 Tripotent linear operator

For a linear operator on a real vector space, the identity $T^3=T$ gives an explicit eigenspace decompositionIndeed, the linear operatorssum to $I$. Their images lie respectively in the eigenspaces for $0$, $1$, and $-1$, since $T{P}_0=0$, $T{P}_{+}=P_{+}$, and $T{P}_{-}=-P_{-}$. On each of these eigenspaces, its corresponding $P$ is the identity and the other two vanish, which proves uniqueness of the decomposition. Thus a finite-dimensional tripotent linear operator admits a diagonalization with diagonal entries in $\{0,1,-1\}$. The condition $T^2\ne T$ is equivalent to the presence of a nonzero $-1$ eigenspace, and $T^2\ne I$ is equivalent to a nonzero kernel.