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This is a repost of the same question on MSE (with no reply/comment): https://math.stackexchange.com/questions/1454314/eigenvalues-of-a-partitioned-self-adjoint-matrix

I would be grateful just for a pointer to a paper or book.

Assume a positive-definite $2\times2$ matrix ($a_3=a_2^*$) $$ A=\pmatrix{a_1 & a_2\\a_3 & a_4} $$ whose eigenvalues are known. Now let's have four different complex $d\times d$ matrices $B_1,B_2,B_3,B_4$ (self-adjoint but not necessarily positive-definite), whose eigenvalues are also known. Can there be said something about the eigenvalues of the following $2d\times 2d$ matrix $$ A'=\pmatrix{a_1B_1 & a_2B_2\\a_3B_3 & a_4B_4} $$ ?

In particular:

Is there an algorithm for calculating the eigenvalues of $A'$ from the eigenvalues of $A$ and $B_i$?

Can it be simplified if $B_2=B_3^*$ where $*$ is conjugate transpose and so $A'$ is self-adjoint?

EDIT: fixed typo in the last sentence ($*$ added)

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    $\begingroup$ I'm afraid the answer is "no", knowledge of the eigenvalues of $B_i$ doesn't help if you seek the eigenvalues of $A'$. $\endgroup$
    – Carlo Beenakker
    Commented Oct 2, 2015 at 19:40
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    $\begingroup$ you might want to think of it like this: knowledge of the eigenvalues of $B_i$ doesn't tell you if $B_i$ has off-diagonal elements or not; but obviously that information is essential if you want to know the eigenvalues of $A'$; so obviously there cannot be "an algorithm for calculating the eigenvalues of $A'$ from the eigenvalues of $A$ and $B_i$". $\endgroup$
    – Carlo Beenakker
    Commented Oct 3, 2015 at 8:12
  • $\begingroup$ True, but with the knowledge of the eigenvalues of $B_i$ comes the knowledge of their eigenvectors (the same for $A$). So that's a lot of useful information that seems should lead at least to an estimate of the eigenvalues of $A'$. $\endgroup$
    – Majordomus
    Commented Oct 3, 2015 at 12:02
  • $\begingroup$ well sure, if you know the eigenvalues and eigenvectors of $B_i$ and $A$, then you have complete knowledge of $A'$ and so you can use any diagonalization algorithm to find its eigenvalues; but that's a trivial (and not particularly useful) statement, is that really your question? $\endgroup$
    – Carlo Beenakker
    Commented Oct 3, 2015 at 17:21
  • $\begingroup$ You are right, I just reacted to the previous comment. I have a complete knowledge of $A$ because it is $2\otimes2$ but $B_i$'s are arbitrary. Probably more accurately, my question was whether there can exist some 'decomposition' of $A'$ or just a 'simplification' in terms of the sublocks... Well, I guess that's it - if the rank of the $B_i$'s is greater than one, I really do not have much to say about $A'$. $\endgroup$
    – Majordomus
    Commented Oct 3, 2015 at 20:28

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