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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 $ ?

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  • $\begingroup$ Inner products aren't linear, they are either sesquilinear (over $\mathbb{C}$) or bilinear (over $\mathbb{R}$). In either case, you computation is correct. Is that all you are asking? $\endgroup$ – Christian Sykes Jun 9 '17 at 19:03
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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.

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  • $\begingroup$ Can I then use the field commutative property and use bilinearity again to get $\langle 2w_1 - v\ | \ 2w_2 - u\rangle = \langle v \ | \ u \rangle$ ? $\endgroup$ – z00x Jun 9 '17 at 20:24
  • $\begingroup$ No. This is true only if $4\langle w_1\ | \ w_2 \rangle -2\langle w_1\ | \ u \rangle - 2\langle v\ | \ w_2 \rangle=0$ $\endgroup$ – Emilio Novati Jun 9 '17 at 20:30

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