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$A=B+BJ$

I am looking for a way to prove that eigenvectors of A and B are same. To do that, I am trying to prove that

the eigenvalues of A is the sum of eigenvalues of B and BJ.

$\lambda_{A+B} = \lambda_1+\lambda_2 : \lambda_1 \in \sigma (A) ,\lambda_2 \in \sigma (B)$

In my case, B and BJ are not commuting. But B,J are symmetric (not diagonal) matrices.

I see the result here that commuting matrices satisfy this. But the matrices above are not commuting and still gives the result. How can it be proved ?

Marc van Leeuwen
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Abhiram V P
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  • What do you even mean by the sum of eigenvalues of $B$ and $BJ$. These are two unordered collections of values, and there are (too) many ways to add an element of one collection to one of the other. – Marc van Leeuwen May 31 '19 at 06:38
  • It holds for commuting matrices. The link given above shows how it is defined. My question is only about extending it to non commuting matrices, with some other constraints maybe. – Abhiram V P May 31 '19 at 06:44
  • For commuting matrices there is a basis of common eigenvectors which allows to order the eigenvalues of the two matrices in a corresponding way. For non commuting matrices such a basis does not exist, so there is no coherent way to do the additions. – Marc van Leeuwen May 31 '19 at 06:50
  • @MarcvanLeeuwen Actually, What I am looking for is to prove that they have common eigenvectors. What are the conditions to be satisfied for having common eigenvectors? – Abhiram V P May 31 '19 at 06:53
  • Having a basis of common eigenvectors is equivalent to (both being diagonalisable and) commuting. Clearly if two matrices are diagonalised on the same basis, then they must commute. Because all diagonal matrices commute with each other. – Marc van Leeuwen May 31 '19 at 06:54
  • @MarcvanLeeuwen So to satisfy that, in my case,$ BBJ=BJB \implies BJ=JB$ Is required. But I have only $BJ=(JB)^T$. – Abhiram V P May 31 '19 at 07:02

1 Answers1

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The statement you wish to prove is just plain wrong. Take $B=\binom{2~~~0}{0~-2}$, $J=\binom{0~~1}{1~~0}$ so that $A=\binom{~2~~~~2}{-2~-2}$. The eigenvalues of $B$ are $\{-2,2\}$, those of $J$ are $\{-1,1\}$, those of $BJ=\binom{~~0~~~2}{-2~~0}$ are complex $(\pm2\mathbf{i})$, and those of $A$ (which is not diagonalisable) are $\{0,0\}$ (one has $A^2=0$). No matter how you want to combine these eigenvalues, they don't add up.

Marc van Leeuwen
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  • I honestly don’t know what is wrong with MSE, but ‘finding a counter example to disprove it’ is all I ever get from here. If it was a general theorem, then I would have found it in some book. There is no need for a human mind to understand and help it. – Abhiram V P May 31 '19 at 14:32
  • The ‘problem’ here is that there is a pattern followed in my case and I am trying to find a mathematical ‘reason’ for that happening. I just cannot see how a ‘counter example’ is going to solve it. I can give counter examples too. What I am looking for is ‘why it is happening’ or ‘what conditions are to be satisfied to give such a trend’. – Abhiram V P May 31 '19 at 14:35
  • I honestly don’t understand what you are expecting from MSE. We are not equipped with paranormal powers. Am I to understand that you have in mind some concrete matrices $B,J,A$ for which what you say is true, but you won't tell us which, yet you want to have an explanation about a property they have that is not a general fact? If you are asking for help with a proof (as many here are) one must assume that you have given people the hypotheses that can be used in such a proof, and that you are asking to show the property in general under those hypotheses. – Marc van Leeuwen May 31 '19 at 15:29
  • ... If there is a counter-example that matches the hypotheses but not the conclusion, then this shows your project is hopeless. That is why people come up with counter-examples, to show you (and other readers) that you are wasting your time. What you cannot expect is for people to guess additional hypotheses that you did not give, but which would allow a proof. Yet sometimes a very kind answerer will say: this is not true in general, but if you assume this or that, then it will be true. In this case somebody might have mentioned commuting as such a hypothesis, but you already dismissed that. – Marc van Leeuwen May 31 '19 at 15:33
  • Commuting matrices follow this. It is given in books. So that itself implies all non commuting matrices is not following it. So finding an example from the pool of those matrices is pointless, as it is obvious. Hence it is not helpful. The doubt was why some non commuting matrices also follow it although the result is for only commuting matrices, hoping to see if there is some other condition which makes it happen. If there was a counter example for commuting matrices following the result, then it would have been useful, as that is supposedly true for general cases. – Abhiram V P May 31 '19 at 16:15