$A$ be a $10*10$ matrix with complex entries s.t. all eigenvalues are non negative real and at least one eigenvalue is positive.

Question

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.

Answer 1

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.

Answer 2

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.