Introduce the $2^{n-1}\times 2^{n-1}$ matrix $B\_n$ recursively as follows: $B\_1(b\_1)=\begin{pmatrix} b\_1\end{pmatrix}$ and
$$B\_n(b\_1,\dots,b\_n)=\begin{pmatrix} B\_{n-1}(b\_1,\dots,b\_{n-1})& b\_nJ\_{n-1}\\ b\_nJ\_{n-1}&B\_{n-1}(b\_1,\dots,b\_{n-1})
\end{pmatrix}.$$
Here $J\_n$ is a $2^{n-1}\times 2^{n-1}$ matrix with $1$'s on the antidiagonal and zeros elsewhere.


*Example.* For $n=2$ and $n=3$, we have
$$B\_2(b\_1,b\_2)=\begin{pmatrix} b\_1&b\_2\\b\_2&b\_1\end{pmatrix}
\qquad\text{and} \qquad
B\_3(b\_1,b\_2,b\_3)=\begin{pmatrix} b\_1&b\_2&0&b\_3\\b\_2&b\_1&b\_3&0\\0&b\_3&b\_1&b\_2\\b\_3&0&b\_2&b\_1\end{pmatrix}.$$



> 
> **Question.** Is there a closed (or interesting) formula for the determinant $\det(B\_n)$?
> 
> 
> 



dyadically recursive matrices: Part I
For a vector $v$ of length $2^{n-2}$, denote by $v'$ the "mirrored" vector $J\_{n-1}v$. 
If $v$ is an eigenvector of $B\_{n-1}$ for the eigenvalue $\lambda$, then $B\_n\binom v{\pm v'}=(\lambda\pm b\_n)\binom v{\pm v'}$. So this gives you all the eigenvalues of $B\_n$ as the sums $ b\_1\pm\cdots\pm b\_n$ (note that $ b\_1$ never has a "$-$" sign) and the eigenvectors as a certain subset (respecting the dyadically recursive structure) of the set of all vectors $ (1,\pm1,...,\pm 1)$ where an even number of minus signs is used. 


The determinant is: $\det(B\_n)=\prod(b\_1\pm b\_2\pm\cdots\pm b\_n)$.


