Exercise 6.4.6

Answers

1.

Observe the fact

T_{1}^{∗} = {(\frac{1}{2} (T + T^{∗}))}^{∗} = \frac{1}{2} (T^{∗} + T) = T_{1}

and

T_{2}^{∗} = {(\frac{1}{2 i} (T - T^{∗}))}^{∗} = - \frac{1}{2 i} (T^{∗} - T) = T_{2} .

2.

Observe that

T^{∗} = U_{1}^{∗} - i U_{2}^{∗} = U_{1} - i U_{2}

. This means that

U_{1} = \frac{1}{2} (T + T^{∗}) = T_{1}

and

U_{2} - \frac{1}{2} (T - T^{∗}) = T_{2} .

3.

Calculate that

T_{1} T_{2} - T_{2} T_{1} = \frac{1}{4 i} (T^{2} - T T^{∗} + T^{∗} T - {(T^{∗})}^{2}) - \frac{1}{4 i} (T^{2} + T T^{∗} - T^{∗} T - {(T^{∗})}^{2})

= \frac{1}{2 i} (T^{∗} T - T T^{∗}) .

It equals to $T_{0}$ if and only if $T$ is normal.

jephianlin

2011-06-27 00:00