Find a unitary matrix given its action on a class of vectors

Question

Im studying quantum computation and have encountered an interesting problem.

I need to find an unitary matrix $U$, such that

$$U\begin{bmatrix}1 \\ 1 \\ 1\\ \vdots \\ -1 \\ \vdots \\ 1\end{bmatrix} = \frac1k\begin{bmatrix}0 \\ 0 \\ 0\\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}$$

where the vector is $n\times1$ and $k$ is a constant (as necessary). Essentially, every $1$ element goes to $0$, and $-1$ goes to $1$. The $-1$ element can be present anywhere in any row the column vector, and the corresponding row in the image vector should have $1$ at that place and $0$ everywhere else.

Can we solve for $U$? If yes,how? (It need not be unique)

Answer 1

The constraints on $U$ can be collected into the single equation $$U(\mathbb 1_n-2I_n)=\frac1kI_n$$ where $\mathbb 1_n$ is an $n\times n$ matrix of all ones. This has the unique solution $$U=\frac1k(\mathbb 1_n-2I_n)^{-1}=\frac1{2k}\left(\frac1{n-2}\mathbb1_n-I_n\right)$$ for $n\gt2$. This $U$ is real and symmetric, so $UU^*=U^2$, which is diagonal only for $n=4$, so it doesn’t seem possible in general to construct a unitary matrix that will perform this mapping.

Answer 2

The matrix cannot be unitary

$$k\begin{bmatrix}1 \\ 1 \\ 1\\ \vdots \\ -1 \\ \vdots \\ 1\end{bmatrix} = U^{-1}\begin{bmatrix}0 \\ 0 \\ 0\\ \vdots \\ 1 \\ \vdots \\ 0\end{bmatrix}$$

$$U^{-1}=k\begin{bmatrix} -1&1&\cdots&1\\ 1&-1&\cdots&1\\ \cdots\\ 1&1&\cdots&-1 \end{bmatrix}$$

$$U=\dfrac{1}{k}\dfrac{1}{2n-4}\begin{bmatrix} -(n-3)&1&\cdots&1\\ 1&-(n-3)&\cdots&1\\ \cdots\\ 1&1&\cdots&-(n-3) \end{bmatrix}$$

For an unitary real matrix $V^{-1}=V^T$, but clearly $U^{-1}\neq U^T$. No solution for the given conditions.