Exercise 2.2.7

Answers

If we have ${({[T]}_{β}^{γ})}_{ij} = A_{ij}$ , this means

T (v_{j}) = \sum_{i = 1}^{m} A_{ij} w_{i} .

And hence we have

aT (v_{j}) = \sum_{i = 1}^{m} a A_{ij} w_{i} .

and thus ${(a {[T]}_{β}^{γ})}_{ij} = a A_{ij}$ .

jephianlin

2011-06-27 00:00