Let V be a finite dimensional subspace 1)Let $T : V \to V$ be a nonzero linear operator. Prove that either $T$ is an isomorphism, or  there exists nonzero $R$, $S$ in $L(V, V)$ such that $RT = O$ and $TS = O$.
2)Let $I(V)$ be the set of all invertible operators on $V$, and let $B(V)$ be the set of all ordered bases of $V$. Determine with proof, a bijection between the sets $I(V)$ and $B(V)$.
For the first part, I tried to prove that $T$ is an isomorphism, but since $T$ is a nonzero linear operator, I couldn't prove injection. For the second one, I could prove that $I(V)$ is bijective, but I couldn't tie it in any reasonable proof.
Please help.
 A: Fix a basis $\left\{e_1,\ldots,e_n\right\}$ for $V$.
1)If $T$ is not an isomorphism, then, since $V$ is finite-dimensional, $T$ is not injective nor surjective.
Since $T$ is not injective, then there exists $x\in V$, $x\neq 0$, such that $Tx=0$. Define $S:V\rightarrow V$ by $S(\sum_{i=1}^n\alpha_i e_i)=\sum_{i=1}^n \alpha_i x$. Then $S$ satisfies the condition you want.
Since $T$ is not surjective, there exists $y\in V\setminus T(V)$. You should know that there exists $R:V\rightarrow R$ linear such that $R|_{T(V)}=0$ and $R(y)=y$ (if not, try to prove this: take a basis for $T(V)$, extend it to a basis of $V$ containing $y$, and define $R$ in this basis). Then $R$ satisfies the condition you want.
On the other hand, if there exists $R$ and $S$ as in the question, then take $x\in V$ such that $Sx\neq 0$. Since $TSx=0$, then $T$ is not injective (this already proves that $T$ is not an isomorphism, so you're done here, but I'm going to do a little more). Take $y\in V$ such that $Ry\neq 0$. Since $RT=0$, then $T$ cannot be surjective.
2)Given a ordered basis $\mathscr{B}=\left\{b_1,\ldots,b_n\right\}$ for $V$ define $T_\mathscr{B}\in L(V,V)$ by $T_\mathscr{B}(\sum_{i=1}^n\alpha_i e_i)=\sum_{i=1}^n\alpha_i b_i$. The association $\mathscr{B}\in B(V)\mapsto T_\mathscr{B}\in I(V)$ gives the bijection you want.
