Question 7 (10 marks) Consider the basis B = {1, 1 ? x, 1 ? x 2} of P2. (a) For

Question

Question 7 (10 marks) Consider the basis B = {1, 1 ? x, 1 ? x 2} of P2. (a) For each of the following polynomials p write the coordinate vector [p]B: (i) p = 1 ? x, (ii) p = 2 ? x ? x 2 , (iii) p = 70 ? x ? x 2 . (b) Show that T : P2 ? P2, T(a + bx + cx2 ) = a ? b ? 69c ? ax + (b + c)x 2 , is a linear transformation. (c) Determine the representing matrix [T]B.

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Question 7 (10 marks) Consider the basis B (1,1 -x,1- of P2 (a) For each of the following polynomials p write the coordinate vector p: (ii) p (iii) p 2--2, 70-r-r2 (b) Show that T: P2P2 T(a + br + cr*) = a-b-69c _ ax + (D+c)2,2, is a linear transformation. (c) Determine the representing matrix [Ts.

Explanation / Answer

7. (a). (i)Since p(x) = 1-x = 0*1+1*(1-x)+0*(1-x2), hence [p]B = (0,1,0)T.

(ii). Since p(x) = 2-x-x2 = 1+(1-x)+(1-x2) , hence, [p]B = (1,1,1)T.

(iii).Let A =

1

1

1

70

0

-1

0

-1

0

0

-1

-1

It may be observed that the entries in the columns of A are the scalar multiples of 1 and the coefficients of x,x2 in the vectors in B and p(x) respectively.

The RREF of A is

1

0

0

68

0

1

0

1

0

0

1

1

Hence, p(x) = 70-x-x2 = 68+(1-x)+(1-x2) so that [p]B = (68,1,1)T.

(b). T: P2? P2 is defined by T(a+bx+cx2) = a-b -69c -ax+(b+c)x2.

Let p(x) = a+bx+cx2 and q(x) = r+sx+tx2 be 2 arbitrary polynomials/vectors in P2 and let k be an arbitrary scalar. Then p(x)+q(x) = a+bx+cx2+r+sx+tx2 = (a+r)+(b+s)x+(c+t)x2 so that T(p(x)+q(x)) =(a+r)-(b+s) -69(c+t) –(a+r)x +( b+s+c+t)x2 = a-b -69c -ax+(b+c)x2+ r-s -69t -rx+(s+t)x2 = T(p(x)+T(q(x). This implies that T preserves vector addition. Also, T(kp(x) = T(ka+kbx+kcx2)= ka-kb -69kc -kax+(kb+kc)x2 = k(a-b -69c -ax+(b+c)x2) = kT(p(x). This implies that T preserves scalar multiplication. Hence T is a linear transformation.

We have T(1) = 1-x, T(1-x) = 1-1-x+x2 = -x+x2 and T(1-x2) = 1-69-x+x2= -68-x+x2. Hence,[T]B=                       [T(1),T(1-x),T(1-x2)] =

1

0

-68

-1

-1

-1

0

1

1

1

1

1

70

0

-1

0

-1

0

0

-1

-1

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