Exercise 6.4.19

Answers

1.

It comes from that

⟨ (T + U) (x), x ⟩ = ⟨ T (x), x ⟩ + ⟨ U (x), x ⟩ > 0

and ${(T + U)}^{∗} = T^{∗} + U^{∗} = T + U$ .

2.

It comes from that

⟨ (cT) (x), x ⟩ = c ⟨ T (x), x ⟩ > 0

and ${(cT)}^{∗} = \bar{c} T^{∗} = cT$ .

3.

It comes from that

⟨ T^{- 1} (x), x ⟩ = ⟨ y, T (y) ⟩ > 0,

where $y = T^{- 1} (x)$ . Note that

{(T^{- 1})}^{∗} T^{∗} = {(T T^{- 1})}^{∗} = I .

So we have ${(T^{- 1})}^{∗} = {(T^{∗})}^{- 1} = T^{- 1}$ .

jephianlin

2011-06-27 00:00