Exercise 6.1.1

Answers

1.: Yes. It’s the definition.
2.: Yes. See the paragraph at the beginning of this chapter.
3.: No. It’s conjugate linear in the second component. For example, in $ℂ$ with standard inner product function we have $⟨ i, i ⟩ = 1$ but not $i ⟨ i, 1 ⟩ = - 1$ .
4.: No. We may define the inner product function $f$ on $ℝ$ to be $f (u, v) = 2 uv$ .
5.: No. Theorem 6.2 does not assume that the dimension should be finite.
6.: No. We may define the conjugate-transpose of any matrix $A$ to be the conjugate of $A^{t}$ .
7.: No. Let $x = (1, 0)$ , $y = (0, 1)$ , and $z = (0, - 1)$ . Then we have $⟨ x, y ⟩ = ⟨ x, z ⟩ = 0$ .
8.: Yes. This means $∥ y ∥ = ⟨ y, y ⟩ = 0$ .

jephianlin

2011-06-27 00:00