How to compute the inverse of a unitary B when it's defined only by B|0⟩=XXXX?

I got confused when reading an article about linear combination of unitary method. It shows its process as the following: I can't figure out how the effect of $$B^{\dagger}$$ is calculated with the limited information of $$B$$.

Here is the article mentioned.

Thanks to you all!

The trick is that you don't need to calculate the inverse of $$B$$. What you really want to evaluate is $$(\langle 0|\otimes I)(B^\dagger \otimes I)(\text{select}(V))(B\otimes I).$$ So, the point is that you only need $$\langle 0|B^\dagger$$ which is the Hermitian conjugate of $$B|0\rangle$$, which you know.
• Thanks for the clear answer! I am getting it like this: since directly deducing the actual form of $W|0\rangle|\psi\rangle$ is hard, we split it into two parts: the part whose first register is $|0\rangle$ and the remaining part. For the part whose first register is $|0\rangle$, we reduce $|0\rangle$ to 1 by multiplying $\langle 0|\otimes I$ in the left, and can see that it can be deduced to $\frac 1s|0\rangle U|\psi\rangle$ in the end. Therefore the remaining part would be some $\sqrt {1-\frac 1{s^2}}|\Phi\rangle$ with $|\Phi\rangle$ orthogonal to $|0\rangle$. Am I right? Jul 17 '20 at 14:42