# 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?

• Let me tell you this has nothing to do with $10$.... you can take any $2\times 2$ or $3\times 3$ matrix... – user87543 Aug 24 '15 at 16:29
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 $B = \frac 12 I$.