Idempotents in $M_2(\mathbb{C})$ Given two idempotents $e,f\in M_2(\mathbb{C})\setminus\{I_2\}$, the sets
$$\{eg^{-1}:g\in GL_2(\mathbb{C}), eg^{-1} \text{ is an idempotent}\}$$
and
$$\{gf:g\in GL_2(\mathbb{C}), gf \text{ is an idempotent}\}$$
are infinite.
Is the following also infinite:
$$\{(eg^{-1},gf):g\in GL_2(\mathbb{C}), gf, eg^{-1} \text{ are idempotents}\}?$$
The motivation behind the question:
It is known that every singular matrix is a product of idempotents. I would like to see if these idempotents can be chosen from some uncountable set. 
 A: If $e=0$ or $f=0$, the answer is clearly affirmative. So, it remains to consider the case where both $e,f$ are nonzero but singular idempotent matrices. With a change of basis, we may assume that $f=\pmatrix{0&0\\ 0&1}$.
A nonzero but singular $2\times 2$ matrix $X$ is idempotent if and only if $\operatorname{tr}(X)=1$. Therefore, $eg^{-1}$ is idempotent iff
$$
\operatorname{tr}(e\ \operatorname{adj}(g))=\det(g)\ne0.\tag{1}
$$
Let $e=\pmatrix{a&b\\ c&d}$ and $g=\pmatrix{1&y\\ 0&1}\pmatrix{p&0\\ q&1}$. Then $gf=\pmatrix{0&y\\ 0&1}$ is idempotent for all $y$ and $(1)$ is equivalent to
$$
a-bq-cy+dp+dqy=p\ne0.\tag{2}
$$
Now there are two possibilities:


*

*$d\ne1$. Then $a$ must be nonzero (otherwise $a+d\ne1$ and $e$ is not idempotent). So, $(2)$ is satisfied by putting $q=0$ and $p=(a-cy)/(1-d)$ with $a-cy\ne0$.

*$d=1$. Then $a=0$ and $(2)$ is equivalent to the two conditions $(y-b)q=cy,\ p\ne0$, which are satisfied by putting $q=cy/(y-b)$ with $y\ne b$ and any $p\ne0$.


In each of the above two cases, there are uncountably infinitely many choices of feasible $y$. As $gf=\pmatrix{0&y\\ 0&1}$ varies with $y$, the last set mentioned in the question is uncountably infinite.
