Exercise 2.6.6

Answers

1.

Calculate directly that

T^{t} (f) (x, y) = fT (x, y) = f (3 x + 2 y, x) = 7 x + 4 y .

2.

Since

β = {(1, 0), (0, 1)}

and

(x, y) = x (1, 0) + y (0, 1)

, we have that

f_{1} (x, y) = x

and

f_{2} (x, y) = y

. So we can find out that

T^{t} (f_{1}) (x, y) = f_{1} T (x, y) = f_{1} (3 x + 2 y, x) = 3 x + 2 y = 3 f_{1} (x, y) + 2 f_{2} (x, y);

T^{t} (f_{2}) (x, y) = f_{2} T (x, y) = f_{2} (3 x + 2 y, x) = x = 1 f_{1} (x, y) + 0 f_{2} (x, y) .

And we have the matrix ${[T^{t}]}_{β^{∗}} = (\begin{matrix} 3 & 1 \\ 2 & 0 \end{matrix}) .$

3.

Since

T (x, y) = (3 x + 2 y, x)

, we can calculate that

{[T]}_{β} = (\begin{matrix} 3 & 2 \\ 1 & 0 \end{matrix})

and

{({[T]}_{β})}^{t} = (\begin{matrix} 3 & 1 \\ 2 & 0 \end{matrix}) .

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

jephianlin

2011-06-27 00:00