Is "bringing down" the conjugation bar ¯¯ into the complex inner product's angled brackets <> allowed

Question

$$\overline{\langle x,y\rangle } = {\langle \overline{x},\overline{y}\rangle } ?$$

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.sstatic.net/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 : )

19/03/24 edit: $\overline{x}$ refers to componentwise complex conjugation. The answer is no, one cannot bring down the conjugation bar.

Counterexample: let $V=\mathbb{C}^n$ and $$ F(v, w)=\bar{v}^t A w $$ for some $A \in M_{n \times n}(\mathbb{C})$. Then $F$ is a sesquilinear form, and $F$ is conjugate symmetric if and only if $A=\bar{A}^t$ as for all $i, j=1, \cdots, n$ we have $$ F\left(e_i, e_j\right)=\bar{e}_i^t A e_j=a_{i j} \text { and } \overline{F\left(e_i, e_j\right)}=\overline{\bar{e}_i^t A e_j}=\bar{a}_{i j} $$ Taking F to be $\langle \cdot, \cdot\rangle$ one replicates a counterexample similar to the one at the end of mechanodroid's answer below. To the Math StackExchange community: thank you for all of the comments and the answer. It was a nice experience reviewing this post and then understanding responses a year later.

Answer 1

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 $$\langle \overline{e_s}, \overline{e_r}\rangle = \langle e_s, e_r\rangle = e_r^tAe_s = A_{rs} \ne \overline{A_{rs}} = \overline{e_r^tAe_s } = \overline{\langle e_s, e_r\rangle}.$$