 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 $v\in \ker T^l\cap \mathrm{im}{T}^l$, write $v=T^{l}w$. 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 $\dim V$. HenceThis proves the finite-dimensional form of the Fitting lemma.

If $\det T=0$, the kernel contains a nonzero vector $u$, and $\mathrm{im}T$ 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, andThus a singular endomorphism has a nonzero two-sided annihilator, as required.

For an idempotent linear map $T$, the decomposition $v=(v-Tv)+Tv$ shows $V=\ker T\oplus \mathrm{im}T$: 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 $\mathrm{diag}(0,I_r)$, with $r=\mathrm{rank}T$. 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 $T_1$ and $T_2$ are similar exactly when their ranks agree, proving the similarity classification of idempotent linear maps.