Show that L is a linear transformation Let P 2 --> P 2 be linear transformation

Question

Show that L is a linear transformation
Let P 2 --> P 2 be linear transformation defined by                      L(at2 + bt + c) = (a+2b)t + (B+c) a.) is -3t2 + 2t - 2 in ker(L)? b.) is 2t + 1 in range(L)? c.) Find basis for ker(L) d.) Find basis for range(L) e.) Is L one to one? f.) is L onto? g.) Find dim of P2 Show that L is a linear transformation
Let P 2 --> P 2 be linear transformation defined by                      L(at2 + bt + c) = (a+2b)t + (B+c) a.) is -3t2 + 2t - 2 in ker(L)? b.) is 2t + 1 in range(L)? c.) Find basis for ker(L) d.) Find basis for range(L) e.) Is L one to one? f.) is L onto? g.) Find dim of P2                      L(at2 + bt + c) = (a+2b)t + (B+c) a.) is -3t2 + 2t - 2 in ker(L)? b.) is 2t + 1 in range(L)? c.) Find basis for ker(L) d.) Find basis for range(L) e.) Is L one to one? f.) is L onto? g.) Find dim of P2

Explanation / Answer

Show that L is a linear transformation Let P 2 --> P 2  be linear transformation defined by                      L(at2 + bt + c) = (a+2b)t + (B+c) THAT IS... L[U] = M* U =V ....WHERE U AND V ARE 2 POLYNOMIALS IN P2. L[A,B.C] = [0,A+2B , B+C] LET M [ D,E,F] AND N[G,H,J] BE ANY 2 ELEMENTS IN U LET K BE ANY SCALAR TO CHECK 1. IF L[M+N]=L[M]+L[N] 2.IF L[K*M]=K*L[M] SINCE U AND V ARE 2 ELEMENTS OF U WE HAVE L[M]=L[D,E,F]=[0,D+2F,E+F] L[N]=L[G,H,J]=[0,G+2J,H+J] L[M]+L[N]= [0,D+2F,E+F]+[0,G+2J,H+J]=[0,D+2E+G+2H, E+F+H+J ] L[M+N]=L[(D+G) , (E+H) ,(F+J) ] = [0, D+G+2(E+H) , E+H+F+J] =[0,D+2E+G+2H, E+F+H+J ]=L[M]+L[N].....OK ========== L[KM]=L[KD,KE,KF]=[0.KD+2KE,KE+KF] KL[M]=K[0,D+2E,E+F]= [0,KD+2KE,KE+KF=L[KM]....OK HENCE THIS IS L.T. a.) is -3t2 + 2t - 2 in ker(L)? KERNEL IS GIVEN BY L[M]=L[D,E,F]=[0,D+2E,E+F] =[0,0,0] 0=0..........OK D+2E=0................D=-2E E+F=0...............E=-F D=-2E=-2[-F]=2F HENCE KERNEL IS GIVEN BY [2F,-F,F] -3t2 + 2t - 2 i  = [-3,2,-2] L[-3,2,-2]=[0,-3+2*2,2-2]=[0,-1,0].....IT IS NOT [0,0,0] ... SO IT IS NOT IN KERNEL OF L b...???? IT IS NOT SHOWING UP PLEASE REPOST INFORMING CRAMSTER Show that L is a linear transformation Let P 2 --> P 2  be linear transformation defined by                      L(at2 + bt + c) = (a+2b)t + (B+c) THAT IS... L[U] = M* U =V ....WHERE U AND V ARE 2 POLYNOMIALS IN P2. L[A,B.C] = [0,A+2B , B+C] LET M [ D,E,F] AND N[G,H,J] BE ANY 2 ELEMENTS IN U LET K BE ANY SCALAR TO CHECK 1. IF L[M+N]=L[M]+L[N] 2.IF L[K*M]=K*L[M] SINCE U AND V ARE 2 ELEMENTS OF U WE HAVE L[M]=L[D,E,F]=[0,D+2F,E+F] L[N]=L[G,H,J]=[0,G+2J,H+J] L[M]+L[N]= [0,D+2F,E+F]+[0,G+2J,H+J]=[0,D+2E+G+2H, E+F+H+J ] L[M+N]=L[(D+G) , (E+H) ,(F+J) ] = [0, D+G+2(E+H) , E+H+F+J] =[0,D+2E+G+2H, E+F+H+J ]=L[M]+L[N].....OK ========== L[KM]=L[KD,KE,KF]=[0.KD+2KE,KE+KF] KL[M]=K[0,D+2E,E+F]= [0,KD+2KE,KE+KF=L[KM]....OK HENCE THIS IS L.T. a.) is -3t2 + 2t - 2 in ker(L)? KERNEL IS GIVEN BY L[M]=L[D,E,F]=[0,D+2E,E+F] =[0,0,0] 0=0..........OK D+2E=0................D=-2E E+F=0...............E=-F D=-2E=-2[-F]=2F HENCE KERNEL IS GIVEN BY [2F,-F,F] -3t2 + 2t - 2 i  = [-3,2,-2] L[-3,2,-2]=[0,-3+2*2,2-2]=[0,-1,0].....IT IS NOT [0,0,0] ... SO IT IS NOT IN KERNEL OF L b...???? IT IS NOT SHOWING UP PLEASE REPOST INFORMING CRAMSTER                      L(at2 + bt + c) = (a+2b)t + (B+c) THAT IS... L[U] = M* U =V ....WHERE U AND V ARE 2 POLYNOMIALS IN P2. L[A,B.C] = [0,A+2B , B+C] LET M [ D,E,F] AND N[G,H,J] BE ANY 2 ELEMENTS IN U LET K BE ANY SCALAR TO CHECK 1. IF L[M+N]=L[M]+L[N] 2.IF L[K*M]=K*L[M] SINCE U AND V ARE 2 ELEMENTS OF U WE HAVE L[M]=L[D,E,F]=[0,D+2F,E+F] L[N]=L[G,H,J]=[0,G+2J,H+J] L[M]+L[N]= [0,D+2F,E+F]+[0,G+2J,H+J]=[0,D+2E+G+2H, E+F+H+J ] L[M+N]=L[(D+G) , (E+H) ,(F+J) ] = [0, D+G+2(E+H) , E+H+F+J] =[0,D+2E+G+2H, E+F+H+J ]=L[M]+L[N].....OK ========== L[KM]=L[KD,KE,KF]=[0.KD+2KE,KE+KF] KL[M]=K[0,D+2E,E+F]= [0,KD+2KE,KE+KF=L[KM]....OK HENCE THIS IS L.T. a.) is -3t2 + 2t - 2 in ker(L)? KERNEL IS GIVEN BY L[M]=L[D,E,F]=[0,D+2E,E+F] =[0,0,0] 0=0..........OK D+2E=0................D=-2E E+F=0...............E=-F D=-2E=-2[-F]=2F HENCE KERNEL IS GIVEN BY [2F,-F,F] -3t2 + 2t - 2 i  = [-3,2,-2] L[-3,2,-2]=[0,-3+2*2,2-2]=[0,-1,0].....IT IS NOT [0,0,0] ... SO IT IS NOT IN KERNEL OF L b...???? IT IS NOT SHOWING UP PLEASE REPOST INFORMING CRAMSTER

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.