Is "bringing down" the conjugation bar ¯¯ into the complex inner product's angled brackets <> allowed $$\overline{\langle x,y\rangle } = {\langle \overline{x},\overline{y}\rangle } ?$$
Hello, first time asker here!
I have looked online for proofs of the above statement. Whilst some stackexchange questions imply that the above statement is true, I have not been able to find a proof.
Why conjugate when switching order of inner product?
Context: I am learning about complex inner product spaces and trying to prove result 7.14 from this question Why does the fact that "Tv is orthogonal to v for all v implies T is the zero operator"? . However the proof does not seem very intuitive to me. Here's an image link since one needs 10 reputation to post images
https://i.stack.imgur.com/XBk1a.png
Before searching for an answer online, I proved via sesquilinearity of the complex inner product that $$\langle T\left( u+v\right) ,(u+v)\rangle =0  \implies \langle Tu,v\rangle =\overline{-\langle u,Tv\rangle }$$ After looking at the image, I thought 'hmm I can replace (u+v) with (u+iv) and that gives me:' $$\langle T\left( u+iv\right) ,(u+iv)\rangle =0  \implies \langle Tu,iv\rangle =\overline{-\langle u,T(iv)\rangle }$$ then perhaps(?) I could do this:
$$ \implies \langle Tu,iv\rangle ={-\langle \overline{u},\overline{T(iv)}\rangle }$$
$$ \stackrel{?}{\implies} \langle Tu,iv\rangle ={-\langle \overline{u},\overline{T}(\overline{iv})\rangle }$$
In the question assigned to me I am not told if T is a matrix or a linear operator, or anything else which is cool and could go into a < , > .
Thank you for reading through this verbose context text, any thoughts would be much appreciated. Anyway,
my understanding is imperfect-I would be very grateful to see a proof (or even a counterexample :o) of
$$\overline{\langle x,y\rangle } = {\langle \overline{x},\overline{y}\rangle }$$
A simple "yes. This is true" would also be helpful : )
 A: As pointed out in the comments, there is no definition of complex conjugation for vectors in a general inner product space.
I assume therefore that you are working on $\Bbb{C}^n$ and with componentwise complex conjugation defined by$$x = (x_1,\ldots, x_n) \implies \overline{x} = (\overline{x_1},\ldots, \overline{x_n}).$$
The result is not true in general. Recall that for any inner product $\langle\cdot,\cdot\rangle$ on $\Bbb{C}^n$ there exists a positive-definite $n\times n$ complex matrix $A \ge 0$ such that
$$\langle x,y\rangle = y^t Ax, \qquad \text{ for all }x,y \in \Bbb{C}^n.$$
Now your result will hold if and only if $A$ is a real matrix. Indeed, suppose $A$ is real. Since complex conjugation commutes with the transpose and matrix multiplication, for all $x,y \in \Bbb{C}^n$ we have
$$\langle \overline{x},\overline{y}\rangle = \overline{y}^t A\overline{x} = \overline{y^t} \overline{Ax} = \overline{y^t Ax} = \overline{\langle x,y\rangle}.$$
Conversely, suppose that $A$ is not real. Then there exists indices $1 \le r,s\le n$ such that $A_{rs} \in \Bbb{C}\setminus \Bbb{R}$. If we denote the canonical vectors in $\Bbb{C}^n$ by $e_1,\ldots, e_n$, we have
$$\langle \overline{e_s}, \overline{e_r}\rangle = \langle e_s, e_r\rangle = e_r^tAe_s = A_{rs} \ne \overline{A_{rs}} = \overline{e_r^tAe_s } = \overline{\langle e_s, e_r\rangle}.$$
