Exercise 6.8.7

Answers

1.
Check that
T^(H)(ax1,x2,y) = H(T(ax1 + x2),T(y))
= H(aT(x1) + T(x2),T(y)) = aH(T(x1),T(y)) + H(T(x2),T(y))
= aT^(H)(x1,y) + T^(H)(x2,y)

and

T^(H)(x,ay1 + y2) = H(T(x,T(ay1 + y2))
= H(T(x),aT(y1) + T(y2)) = aH(T(x),T(y1)) + H(T(x),T(y2))
= aT^(H)(x,y1) + T^(H)(x,y2).
2.
Check that
T^(cH1 + H2)(x,y) = (cH1 + H2)(T(x),T(y))
= cH1(T(x),T(y)) + H2(T(x),T(y))
= [cT^(H1) + T^(H)](x,y).
3.
Suppose T is injective and surjective. If H is an nonzero bilinear form with H(x1,y1)≠0 for some x1,y1 ∈ W and T^(H) is the zero bilinear form, we may find x0,y0 ∈ V such that T(x0) = x1 and T(y0) = y1 since T is surjective. Thus we’ll have
0 = T^(H)(x0,x1) = H(x,y)≠0,

a contradiction. This means that T^ is injective. On the other hand, since T is an isomorphism, the inverse of T exists. Then for each H ∈B(V ), we can define

H0(x,y) := H0(T−1(x),T−1(y))

such that

T^(H0) = H.
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2011-06-27 00:00
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