Exercise 6.4.11

Answers

1.

We prove it by showing the value is equal to its own conjugate. That is,

\bar{⟨ T (x), x ⟩} = ⟨ x, T (x) ⟩

⟨ x, T^{∗} (x) ⟩ = ⟨ T (x), x ⟩ .

2.

As Hint, we compute

0 = ⟨ T (x + y), x + y ⟩

= ⟨ T (x), x ⟩ + ⟨ T (x), y ⟩ + ⟨ T (y), x ⟩ + ⟨ T (y), y ⟩

= ⟨ T (x), y ⟩ + ⟨ T (y), x ⟩ .

That is, we have

⟨ T (x), y ⟩ = - ⟨ T (y), x ⟩ .

Also, replace $y$ by $iy$ and get

⟨ T (x), iy ⟩ = - ⟨ T (iy), x ⟩

and hence

- i ⟨ T (x), y ⟩ = - i ⟨ T (y), x ⟩ .

This can only happen when

⟨ T (x), y ⟩ = 0

for all $x$ and $y$ . So $T$ is the zero mapping.

3.

⟨ T (x), x ⟩

is real, we have

⟨ T (x), x ⟩ = ⟨ x, T (x) ⟩ = ⟨ T^{∗} (x), x ⟩ .

This means that

⟨ (T - T^{∗}) (x), x ⟩ = 0

for all $x$ . By the previous argument we get the desired conclusion $T = T^{∗}$ .

jephianlin

2011-06-27 00:00