Let $A \in \mathbb{R}^{n \times n}$ be a bidiagonal matrix with non zero elements on its diagonal and super diagonal. Let $B$ be defined as 
$$B=\begin{bmatrix}0&{A} \\{A}^T &0 \end{bmatrix}$$.


Find permutation matrix $P$ such that $C=PBP^T$ is a tridiagonal matrix.


I noticed that permutation matrices with only row exchanges did not satisfy this condition. I am unable to find a way to determine a non-symmetric permutation matrix that converts $B$ into a tridiagonal matrix.



Using permutation matrix to convert a matrix into tridiagonal matrix
The permutation matrix $P$ corresponding to the permutation
$$ \sigma \colon \begin{cases} i \mapsto 2i & \text{ if } i \in [1,n] \\ i \mapsto 2i - 2n - 1 & \text{ if } i \in [n+1,2n] \end{cases} $$
does the job.


**Edit:**
As requested, I am providing some details. Notice that the original matrix $B$ has zero diagonal, which implies that $C$ also has zero diagonal, i.e., it has a non-zero subdiagonal and a non-zero superdiagonal but zeros everywhere else. There are precisely $4n-2$ non-zero entries in $B$ and also precisely $4n-2$ non-zero entries in the matrix $C$ we are looking for. We have (for $n=4$, for example)
$$ B = \begin{pmatrix}
&&&& \color{red}{\*} & \color{red}{\*} && \\
&&&&& \color{red}{\*} & \color{red}{\*} & \\
&&&&&& \color{red}{\*} & \color{red}{\*} \\
&&&&&&& \color{red}{\*} \\
 \color{blue}{\*} &&&&&&& \\
 \color{blue}{\*} & \color{blue}{\*} &&&&&& \\
& \color{blue}{\*} & \color{blue}{\*} &&&&& \\
&& \color{blue}{\*} & \color{blue}{\*} &&&&
\end{pmatrix}$$
and the permutation $\sigma$ transforms $B$ into the matrix
$$ C = \begin{pmatrix}
& \color{blue}{\*} &&&&&& \\
\color{red}{\*} && \color{red}{\*} &&&&& \\
& \color{blue}{\*} && \color{blue}{\*} &&&& \\
&& \color{red}{\*} && \color{red}{\*} &&& \\
&&& \color{blue}{\*} && \color{blue}{\*} && \\
&&&& \color{red}{\*} && \color{red}{\*} & \\
&&&&& \color{blue}{\*} && \color{blue}{\*} \\
&&&&&& \color{red}{\*} & \\
\end{pmatrix}.$$


