 Solution

For this tripotent linear operator, the relation $T^3=T$ gives $T(T^2-I)=0$. Since $T^2\ne I$, there is $v$ with $(T^2-I)v\ne 0$, and that nonzero vector belongs to $\ker T$. Hence $n_T\ge 1$. Also $T\ne 0$, because $T=0$ would imply $T^2=T$. The rank-nullity theorem in dimension three therefore givesNow suppose $n_T=2$. Then $\mathrm{im}T$ is a line. On this line, $T^2$ is the identity operator: if $y=Tx$, then $T^{2}y=T^{3}x=Tx=y$. Consequently $T$ acts on this line by a scalar $c$ satisfying $c^2=1$. If $c=1$, then $T^{2}x=Tx$ for every $x$, contrary to $T^2\ne T$. Thus $c=-1$.

Choose a nonzero $w\in \mathrm{im}T$, so $Tw=-w$. It does not belong to $\ker T$. A basis $k_1,k_2$ of $\ker T$, followed by $w$, is therefore a basis of $V$, in which