I want to understand the linearity of an inner product. Let's say I have a linear operator $S_U:V\rightarrow V$, $\forall v\in V \ S_U(v)=2w-v$ such that $w$ is the orthogonal projection of $v$ onto $U\subset V$ .
I am trying to calculate $\langle S_U(v)\ | \ S_U(u)\rangle$ for two vectors $u,v \in V$.
by definition of $S_U$ I can write $\langle 2w_1 - v\ | \ 2w_2 - u\rangle$ such that $w_1$ is the orthogonal projection of $v$ onto $U$ and $w_2$ is the orthogonal projection of $u$ onto $U$.
Then, by lineraity of the inner product can I write:
$\langle 2w_1 - v\ | \ 2w_2 - u\rangle = \langle 2w_1\ | \ 2w_2 \rangle - \langle v\ | \ 2w_2 \rangle -\langle 2w_1\ | \ u \rangle +\langle v\ | \ u \rangle $ ?
The inner product is a bilinear function. This means that it is linear in both its two arguments, so we have: $$ \langle 2w_1 - v\ | \ 2w_2 - u\rangle = \langle 2w_1\ | \ 2w_2-u \rangle - \langle v\ | \ 2w_2-u \rangle= \langle 2w_1\ | \ 2w_2 \rangle -\langle 2w_1\ | \ u \rangle - \langle v\ | \ 2w_2 \rangle +\langle v\ | \ u \rangle $$ as you have found.