Let A be a 10×10 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.
i know that AB – BA is never equal to identity matrix.But how should i prove it in general? is there any formula or trick used, either it will take hell lot of time by hit and trial method.
Best Answer
This why knowing the proof of some fact is often much more useful than knowing the fact itself. You know that $AB-BA = I_n$ has no solution, but why? Because when you apply the trace operator, $$\operatorname{tr}(AB-BA) = \operatorname{tr}(AB) - \operatorname{tr}(BA) = \operatorname{tr}(AB) - \operatorname{tr}(AB) = 0$$ whereas $\operatorname{tr}(I_n) = n \neq 0$.
You're (correctly) guessing that this will useful here. So here what you can try to do is to apply the trace operator to each equation and see if you reach a contradiction. You're also given information about the eigenvalues of $A$: they're all nonnegative and one of them is positive. If you recall, the trace of a matrix is the sum of its eigenvalues (with multiplicity). This implies that $\operatorname{tr}(A) > 0$.
But if statement B were true, you would find $$\operatorname{tr}(A) = \operatorname{tr}(AB-BA) = 0,$$ a contradiction. So there cannot exist a matrix $B$ such that $AB-BA = A$. I leave it to you to find examples for all the other statements A, C, D.