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Let $A$ be a $10 \times 10$ matrix with complex entries s.t. all eigenvalues are non negative real and at least one eigenvalue is positive. Then which of the following statements is always false?

A. there is a matrix $B$ s.t. $AB-BA=B$

B.there is a matrix $B$ s.t. $AB-BA=A$

C.there is a matrix $B$ s.t. $AB+BA=A$

D.there is a matrix $B$ s.t. $AB+BA=B$

I am new comer in Liner algebra. I have studied finite dimensional vector space, eigen value eigen vector from a book of G. Strang. I have found this in a competitive exam. I have no idea how to tackle this question. Can anybody help to solve this problem.

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    $\begingroup$ A. and D. are always true with $B$ the null matrix. C is true with $B=(1/2)I$, where $I$ is the identity matrix. $\endgroup$
    – enzotib
    Dec 15 '14 at 19:06
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Condition B is impossible to fulfill. To see this, take the trace of the equation. As $tr(AB)=tr(BA)$, if follows that in order to fulfill B, the trace of $A$ has to be zero.

Now, the trace is the sum of the eigenvalues. By the assumptions, all eigenvalues are non-negative, with at least one of them is positive. Hence, the trace of $A$ is positive, and a matrix $B$ fulfilling B does not exist.

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B can never be true, There exists an inverse of $A$, $A^{-1}$ (This step is wrong. For example the null matrix staisfies the conditions of the question, and is clearly non-invertible. For a correct answer, refer to the answer of daw)

multiply equation b) with $A^{-1}$ from the left, we are left with:

$B-A^{-1} B A= I$ with $I$ the identity. hence

$B-I=A^{-1} B A$

Take the trace on the left and righthand side. Matrices may be permuted cyclicly under the trace (i.e. $Tr(AB)=Tr(BA)$ for all $A$ and $B$) so we get:

$Tr(B-I)=Tr(B)$ And this equation can never be satisfied.

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    $\begingroup$ $A$ is not necessarily invertible from the assumptions. $\endgroup$
    – daw
    Dec 15 '14 at 19:51
  • $\begingroup$ you are perfectly right, excuse me. as I am new to this forum, is it customary to delete my answer? $\endgroup$
    – damazter
    Dec 15 '14 at 19:54
  • $\begingroup$ @damazter: Even better is to edit your answer. I have done it for you this time. Feel free to delete the answer though, if you are not happy with my edit. $\endgroup$
    – TonyK
    Dec 15 '14 at 19:57

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