I need to find an inner product in $\mathbb{C}^2$ such that $\{(1,i),(-1,i)\}$ is orthonormal.
I tried the following:
I know that $\langle(1,i),(1,i)\rangle=\langle(-1,i),(-1,i)\rangle=1$ and $\langle(1,i),(-1,i)\rangle=0$.
So I say that $v=(x,y)=\alpha(1,i)+\beta(-1,i)$ and $w=(x',y')=\alpha'(1,i)+\beta'(-1,i)$.
Solving $(x,y)=(\alpha-\beta,\alpha i+\beta i)$, I get that $\alpha=\frac{x-yi}{2}$ and $\beta=\frac{-x-yi}{2}$, and in the same way we get the value of $\alpha'$ and $\beta'$.
So I get that the inner product is $\langle(v,w)\rangle=\langle(x,y),(x',y')\rangle=\overline{'}+\overline{'}=\left(\frac{x-yi}{2}\right)\left(\frac{x'+y'i}{2}\right) +\left(\frac{-x-yi}{2}\right)\left(\frac{-x'+y'i}{2}\right)$
But when I try it, it gives me that $\langle(1,i),(-1,i)\rangle=1$
Do you know what I'm doing wrong? Thanks!
The variables $x, y, x'$ and $y'$ are potentially complex (perhaps use $z$'s and $w$'s next time!), and this needs to be taken into account when you take conjugates in the final step. We have $$\overline{\alpha'} = \overline{\frac{x'-iy'}{2}} = \frac{\overline{x'} + i \overline{y'}}{2}$$ and $$\overline{\beta'} = \overline{\frac{-x'-iy'}{2}} = \frac{-\overline{x'} + i \overline{y'}}{2},$$ which gives us \begin{equation} \begin{split} \alpha \overline{\alpha'} + \beta \overline{\beta'} &= \bigg(\frac{x-iy}{2} \bigg) \bigg(\frac{\overline{x'} + i \overline{y'}}{2} \bigg) + \bigg(\frac{-x-iy}{2} \bigg) \bigg(\frac{-\overline{x'} + i \overline{y'}}{2} \bigg)\\ & = \frac{1}{4}\big(2x\overline{x'} + 2y\overline{y'}\big) \end{split} \end{equation}