distribution in an inner product (inner product spaces)

Question

Sorry for the dumb question.

Suppose I'm in a general inner product space. How would I compute something like the following?

$$\langle x - \alpha y, x - \alpha y\rangle$$

where $\alpha$ is a complex scalar. Is the following right?

$$\begin{align} \langle x - \alpha y, x - \alpha y\rangle &= \langle x, x\rangle + \langle x, -\alpha y\rangle + \langle -\alpha y, x\rangle + \langle -\alpha y, -\alpha y\rangle \\ &= \|x\|^2 + \Re({-\alpha})\langle x, y\rangle + \Re{(-\overline{\alpha}})\langle x, y\rangle + |-\alpha|^2\|y\|^2 \\ &= \|x\|^2 + 2\Re{(-\alpha)}\langle x, y\rangle + |\alpha|^2\|y\|^2 \end{align}$$

It is mostly the negative sign that is throwing me off. I wasn't sure whether I should use minus instead of the plus between my terms.

Answer 1

$$ <x,y> = \overline{<y,x>}\\ <ax,y> = a<x,y> \\ <x+y,z> = <x,z> + <y,z> $$

Thus

$$ <x-ay, x-ay> = <x,x-ay> + <-ay, x - ay > = <x, x - ay > -a <y,x - ay> = \\ =\overline{<x - ay, x>} - a \overline{<x-ay,y>} = \overline{<x,x> + <-ay,x> } - a (\overline{<x,y> + <-ay,y>}) = \overline{<x,x>} + \overline{<-ay,x>} -a(\overline{<x,y>} + \overline{<-ay,y>}) = \\ = \|x\|^2 - \overline{a}<x,y> -a<y,x> + |a|^2\|y\|^2 $$

Answer 2

The "real part" should appear after summing the two central terms because such a sum has the form $z+\overline{z}$ which equals to $2\Re(z)$ (in your case you have this for $z=\overline{-\alpha}\langle x,y\rangle$).

Here is the step-by-step:

$$\begin{align} \langle x - \alpha y, x - \alpha y\rangle &= \langle x, x\rangle + \langle x, -\alpha y\rangle + \langle -\alpha y, x\rangle + \langle -\alpha y, -\alpha y\rangle \\ &= \langle x, x\rangle + (\overline{-\alpha})\langle x, y\rangle + (-\alpha)\langle y, x\rangle + (-\alpha)(\overline{-\alpha})\langle y, y\rangle \\ &= \langle x, x\rangle + (\overline{-\alpha})\langle x, y\rangle + (-\alpha)\overline{\langle x, y}\rangle + (-\alpha)(\overline{-\alpha})\langle y, y\rangle \\ &= \|x\|^2 + (\overline{-\alpha})\langle x, y\rangle + \overline{(\overline{-\alpha})\langle x, y\rangle} + |-\alpha|^2\|y\|^2 \\ &= \|x\|^2 + 2\Re\left((-\alpha)\langle x, y\rangle\right) + |\alpha|^2\|y\|^2 \end{align}$$