Exercise 2.6.7

Answers

1.

Calculate directly that

T^{t} (f) (a + bx) = fT (a + bx) = f (- a - 2 b, a + b) = - 3 a - 4 b .

2.

Since

β = {1, x}

and

a + bx = a \times 1 + b \times x

, we have that

f_{1} (a + bx) = a

and

f_{2} (a + bx) = b

. And since

γ = {(1, 0), (0, 1)}

and

(a, b) = a (1, 0) + b (0, 1)

, we have that

g_{1} (a, b) = a

and

g_{2} (a, b) = b

. So we can find out that

T^{t} (g_{1}) (a + bx) = g_{1} T (a + bx) = g_{1} (- a - 2 b, a + b) = - a - 2 b

= - 1 \times g_{1} (a, b) + (- 2) \times g_{2} (a, b);

T^{t} (g_{2}) (a + bx) = g_{2} T (a + bx) = g_{2} (- a - 2 b, a + b) = a + b

= 1 \times g_{1} (a, b) + 1 \times g_{2} (a, b) .

And we have the matrix ${[T^{t}]}_{γ^{∗}}^{β^{∗}} = (\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}) .$

3.

Since

T (a + bx) = (- a - 2 b, a + b)

, we can calculate that

{[T]}_{β}^{γ} = (\begin{matrix} - 1 & - 2 \\ 1 & 1 \end{matrix})

and

{({[T]}_{β}^{γ})}^{t} = (\begin{matrix} - 1 & 1 \\ - 2 & 1 \end{matrix}) .

So we have that ${[T^{t}]}_{γ^{∗}}^{β^{∗}} = {({[T]}_{β}^{γ})}^{t}$ .

jephianlin

2011-06-27 00:00