A complex matrix with real eigenvalues Let  $A$  be  a  $10\times 10$ matrix  with complex entries  and  all eigenvalues  non-negative  real  numbers  and  at  least one  eigenvalue strictly positive . Then there  exist  a  matrix  $B$  such  that 
$A$. $AB-BA=B$
$B$. $AB-BA=A$
$C$. $AB+BA=A$
$D$. $AB+BA=B$
Complex  matrix  with  real  eigenvalues  , is  that  Hermitian  then? 
Some  lead  please  I  am clueless.

Of course A and D must be true: we can take $B = 0$.

Note that $\operatorname{trace}(AB - BA) = 0$ for any $A,B$ (why)?  So, B can only be true if $A$ has a trace of $0$, which is necessarily not the case from the premise of the question.

The answer to C is yes. Try $B = \frac 12 I$.
