Let $V$ be finite dimensional v.s. and $0 \ne T\in \mathscr L(V)$ , then $\exists$ $S \in \mathscr L(V)$ such that $0 \ne T \circ S$ is idempotent If $V$ is a finite dimensional vector space and $T \ne0$ is a linear operator on $V$ , then how may we prove that there is a linear operator $S$ on $V$ such that $T\circ S$ is non-zero and idempotent?
 A: One can easily ensure $T\circ S\circ T=T$, from which $T\circ S\neq0$ and $(T\circ S)^2=T\circ S$ follow. Here's  how.
Choose a basis $b_1,\cdots,b_k$ of $\ker T$, and extend by $b_{k+1},\cdots,b_n$ to a basis of $V$. Then put $e_i=T(b_{i+k})$ for $i=1,\ldots,n-k$, which are linearly independent, since if some linear combination of those $e_i$ is zero, then the corresponding linear combination of the $b_{i+k}$ is in $\ker T$, which given the choices made is only possible for the trivial linear combination. One can then extend by $e_{n-k+1},\ldots,e_n$ to a basis of$~V$. Now define $S$ on the basis $e_1,\ldots,e_n$ by $S(e_i)=b_{i+k}$ for $i\leq n-k$, and $S(e_i)$ arbitrary (for instance zero) for $i>n-k$. One immediately checks $(T\circ S\circ T)(b_i)=T(b_i)$ for all $i$, whence $T\circ S\circ T=T$ by linearity.
Note that this also works for any linear $T:V\to W$ giving $S:W\to V$; in the argument one just needs to adjust the number of basis vectors $e_i$ to the dimension of$~W$ rather than$~V$.
A: Hint: Suppose $(c,v)$ is an eigenpair of $T$. Consider $T \circ P_v$, where $P_v$ is the orthogonal projector onto the span of $v$. When $c=1$ or $c=0$, I claim this is idempotent. How do you fix it otherwise?
