$A=B+BJ$
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 ?
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.