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Question:

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 $$AB+BA=B$$

Discussion:

We know that for two square matrix $$A$$ and $$B$$ of same size, $$Tr(AB)=Tr(BA)$$ ($$TrM$$=Trace of $$M$$ ).

In other words, $$Tr(AB-BA)=0$$ for any two square matrices of the same size.

Since trace of a square matrix is also the sum of its eigenvalues, and $$A$$ has all eigenvalues non-negative with at least one positive eigenvalue, we have $$Tr(A) > 0$$. Taking trace of both sides of $$AB-BA=A$$ we get a contradiction. So there does not exist any $$B$$ such that $$AB-BA=A$$.

Take $$B=0$$ the 10×10 zero-matrix. Then $$AB-BA=B$$ is satisfied. So is $$AB+BA=B$$.

Take $$B=\frac{1}{2}I$$, where $$I$$ is the 10×10 identity matrix. Then $$AB+BA=A$$ is satisfied.