Exercise 2.4.19

Answers

1.

It would be

(\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) .

2.

We may check that

L_{A} ϕ_{β} (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = (\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}) (\begin{matrix} 1 \\ 2 \\ 3 \\ 4 \end{matrix}) = (\begin{matrix} 1 \\ 3 \\ 2 \\ 4 \end{matrix})

and

ϕ_{β} T (\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}) = ϕ_{β} (\begin{matrix} 1 & 3 \\ 2 & 4 \end{matrix}) = (\begin{matrix} 1 \\ 3 \\ 2 \\ 4 \end{matrix}) .

So they are the same.

jephianlin

2011-06-27 00:00