**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.

## No comments, be the first one to comment !