Eigen value and transpose of the $Matrix$ 
Let $A$ and $B$ are linear operator on a finite dimensional vector space $V$ over $\mathbb R$ such that $(AB)^2 = AB$ and $BA$ is invertible , then which of the following are true ?

*

*$AB = BA$ on $V$.


*$ Tr (A)$ is non zero


*$0$ is an eigen value of $B$


*$1$ is an eigen value of$A$

Since $BA$ is invertible , then $A$ and $B$ are invertible , and eigen value of $AB$ are $0$ and $1$ and $AB$ is dioganlizabe , because minimal polynomial of $AB$ is $x^2 -x = 0 $, and we know that the eigen value of $AB$ and $BA$ are same counted with the multiplecity of non -zeroeigen value and $BA$ is invertible . So the eigen value of $BA$ is $1$ with multiplecity $n$.
I am confused. my concept is contradicting.please give a me way how to solve.
Thank you.
 A: $A$ and $B$ are square matrices. Since $BA$ is invertible, $A$ and $B$ are both invertible, because $\det(B)\det(A)=\det(BA)\neq 0$.
Thus $ABAB=AB$ implies $AB=I$, and $B$ is the inverse of $A$, thus $AB=BA=I$.
Since $B$ is invertible, it does not have $0$ as an eigenvalue.
You can't say anything about (2) and (4) because all invertible matrices $A$ satisfy your conditions. And of course, not all invertible matrice satisfy (2) or (4), though some do (for example the identity matrix).
A trivial counter example to (2) is a permutation matrix associated to a derangement (thus only $0$s on the diagonal). A trivial counter example to (4) is any multiple $aI$ of the identity matrix with $a\neq 1$.
A: can we do this?  it looks like $AB = I.$ here is the reason. we have $$I = (AB)^2 =  ABAB = AB$$ multiplying by $A$ on the right and $(BA)^{-1}$ on the left  gives us $$ABA(BA)(BA)^{-1}=A(BA)(BA)^{-1}\to ABA = A \to (BA)^2 = (BA) $$ and that $BA$ is invertible implies that only eigenvalues of $BA$ is $1.$
i think $(BA)^2 = BA$ and $1$ is the only eigenvalue must imply  that $$BA = I $$ which in turn give $AB = I.$
