Consider the transformation T : P2 P2 defined by T(p(z)) = (5s + 1)p,(z) + 2p(x)
ID: 3116542 • Letter: C
Question
Consider the transformation T : P2 P2 defined by T(p(z)) = (5s + 1)p,(z) + 2p(x). (a) Compute T(az2 +bc) b) Determine a basis for the image of T. (c) Is T an onto transformation? (d) Determine a basis for the kernel of T (e) Is T a one to one transformation? (f) Is T an isomorphism? (g) Determine the matrix representing T under the basis {1, z, · h) Does there exist a basis B of P2 such that the matrix representing T under the basis B is a diagonal matrix D? If so, then find then find the basis B and diagonal matrix D. If not, then explain why notExplanation / Answer
(a).If p(x) = ax2+bx+c, then p’(x) = 2ax+b so that (5x+1)p’(x) +2p(x) = (5x+1)(2ax+b)+2(ax2+bx+c) = 12ax2+(2a+7b)x+(b+2c). Hence T(ax2+bx+c)= 12ax2+(2a+7b)x+(b+2c).
(b). {x2,x,1} is a basis for the image of T.
(c ). Let 12a = a1, 2a+7b = b1 and b+2c = c1. Then a = a1/12 , b = (6b1-a1)/42 and c = (21c1-6b1+a1)/21 so that an arbitrary polynomial ax2+bx+c in P2 has a pre-image (a/12)x2 + [(6b-a)/42]x + (21c-6b+a)/21 in P2. Hence T is onto.
(d). The kernel of T is the set of solutions to the eqquation T(X) = 0 . If X = ax2+bx+c, then this equation chages to 12ax2+(2a+7b)x+(b+2c) = 0 so that a = 0, 2a+7b = 0, or, 7b = 0 or, b = 0 and b+2c = 0 , or, c = 0. Then X = 0. Hence, {0} is the basis for ker(T).
(e). Let p(x) = ax2+bx+c and q(x) = dx2+ex+f. Now, if T(p(x)) = T(q(x), then 12ax2+(2a+7b)x+(b+2c)= 12dx2+(2d+7e)x+(e+2f). Then 12a = 12d or, a = d, 2a+7b = 2d+7e or, 2a+7b = 2a +7e or, 7b = 7e or, b = e and b+2c = e+2f or, b+2c + b+2f or, 2c= 2f or, c = f. Hence, p(x) = q(x). Thus, T is one-to-one.
(f). Since T is both one-to-one and onto, hence T is an isomorphism.
(g). We have T(1) = 2+0x+0x2 (a=b=0,c =1), T(x) = 7x+1 = 1+7x+0x2 (a=c=0,b =1)and T(x2)=12x2 +2x =0+2x+12x2 (a=1,b=c =0). Hence the matrix of T under the basis {1,x.x2} is
2
1
0
0
7
2
0
0
12
(h). We have T(ax2+bx+c)= 12ax2+(2a+7b)x+(b+2c). Also, T(1) = 2+0x+0x2, T(x-1/2) = 7x = 0+7x+0x2 and T(x2-2x/7+1/7) = 12x2 = 0*1+0x +12x2 so that the matrix of T under the basis B ={1, x-1/2, x2-2x/7+1/7) is D=
2
0
0
0
7
0
0
0
12
which is a diagonal matrix.
2
1
0
0
7
2
0
0
12
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