I am reading "Introduction to Linear Algebra" (in Japanese) by Kazuo Matsuzaka.
There is the following problem(problem 11 on p.225) in this book:
Let $V$ be a finite-dimensional vector space.
Let $F$ be a linear map on $V$.Prove that $F = GP$ for some projection $P$ on $V$ and for some non-singular linear map $G$ on $V$.
A linear map $P$ on $V$ is a projection if and only if $P = P^2$.
My attempt is here:
If $F$ is a non-singular linear map on $V$, then $F = F I$.
If $F$ is a singular linear map on $V$, ...
One hint here is that, whenever such a decomposition is possible, $\operatorname{ker} F = \operatorname{ker} P$, since $G$ is non-singular. Also, projections are determined uniquely by their kernel and their image, so we simply need to find the image of $P$. As it turns out, we don't really need to find the image, so much as nominate one, and let our choice of $G$ do the rest.
Let $W$ be complementary to $\operatorname{ker} F$, i.e. $W \oplus \operatorname{ker} F = V$. Define $P$ to be the projection along $\operatorname{ker} F$ onto $W$. Note that $F$ is injective when restricted to $W$. Hence, if $w_1, \ldots, w_k \in W$ form a basis for $W$, then $Fw_1, \ldots, Fw_k$ is linearly independent in $V$.
Let $u_{k+1}, \ldots, u_n$ be a basis for $\operatorname{ker} F$. Then $w_1, \ldots, w_k, u_{k+1}, \ldots, u_n$ of $V$ is a basis. Extend $Fw_1, \ldots, Fw_k$ to a basis $Fw_1, \ldots, Fw_k, u'_{k+1}, \ldots, u'_n$ arbitrarily. Define the unique linear map $G : V \to V$ by mapping the former basis to the latter, in order. Then, as $G$ maps one basis to another, $G$ is non-singular.
We just need to show $F = GP$. We do this by showing they are equal on a basis. We have $Pw_i = w_i$, since $w \in W$, the image of the projection $P$. Further, $GPw_i = Gw_i = Fw_i$, by definition of $G$. We also have $GPu_i = G0 = 0 = Fu_i$, since $u_i \in \operatorname{ker} F = \operatorname{ker} P$. Thus, $F = GP$ as required.