† While trying to prove that $$(A+B)^†=A^†+B^†$$ I have stumbled accross a self proof that seems to validly suggest that $A^† = A^*$ This intuitively seems false but I cannot find where in my proof my mathematical reasoning has failed.
I would like an answer to the question of if I have gone wrong somewhere or if that the adjoint of an operator is indeed equivalent to its complex conjugate within a quantum information viewpoint.
The proof is as follows:
$$\left<v\right|A^†\left|w\right> = \left(\left<w\right|A\left|v\right>\right)^*=\left<w\right|^* A^* \left|v\right>^*= \left<v\right| A^* \left|w\right>$$
Where $^*$ is to notify complex conjugate and $^†$ is to signify the adjoint. This clearly implies that $A^† = A^*$.
I have used the quantum informational identities that $\left<v\right|A^†\left|w\right> = (\left<w\right|A\left|v\right>)^*$ and $ \left<w\right|^* = \left|w\right>$
The answer was explicitly given by Luftbahnfahrer, whereby $\left<v\right| \neq \left|v\right>^*$ and the correct equality is $\left<v\right| = \left|v\right>^\dagger$. The confusion arose because within the course vector notation is rarely used and it is more common to express a qbit as $\left|\psi\right> = \alpha\left|0\right> + \beta \left|1\right>$, this makes the transpose implicit once all ket's are converted to bra's and therefore only a complex conjugate must be taken to obtain the equality from there; i.e., $ \left< \psi \right| = \alpha^{*} \left< 0 \right| + \beta^{*} \left< 1 \right| $
It seems to be a defining property that the associated bra to a ket is the hermitian adjoint of the ket, (i.e. the complex conjugate of the ket expressed as a row vector).
In any case in light of this new information the proof is false, as it should be. Thanks Luftbahnfahrer!