I am having some trouble with the inner product and the Gram-Schmidt process for complex vectors as I am trying to learn it on my own. This is mainly due to the discrepancy with my text book and what I found online.
1) inner product:
if you refer to: What is the dot product of complex vectors?
you would notice that the author of the answer wrote that $$\mathbf{A} \cdot \mathbf{B} = \sum_i a_i \bar{b_i} $$
Where I understand that this implies that: $$\mathbf{A} \cdot \mathbf{B} = \mathbf{A} \cdot \mathbf{B}^* $$
Where $\mathbf{B}^*$ is the complex conjugate of the matrix B
My textbook however, wrote this:
$$\langle \mathbf{A} \mid \mathbf{B} \rangle = \sum_{i=1}^N a_i^*b_i $$
which kind of contradicts what i found online. So who is correct?
2) The Gram-Schmidt for complex vectors
My textbook writes that if the vectors $\mathbf{x_i}$ are not mutually orthogonal, and we want to construct new vectors $\mathbf{z_i}$ that are orthogonal, the process is:
$$z_1 = x_1$$ $$z_2 = x_2 - [(\hat{z_1}^\dagger)x_2]\hat{z_1} $$ $$...$$
for $z_2$, I understand that:
$$[(\hat{z_1}^\dagger)x_2]\hat{z_1} = \frac{{z_1^\dagger}{x_2}}{\|z_1\|^2} z_1 = \frac{\langle \mathbf{z_1}\mid\mathbf{x_2}\rangle}{\langle \mathbf{z_1}\mid\mathbf{z_1} \rangle} \mathbf{z_1}$$
But if we refer to wikipedia: https://en.wikipedia.org/wiki/Gram–Schmidt_process
We can see that wiki clearly defines
$$\mathbf{u_2} = \mathbf{v_2} - \operatorname{proj}_\mathbf{u_1} (\mathbf{v_2})=\mathbf{v_2}-\frac{\langle \mathbf{v_2} \mid \mathbf{u_1} \rangle}{\langle \mathbf{u_1} \mid \mathbf{u_1} \rangle} \mathbf{u_1}$$
which in my textbook example, would be:
$$\mathbf{z_2} = \mathbf{x_2} - \operatorname{proj}_\mathbf{z_1} (\mathbf{x_2})=\mathbf{x_2}-\frac{\langle \mathbf{x_2} \mid \mathbf{z_1} \rangle}{\langle \mathbf{z_1} \mid \mathbf{z_1} \rangle} \mathbf{z_1}$$
So you can see, if both the wiki and my textbook is correct, then this implies that:
$${\langle \mathbf{z_1}\mid\mathbf{x_2}\rangle} = {\langle \mathbf{x_2}\mid \mathbf{z_1} \rangle} $$
But this does not make sense! Because:
$${\langle \mathbf{z_1}\mid\mathbf{x_2}\rangle} = z_1^\dagger x_2 \neq x_2^\dagger z_1 = {\langle \mathbf{x_2} \mid \mathbf{z_1} \rangle} $$
For both questions, I don't think anyone is wrong, rather, I feel that I may be missing something.
Sorry if both are stupid questions, but thanks in advance!