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.

  • 1
    $\begingroup$ Let me tell you this has nothing to do with $10$.... you can take any $2\times 2$ or $3\times 3 $ matrix... $\endgroup$
    – user87543
    Aug 24 '15 at 16:29
  • 1
    $\begingroup$ It is not necessarily Hermitian $\endgroup$ Aug 24 '15 at 16:32

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 $B = \frac 12 I$.


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