Which of the following statement is true about complex matrices? Let A be a 10×10 matrix with complex entries such that all its eigenvalues are non-negative real numbers, and at least one eigenvalue is positive.
Which of the following statements is always false ?
A. There exists a matrix B such that AB − BA = B
B. There exists a matrix B such that AB − BA = A
C. There exists a matrix B such that AB + BA = A
D. There exists a matrix B such that AB + BA = B.
i know that AB - BA is never equal to identity matrix.But how should i prove it in general? is there any formula or trick used, either it will take hell lot of time by hit and trial method.
 A: This why knowing the proof of some fact is often much more useful than knowing the fact itself. You know that $AB-BA = I_n$ has no solution, but why? Because when you apply the trace operator,
$$\operatorname{tr}(AB-BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = \operatorname{tr}(AB) - \operatorname{tr}(AB) = 0$$
whereas $\operatorname{tr}(I_n) = n \neq 0$.
You're (correctly) guessing that this will useful here. So here what you can try to do is to apply the trace operator to each equation and see if you reach a contradiction. You're also given information about the eigenvalues of $A$: they're all nonnegative and one of them is positive. If you recall, the trace of a matrix is the sum of its eigenvalues (with multiplicity). This implies that $\operatorname{tr}(A) > 0$.
But if statement B were true, you would find
$$\operatorname{tr}(A) = \operatorname{tr}(AB-BA) = 0,$$
a contradiction. So there cannot exist a matrix $B$ such that $AB-BA = A$. I leave it to you to find examples for all the other statements A, C, D.
