What is the name for the type of matrices?

Question

Let $ K $ be a field. We can recursively define matrices as $ M_{a} = (a)$ for any $ a\in K $ and $$ M_{a_1, \cdots, a_{2^i}} = \begin{pmatrix} M_{a_1, \cdots, a_{2^{i-1}}} & M_{a_{2^{i-1} +1}, \cdots, a_{2^i}}\\ M_{a_{2^{i-1} +1}, \cdots, a_{2^i}} & M_{a_1, \cdots, a_{2^{i-1}}}\\ \end{pmatrix} $$ when $ i>0 $ and $a_j\in K$.
What is the name for the type of matrices? Let $ a_1, a_2, \cdots, a_{2^n} $ and $ b_1, b_2, \cdots, b_{2^n} $ be two list of elements in $ K $. Is there a formula for the eigenvalues of $$ M_{a_1, a_2, \cdots, a_{2^n}} - \operatorname{diag} ( b_1, b_2, \cdots, b_{2^n})? $$

Answer 1

As for the eigenvalues there is neat trick I learned once: A matrix of the form $\left(\begin{smallmatrix} A & B \\ B & A\end{smallmatrix}\right)$ is conjugate (by $\left(\begin{smallmatrix} I & I \\ I & -I \end{smallmatrix}\right)$) to $\left(\begin{smallmatrix} A+B & \\ & A-B\end{smallmatrix}\right)$. Therefore there is a recursion for the (multi)set of eigenvalues of your matrix.

EDIT: I just realized that the diagonal matrix you're subtracting messes up this special form of the $M_a$ so that what I described leads only to a solution for the special case of a scalar matrix $b_1=...=b_{2^n}$. Maybe you still can use it anyway.

Answer 2

$\let\eps\varepsilon$Making Johannes' answer more exact, one may see that all matrices $M_{a_1\dots a_{2^n}}$ have a common eigenbasis. Namely, for every $\eps_0,\dots,\eps_{n-1}\in\{0,1\}$ one may choose $\mathbf e_i=(e_{i1},\dots,e_{in})$ where $e_{ij}=(-1)^{\eps_0p_{j0}+\dots+\eps_{n-1}p_{j,n-1}}$ and $j-1=\overline{p_{j,n-1}\dots p_{j0}}$ is the binary expansion of $j-1$. The corresponding eigenvalue is $\sum_{j=1}^{2^n}a_j(-1)^{\eps_0p_{j0}+\dots+\eps_{n-1}p_{j,n-1}}$. (The transition matrix is a known Hadamard matrix of order $2^n$.)