Please suggest a proof, or some article, if it is right.
Of course, H,V,C are finite dimensional square matrix(non-negative), C is not zero and [,] represent the commutation.
It seems to me that the following two 3\times3 matrices form a counterexample to your suggestion:
000
H = 100
000
----------------------------------
000
V = 000
010
Thank you. The example of Vladimir is nice. But I want H to have only one eigenvalue 0 (like the generators of a Markov process). I was trying for a perturbation on Markov dynamics . I mean I had additional constraints, dxd matrix H has rank (d-1). For me choice of V by Vladimir is fine.
Example : 3x3 H = { -(a+b), a,b}, {c, -(c+d), d}, {e,f, -(e+f)} } ; a,b,c,d,e,f are nonnegative. I have not been able to construct one V for this H yet :(
In my example you can add to H any matrix A that commutes with V and C. The the commutator [H+A,V] of H + A and V will remain the same C as before, and consequently will commute with H+A and with V. With this trick you can try to get examples with additional properties.
Unfortunately if I take A = { -(a+b), a,b}, {c, -(c+d), d}, {e,f, -(e+f)} } to commute with V and H (above): one gets a,b,c,d=0 and f= - e. The transition rates can not be -ve and I am then forced to get f=0=e, which means A=0. The same happens for 4x4.
Thank you Vladimir for the example, and for teaching me the way to think about it.PK
Dear P.K.,
you have right, with these additional conditions the "small perturbation" idea from my previous message does not work.
Vladimir
If the following is true, something strange happens physically; this forces me to think that following should not be possible-
1) H = Markov generator (non-diagonal elements >=0, diagonal = - "RowSum" )
2) V = any non negative matrix 3) H,V are finite dimensional
then [H,V] can not commute with V and H simultaneously unless [H,V]=0.
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