2
$\begingroup$

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 $$ M_{a_1, a_2, \cdots, a_{2^n}} - \operatorname{diag} ( b_1, b_2, \cdots, b_{2^n})? $$

$\endgroup$
2
  • $\begingroup$ Please add at least one $2\times 2$ example to help us. $\endgroup$ Mar 26, 2014 at 1:00
  • $\begingroup$ You may regard $M_{a_1,\dots,a_{2^n}}$ as the `addition table' for $n$-dimensional vector space over $\mathbb F_2$. $\endgroup$ Mar 26, 2014 at 8:30

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$\let\eps\varepsilon$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 $2^n$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.