- For complex vector spaces, i.e. vector spaces with scalars from the field $C $of complex numbers, inner products must have slightly different properties.
To see why, consider the following vectors in $C^2$ : $u = (1 + i, 2 − i)$, $ v = (2 − 3i, 4 + i)$ .
(a) Show that the usual dot product for $R^2$ fails to produce a real positive result for $<u, u>$.
(b) For this reason, we define the inner product for $C^ 2$ as $<u, v> = \overline u_1v_1 + \overline u_2v_2$ where $\overline u$ denotes the complex conjugate of u. Show that using this inner product $<u, u>$ is now real and positive. However show that the symmetry axiom $<u, v>$ = $<v, u>$ is now not satisfied. For this reason, the symmetry axiom for complex inner product spaces becomes: $<u, v>$ = complex conjugate of($<v, u>$) which reduces to the usual axiom for real vector spaces.
(c) The angle θ between two vectors u and v is defined by ||u||||v|| cos θ = R(< u, v >), where R(< u, v >) denotes the real part of the inner product. Find the angle between u and v.
I've posted the entire question to give a background of it. Questions (a) and (b) are relatively easy and straightforward but how would one approach question (c)?
One approaches this problem by doing exactly what it says. Calculate these values: $$\|u\| = \sqrt{\langle u, u \rangle} = \sqrt{\bar u_1 u_1 + \bar u_2 u_2}$$ $$\|v\| = \sqrt{\langle v, v \rangle} = \sqrt{\bar v_1 v_1 + \bar v_2 v_2}$$ $$\langle u, v \rangle = \bar u_1 v_1 + \bar u_2 v_2$$
Next take the real part of the last one and divide it by the other two. That ratio is $\cos \theta$. Take the inverse cosine, and you're done.