Let be $P\in \mathbb{R}^{n\times n}$ of rank $r\leq n$ and idempotent, i.e. $PP=P$. I want to show that there exist matrices $A\in \mathbb{R}^{n\times r}$ and $B\in \mathbb{R}^{r\times n}$ such that $P=AB$ and $BA=I_r$.
I know that P is diagonizable and one idea was to consider the matrix $C\in \mathbb{R}^{n\times r}$ where the columns of $C$ are $r$ independent eigenvectors to the eigenvalue $1$ of $P$. We know that they exist because $P$ is idempotent. Another usefull thing might be that $(C'C)^{-1}C'$ is a right-inverse of $C$. But I couldn't find a proof so far. A left-inverse would be more helpful but I think it does not exist. I would be glad for any help.
Like you said, $P$ is diagonalizable, and in such conditions it is helpful to check first what you would do if $P$ was diagonal.
In that case: $$D=\left(\matrix{I_r &0\\0&0}\right)=\underbrace{\left(\matrix{I_r \\0}\right)}_{\text{call this $A_0$}}\cdot \overbrace{\left(\matrix{I_r &0}\right)}^{\text{call this $B_0$}}$$ where the $0$ stand for submatrices with the relevant dimensions (and full of zeroes).
Now $P$ is not actually equal to $D$: \begin{align*}P&=QDQ^{-1}\\&=(QA_0)(B_0Q^{-1})\end{align*}
with $$(B_0Q^{-1})(QA_0)=I_r$$