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.