Exercise 2.5.8

Answers

This is similar to the proof of Theorem 2.23 since

{[T]}_{β^{'}}^{γ^{'}} = {[I_{W}]}_{γ}^{γ^{'}} {[T]}_{β}^{γ} {[I_{V}]}_{β^{'}}^{β} = P^{- 1} {[T]}_{β}^{γ} Q .

jephianlin

2011-06-27 00:00