# SOLUTION: Let S be the basis {1,t,t^2} of P2. Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of P2 and find the coordinates of 1,t,t^2 in this new basis.

Algebra ->  Algebra  -> College  -> Linear Algebra -> SOLUTION: Let S be the basis {1,t,t^2} of P2. Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of P2 and find the coordinates of 1,t,t^2 in this new basis.       Log On

 Algebra: Linear Algebra (NOT Linear Equations) Solvers Lessons Answers archive Quiz In Depth

 Question 26765: Let S be the basis {1,t,t^2} of P2. Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of P2 and find the coordinates of 1,t,t^2 in this new basis. Answer by venugopalramana(3286)   (Show Source): You can put this solution on YOUR website!Let S be the basis {1,t,t^2} of P2.BETTER CALL THIS Q2 TO AVOID CONFUSION WITH P2 BELOW Set P1= 2+t^2 P2= 1-t+t^2 P3= 3-t+t^2 in P2. Show that P1,P2,P3 isa basis of Q2 and find the coordinates of 1,t,t^2 in this new basis. TO SHOW THAT P1,P2,P3 IS A BASIS,WE NEED TO PROVE THAT THEY ARE INDEPENDENT. THAT IS IF AP1+BP2+CP3=0...THEN A=B=C=0 AP1+BP2+CP3=A(2+T^2)+B(1-T+T^2)+C(3-T+T^2)=0 T^2(A+B+C)-T(B+C)+(2A+B+3C)=0.....................I..........HENCE A+B+C=0 ............................II B+C=0...............................III 2A+B+3C=0...........................IV EQN.II - EQN.III..GIVES.....A=0......V...SUBSTITUTING IN IV,WE GET B+3C=0...............................VI EQN.VI-EQN.III...GIVES....2C=0...OR...C=0..WHICH ON SUBSTITUTING IN EQN.I GIVES B==0...SO A=B=C=0...HENCE P1,P2,P3 ARE INDEPENDENT.HENCE THEY FORM A BASIS. NOW TO WRITE Q2 IN THE EW BASIS WE EQUATE EQN.I TO Q2 AND FIND A,B,C. T^2(A+B+C)-T(B+C)+(2A+B+3C)=Q2=1+T+T^2...HENCE A+B+C=1.....................VII B+C=-1......................VIII 2A+B+3C=1...................IX PROCEEDING AS BEFORE WE GET A=2,B=0,C=-1..HENCE Q2=1+T+T^2=2(2+T^2)-(3-T+T^2)