Math 1231 Assignment 2019 T2.docx - Math 1231 Assignment...

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Math 1231 Assignment 2019 T2z5258929 Steven SokQuestion 2.The functionT:R3→R2is defined by :T:(x1x2)=(x25x1x2x1)for all(x1x2)R3.We are required to show that the functionTis linear. Since the domainR3andcodomainR2are both vector spaces, we can prove that this function is linear by checkingthe addition and scalar multiplication conditions.Addition Condition.For a vectorv=(v1v2v3)andu=(u1u2u3)R3, we havev+u=(v1+u1v2+u2v3+u3), and hencev(¿¿2+u2)T(v+u)=(¿5(v1+u1)(v2+u2)v1+u1)So,T(v+u)=(v2+u25v1+5u1v2u2v1+u1)T(v+u)=((v2)+(u2)(5v1v2)+(5u1u2)v1+u1)T(v+u)=(v25v1v2v1)+(u25u1u2u1)T(v+u)=T(v)+T(u)Thus, the addition condition is satisfied.
Scalar Multiplication Condition.For a vectorv=(v1v2v3)R3, we haveλv=(λv1λv2λv3)R3, λR, and hence2λ v¿¿¿5(λv1)(λ v2)¿(¿(λ v2)¿¿)T(λ v)=¿T(λ v)=λ(v25v1v2v2)T(λ v)=λT(v)Thus, the scalar multiplication condition is also satisfied and since it both conditions aresatisfied, It can be concluded that the functionT:R3→R2is defined by :T:(x1x2)=(x25x1x2x1)for all(x1x2)R3is linear.Question 3.We are required to show that the functionS={xR4:3x1x2+2x3+3x4=0}is a subspaceofR4.To show that the function is a subspace ofR4we can utilize the subspace theorem,firstcheck that the function satisfies the subspace theorem, which states that a function is asubspace if it satisfies the following conditions:i)It contains the zero element,0Sii)S is closed under vector additioniii) S is closed under scalar multiplicationi) Existence of Zero VectorSubstitute the vector(0000)into the function, so
3(0)(0)+2(0)+3(0)=00=0LHS = RHSHence the vector(0000)exists inS ,and the first condition is satisfied.ii) Closed Under Addition.Let vectorsv ,uS. Then, by substituting the vectors into the functionS,v=(v1v2v3v4)andu=(u1u2u3u4)we can derive the following equations,3v1v2+2v3+3v4=0and,3u1u2+2u3+3u4=0But,v+u=(v1+u1v2+u2v3+u3v4+u4)and we want to determine if this vector lies on the plane to prove

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Term
Three
Professor
Mak

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