dyadically recursive matrices: Part I

Question

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 $\det(B_n)$?

Answer 1

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: $\det(B_n)=\prod(b_1\pm b_2\pm\cdots\pm b_n)$.

Answer 2

Yes and No. On the one hand, you can prove by a simple induction that $B_n$ commutes with $J_n$. Now, $B_n$ appears to be given blockwise, where all the blocks commute with each other. This allows you to write that the determinant of $B_n$ is the ``determinant of determinants'' (this is false for non-commuting blocks): $$\det B_n=(\det B_{n-1})^2-(\det b_nJ_{n-1})^2=(\det B_{n-1})^2-b_n^{2^{n-1}}.$$ On the other hand, it seems difficult to exploit this relation to give a closed form for $\det B_n$. Recall that even the iteration $z\mapsto z^2-a$ has outstanding complexity, which yields Julia and Fatou sets.