Let $V$ be a finite dimension unitary space. Let $T$ be a non negative transformation from $V$ to $V$.
Assuming there exists a natural $n$ such that $T^n = I$.
Does $T = I $ ?
I have a hard time finding a counter example so I tend to believe it is true, but I also have a hard time proving it. Any ideas?
Since $T^n=I$, any eigenvalue $\lambda$ satisfies $\lambda^n=1$, and since you are assuming the eigenvalues are all non-negative real numbers, the only possible eigenvalue is 1. Since its eigenvalues are real, $T$ may be put into its Jordan form $J$ over $\mathbb{R}$, which still satisfies $J^n=I$ so it must be diagonalizable. But it only has the eigenvalue $1$ so it must be the identity.
Being unitary isn't required.