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Apr 13, 2017 at 12:19 history edited CommunityBot
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Oct 3, 2015 at 20:28 comment added Majordomus 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 $A'$.
Oct 3, 2015 at 17:21 comment added Carlo Beenakker 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?
Oct 3, 2015 at 12:02 comment added Majordomus 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 $A'$.
Oct 3, 2015 at 8:12 comment added Carlo Beenakker 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 $B_i$".
Oct 2, 2015 at 19:40 comment added Carlo Beenakker I'm afraid the answer is "no", knowledge of the eigenvalues of $B_i$ doesn't help if you seek the eigenvalues of $A'$.
Oct 2, 2015 at 17:17 history edited Majordomus CC BY-SA 3.0
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Oct 2, 2015 at 17:10 history edited Majordomus CC BY-SA 3.0
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Oct 2, 2015 at 15:51 history asked Majordomus CC BY-SA 3.0