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Let $A$ be a $10\times 10$ 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 just need a hint to start thinking.

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B is always false. The trace of $AB-BA$ is always zero, but the trace of $A$, which is the sum of its eigenvalues, is positive.

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Edit: Since I read the question wrong, I will try a different way to at least show which answer you should suspect to be false. Namely, I will show that options A, C, and D are sometimes possible.

For A, let $B$ be the zero matrix. This shows that A holds for any choice of matrix $A$.

For C, Let $B$ be the diagonal matrix with diagonal entries equal to $\frac{1}{2}$. This shows C is true for any matrix $A$.

For D, let $A$ be the matrix $B$ above. This choice of $A$ meets the hypothesis, and for this choice of $A$, D holds for any other matrix $B$, so D is at least sometimes true.

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  • $\begingroup$ Please tell something more. I'm really stuck and confused about it. $\endgroup$ – Katherine Jul 17 '15 at 16:09
  • $\begingroup$ @Katherine If $B$ commutes with $A$, then $AB+BA=AB+AB=2AB$. What kind of matrices commute with all other matrices? $\endgroup$ – TomGrubb Jul 17 '15 at 16:11
  • $\begingroup$ scalar matrices, but I still dont get the point. How can I say that no such matrix B exists. You may reveal the answer. $\endgroup$ – Katherine Jul 17 '15 at 16:16
  • $\begingroup$ @Katherine Oh god, I thought it said always true.....sorry! $\endgroup$ – TomGrubb Jul 17 '15 at 16:16
  • $\begingroup$ I got it, since trace(AB)=trace(BA) implies option B is false. Thanks for help $\endgroup$ – Katherine Jul 17 '15 at 16:18

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