I am currently working on quantum finite automata, when I came across a paper by Ambainis and Nahimovs. I am having some struggle with the "theorem from linear algebra" stated by them in Theorem 1 (the $|q_i \rangle$ are unit vectors):
Let $\alpha_1,...,\alpha_m$ be such that $\|\alpha_1 \|^2 + ... + \|\alpha_m \|^2 = 1$. Then,
- there is a unitary transformation $U_1$ such that $U_1 |q_1 \rangle$ = $\alpha_m |q_m \rangle$ + ... + $\alpha_m |q_m \rangle$.
- there is a unitary transformation $U_2$ such that, for all $i \in \lbrace 1,...,m \rbrace$, $U_2 |q_i \rangle$ is equal to $\alpha_i |q_1 \rangle$ plus some combination of $|q_2 \rangle,...,|q_m \rangle$.
In the second case, we also have $U_2 (\alpha_i |q_1 \rangle + ... + \alpha_m |q_m \rangle) = |q_1 \rangle$.
While the first part is clear to me, I do not see why/how the second statement is conform to their subsequent remark. Writing $U_2 |q_i \rangle = \alpha_i |q_1 \rangle + \sum_{k=2}^m \beta^{(i)}_k |q_k \rangle$, $U_2$ should be unitary if $\sum_{k=2}^m | \beta^{(i)}_k |^2 = 1 - | \alpha_i |^2$ for all $i \in \lbrace 1,...,m \rbrace$. Therefore I think, it is always possible to define $U_2$ in this way indeed.
With that in mind, one would have: $$ U_2 (\alpha_i |q_1 \rangle + ... + \alpha_m |q_m \rangle) = \sum_{i=1}^m (| \alpha_i |^2 |q_1 \rangle) + \sum_{i=1}^m \alpha_i \sum_{k=2}^m | \beta^{(i)}_k |^2 \\ = |q_1 \rangle + \sum_{i=1}^m \alpha_i \sum_{k=2}^m | \beta^{(i)}_k |^2. $$ But this would contradict their remark, unless the latter term in the above equation is always zero (otherwise $U_2$ would not even be unitary). If so, is it generally possible to define the $\beta^{(i)}_k$ in this way? I got some results for the case that m=2 by solving a system of linear equations but this procedure seems not practical for higher dimensions. Or is there a contradiction?