Problem 6 Let op: VV be a linear map (also called "self-map" or endomorphism) of a vector space V. Given any real polynomial Pag+a+a2x²++ add, we can define the linear map P(): VV as aqldy +14+244 +...+ad popo...oy. d-times For any ≥0, we will denote pop0...04 = 4° (400 = Idy and 4°1 = 4). (-times As a reminder: for any polynomials P,Q € R[x], Q(p) P(4) P() Q() and (P+Q)() = P(x)+Q(4). = (QP)(x) = (PQ)(4) = (1) Given a polynomial P, show that for any E Ker(P()), () Є Ker(P()). hint: for a vector w being in the kernel of a linear map just means (w)=0+ try to show that and P(p) commute. So induces a linear self-map of the vector subspace Ker(P()). (2) Let ab be real numbers, we recall that 2- (a+b)x+ab = (-a)(x - b). We want to prove that Ker(p2 (a) Simplify (a+b)+ abldy) = Ker(-ald) Ker( - bld) ((-a) - (x - b)). (b) Apply the equality of polynomials obtained in the previous question to p. (c) Show that if Ker(2-(a+b)+abldy), then (-ald)(v) € Ker(p-bldy). hint: meaning of being in the kernel of a linear map? + use that polynomial multi- plication corresponds to composition, when applied to a linear map (d) Show that if Є Ker(p2-(a+b)+abldy), then (p-bld)(v) € Ker(p-aldy). (e) Now prove that Ker(2- (a + b) + abldy) = Ker(y-ald) + Ker( - bld). hint: double inclusion as usual. One inclusion uses (2) (b) and for the other, one can use the fact that if U, W, H are vector subspaces of V and UC H and WCH; then as H is closed under addition U+WCH.
Problem 6 Let op: VV be a linear map (also called "self-map" or endomorphism) of a vector space V. Given any real polynomial Pag+a+a2x²++ add, we can define the linear map P(): VV as aqldy +14+244 +...+ad popo...oy. d-times For any ≥0, we will denote pop0...04 = 4° (400 = Idy and 4°1 = 4). (-times As a reminder: for any polynomials P,Q € R[x], Q(p) P(4) P() Q() and (P+Q)() = P(x)+Q(4). = (QP)(x) = (PQ)(4) = (1) Given a polynomial P, show that for any E Ker(P()), () Є Ker(P()). hint: for a vector w being in the kernel of a linear map just means (w)=0+ try to show that and P(p) commute. So induces a linear self-map of the vector subspace Ker(P()). (2) Let ab be real numbers, we recall that 2- (a+b)x+ab = (-a)(x - b). We want to prove that Ker(p2 (a) Simplify (a+b)+ abldy) = Ker(-ald) Ker( - bld) ((-a) - (x - b)). (b) Apply the equality of polynomials obtained in the previous question to p. (c) Show that if Ker(2-(a+b)+abldy), then (-ald)(v) € Ker(p-bldy). hint: meaning of being in the kernel of a linear map? + use that polynomial multi- plication corresponds to composition, when applied to a linear map (d) Show that if Є Ker(p2-(a+b)+abldy), then (p-bld)(v) € Ker(p-aldy). (e) Now prove that Ker(2- (a + b) + abldy) = Ker(y-ald) + Ker( - bld). hint: double inclusion as usual. One inclusion uses (2) (b) and for the other, one can use the fact that if U, W, H are vector subspaces of V and UC H and WCH; then as H is closed under addition U+WCH.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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