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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)

**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 10x10 zero-matrix. Then (AB-BA=B) is satisfied. So is (AB+BA=B).

Take (B=\frac{1}{2}I), where (I) is the 10x10 identity matrix. Then (AB+BA=A) is satisfied.

**What is this topic:**Linear Algebra**What are some of the associated concept:**Trace,Eigenvalue**Book Suggestions:**Linear Algebra done Right by Sheldon Axler

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