Exercise 6.1.30

Answers

First we may observe that the condition for norm on real vector space is loosen than that on complex vector space. So naturally the function ∥⋅∥ is still a norm when we regard V as a vector space over . By Exercise 6.1.27, we’ve already defined a real inner product [⋅,⋅] on it since the parallelogram law also holds on it. And we also have

[x,ix] = 1 4[x + ix2 x ix2] =
= 1 4[x + ix2 (i)(x + ix)2] = 1 4[x + ix2 | i|(x + ix)2] = 0.

So by Exercise 6.1.29 we get the desired conclusion.

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2011-06-27 00:00
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