Exercise 6.5.8

Answers

Exercise 6.4.10 says that $({(T - iI)}^{- 1}) * = T + iI$ . So check that

{[(T + iI) {(T - iI)}^{- 1}]}^{∗} (T + iI) {(T - iI)}^{- 1}

= {({(T - iI)}^{- 1})}^{∗} {(T + iI)}^{∗} (T + iI) {(T - iI)}^{- 1}

= {(T + iI)}^{- 1} (T - iI) (T + iI) {(T - iI)}^{- 1}

= {(T + iI)}^{- 1} (T + iI) (T - iI) {(T - iI)}^{- 1} = I .

Using Exercise 2.4.10, we get that the operator is unitary.

jephianlin

2011-06-27 00:00