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 Incoming links: Real angle in a complex inner product space

Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 11 / iv / Solution 2026-10-25

Using the Hermitian inner product, the scalar product is

1 - i + 2 i^{2} = - 1 - i

. The two squared norms are

6

and

4

. The specified real angle in a complex inner product space therefore has

cos θ = \frac{Re ( - 1 - i )}{6 4} = - \frac{1}{2 6} .

The real part is essential: the complex scalar product itself need not be real.

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Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 11 / v / Solution 2026-10-25

Conjugating the first vector gives

⟨ a, b ⟩ = i e^{- i x} + (1 - i) e^{in x}, Re ⟨ a, b ⟩ = sin x + cos (n x) + sin (n x) .

Their squared norms are

3

and

2

, independently of

x, n

. Thus the specified real angle in a complex inner product space obeys

cos θ = \frac{sin x + cos ( n x ) + sin ( n x )}{6} .

The complex Cauchy-Schwarz inequality ensures that this real quotient lies in

[- 1, 1]

.

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