Exercise 2.5.5

Answers

We have that

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

and

{[T]}_{β^{'}} = {[I]}_{β}^{β^{'}} [{[T]}_{β} {[I]}_{β^{'}}^{β}

= (\begin{matrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{matrix}) (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix})

= (\begin{matrix} \frac{1}{2} & - \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{matrix}) .

jephianlin

2011-06-27 00:00