Let V Span ex e2x e3x C 0 1 Let L V rightarrow V be given b

Let V = Span {e^x, e^2x, e^3x} C [0, 1]. Let L: V rightarrow V be given by Lf(x) = f\'(x). SHOW that L is invertible. FIND the matrix M of L respect to the basis B = {v_1 = e^x, v_2 = e^2x, v_3 = e^3x}. FIND the matrix R of L^-1 with respect to B. FIND the matrix M_2 of L relative to the new basis C = {u_1 = e^x + e^2x + u_2 = e^2x + e^2x + e^3x, v_3 = e^x + e^3x}. FIND the transition matrix T such that T^-1 MT = M_2.

Solution

Let the minimal polynomial for T over F be m = aà + aèx + ~ ~ ~ + an-1xn-1 + xn . We first assume that aà 0. Since m is the minimal polynomial for T, we have m(T) = aà1 + aèT + ~ ~ ~ + an-1Tn-1 + Tn = 0 and hence multiplying by aàî and using Theorem 7.2 yields 0 = 1+ a0 !1 T (a11+ a2T +!+ an!1T n!2 +T n!1) or 1 = T[!a0 !1(a11+ a2T +!+ an!1T n!2 +T n!1)] = [!a0 !1(a11+ a2T +!+ an!1T n!2 +T n!1)]T !!. This shows that Tî = -aàî(aè1 + aìT + ~ ~ ~ + an-1 Tn-2 + Tn-1), and hence T is invertible.

Now suppose T is invertible, but that aà = 0. Then we have 0 = a1T + a2T 2 +!+ an!1T n!1 +T n = (a11+ a2T +!+ an!1T n!2 +T n!1)T !!. Multiplying from the right by Tî yields 0 = aè1 + aìT + ~ ~ ~ + an-1Tn-2 + Tn-1 and hence T satisfies the polynomial p = aè + aìx + ~ ~ ~ + an-1 xn-2 + xn-1 F[x]. But deg p = n - 1 < n which contradicts the definition of m as the minimal polynomial. Therefore we must have aà 0.