptions of Span(X):r+s+t r+t r r,s,t∈R =Span(X)=R3.(A priori,it is not obvious that the left hand side of this equation gives R3.)7.The vector space P2consists of all polynomials of degree less than or equal to2.Let X={p1,p2,p3}⊂P2,where p1(x)=x+2,p2(x)=3−x,p3(x)=x2−x,for all x∈R.a)Show that X spans P2.(That is,show that Span(X)=P2.)b)Let p4(x)=2x2−x−3.Show that{p2,p3,p4}does not span P2.Solutiona)We must show that every vector in P2can be written as a linear combination of p1,p2,p3.That is,if p is an arbitrary element of P2,such that p(x)=α+βx+γx2say,then we have to show that p(x)=ap1(x)+bp2(x)+cp3(x),for some a,b,c∈R.Expanding this equation we see that we require thata(x+2)+b(3−x)+c(x2−x)=α+βx+γx2.Looking at the constant term and the coefficients of x and x2we obtain the following three equations:2a+3b=αa−b−c=βc=γTo solve for a,b and c we apply Gaussian elimination to the augmented matrix:230α1−1−1β001γ R1←→R2−−−−−→ 1−1−1β230α001γR2:=R2−2R1−−−−−−−→ 1−1−1β052α−2β001γHence,c=γ,b=15 α−2β−2c =15 α−2β−2γ and a=β+b+c=β+15 α−2β−2γ +γ=15α+35γ+35β.In particular,we can always solve for a,b and c,so Span(X)=P2as required.b)To show that Span{p2,p3,p4}=P2we need tofind an element of P2which is not a linearcombination of these three polynomials.Once again,let p be an arbitrary element of P2, where p(x)=α+βx+γx2.Then p belongs to Span{p2,p3,p4}if and only if we can find a,b,c∈R such that ap2(x)+bp3(x)+cp4(x)=p(x);that isa(3−x)+b(x2−x)+c(2x2−x−3)=α+βx+γx2.Looking at the constant term and the coefficients of x and x2gives us the following three equations:3a−3c=α−a−b−c=βb+2c=γWe again reduce the associated augmented matrix to row echelon form:30−3α−1−1−1β012γ R1←→R2R1:=R1×−1−−−−−−−→ 111−β30−3α012γR2:=R2−3R1−−−−−−−→ 111−β0−3−6α+3β012γR2:=−13R2−−−−−−→ 111−β012−13α−β012γR3:=R3−R2−−−−−−−→ 111−β012−13α−β00013α+β+γTherefore,p belongs to the span of{p2,p3,p4}if and only if13α+β+γ=0.In particular,if we take p such that p(x)=1(or1−3x2,or1+5x+7x2,or...)then p does not belong to Span{p2,p3,p4}.Notice that p4(x)=2x2−x−3=2(x2−x)+(x−3)=2p3(x)−p2(x).We will see later why this means that Span{p2,p3,p4}=P2.8.Suppose that V is a vector space and that v∈e the vector space axioms to show that:a)0×v=0.b)−1×v is the negative of v.Solutiona)Using the vector space axioms we have0×v=0+0×v by A4,=−v+v+0×v by A5,=−v+(1+0)×v by S3,=−v+1×v=−v+v by S5,=0by A5.b)Using the vector space axioms,v+(−1)×v=1×v+(−1)×v,by S5,=(1−1)×v by S3,=0×v=0by part(a).Therefore,−1×v is the negative of v.。