= Solution
The ascending chain $\ker T\subseteq\ker T^2\subseteq\cdots$ must stabilize because $V$ is finite-dimensional. Once $\ker T^l=\ker T^{l+1}$, it stays stable: if $T^{l+2}v=0$, then $Tv\in\ker T^{l+1}=\ker T^l$, so $v\in\ker T^{l+1}=\ker T^l$; induction repeats the argument. Choose a positive such $l$, giving $\ker T^{2l}=\ker T^l$.
If $v\in\ker T^l\cap\operatorname{im}T^l$, write $v=T^lw$. Then $T^{2l}w=0$, so $w\in\ker T^l$ and $v=0$. By the <rank-nullity theorem>, the dimensions of those two subspaces sum to $\dim V$. Hence
$$
\boxed{V=\ker T^l\oplus\operatorname{im}T^l.}
$$
This proves the finite-dimensional form of the <Fitting lemma>.
If $\det T=0$, the <kernel> contains a nonzero <vector> $u$, and $\operatorname{im}T$ is a proper <vector subspace>. Extend a <basis> of that <image of a linear map> to a <basis> of $V$ and choose a nonzero <linear functional> $\lambda$ which vanishes on the former <basis>. Define $S(v)=\lambda(v)u$. It is a nonzero <endomorphism>, since $\lambda$ takes a nonzero value somewhere, and
$$
TS(v)=\lambda(v)Tu=0,\qquad ST(v)=\lambda(Tv)u=0.
$$
Thus \b[a <singular endomorphism has a nonzero two-sided annihilator>], as required.
For an <idempotent linear map> $T$, the decomposition $v=(v-Tv)+Tv$ shows $V=\ker T\oplus\operatorname{im}T$: the first term is killed by $T$, the second is fixed, and a <vector> both fixed and killed is zero. An adapted <basis> makes its <matrix> $\operatorname{diag}(0,I_r)$, with $r=\operatorname{rank}T$. Consequently equal ranks give identical adapted <matrices> and therefore a change of <basis> relating $T_1,T_2$. Conversely, similar <matrices> have the same rank because pre- and postmultiplication by <invertible matrices> preserve rank. Hence \b[$T_1$ and $T_2$ are similar exactly when their ranks agree], proving the <similarity classification of idempotent linear maps>.
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