mid1042012_solutions - Sample solutions to a past MAT224...

This preview shows page 1 - 2 out of 6 pages.

Sample solutions to a past MAT224 midterm(held on March 1, 2012)Z. Amir-Khosravi1.First we find the kernel ofT:p(x)ker(T)p(x-1) =p(x+ 1).Letp(x)ker(T). Settingy=x-1, we get thatp(x) isperiodic:p(y) =p(x-1) =p(x+ 1) =p(y+ 2).(1)We show that the only possiblep(x) with this property are constant poly-nomials:Supposep(x)ker(T) is non-constant.Thenp(x) has a complex rootα.Settingy=αin equation (1) we get0 =p(α) =p(α+ 2).That meansα+ 2 is also a root. But now settingy=α+ 2 in (1), we get that(α+ 2) + 2) =α+ 4 is also a root. Repeating this process, we getα, α+ 2, α+ 4, α+ 6, ...are infinitely many distinct roots of the non-constant polynomialp(x). This isimpossible, so the assumption that there is a non-constantp(x)ker(T) mustbe wrong.Ifp(x) is constant, i.e. ifp(x)cfor some real numberc, we havep(x-1) =c=p(x+ 1)p(x)ker(T).This shows that ker(T) consists of all constant polynomials. Then{1}is abasis for ker(T).Since{1, x, x2, x3, x4, x5}is basis forP5(R), the image ofTis spanned by{T(1), T(x), T(x2), T(x3), T(x4), T(x5)}.But sinceT(1) = 0, it contributes

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture