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)
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