Exercise 2.3.3

Answers

1.

We can calculate that

{[U]}_{β}^{γ} = (\begin{matrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & - 1 & 0 \end{matrix})

and

{[T]}_{β} = (\begin{matrix} 2 & 3 & 0 \\ 0 & 3 & 6 \\ 0 & 0 & 4 \end{matrix})

and finally

{[UT]}_{β}^{γ} = (\begin{matrix} 2 & 6 & 6 \\ 0 & 0 & 4 \\ 2 & 0 & - 6 \end{matrix}) .

2.

We can calculate

{[h (x)]}_{β} = (\begin{matrix} 3 \\ - 2 \\ 1 \end{matrix})

and

{[U (h (x))]}_{β} = {[U]}_{β}^{γ} {[h (x)]}_{β} (\begin{matrix} 1 \\ 1 \\ 5 \end{matrix}) .

jephianlin

2011-06-27 00:00