= Tripotent linear operator
{title2=$T^3=T$}
For a <linear operator> on a real <vector space>, the identity $T^3=T$ gives an explicit <eigenspace> decomposition
$$
V=\ker T\oplus\ker(T-I)\oplus\ker(T+I).
$$
Indeed, the <linear operators>
$$
P_0=I-T^2,\qquad P_+=\frac{T^2+T}{2},\qquad P_-=\frac{T^2-T}{2}
$$
sum to $I$. Their images lie respectively in the <eigenspaces> for $0$, $1$, and $-1$, since $TP_0=0$, $TP_+=P_+$, and $TP_-=-P_-$. 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>.
Back to article page
