Dimension of endomorphisms subspaces Let $V$ an $\mathbb R$-vector space with finite dimension $n$ ($n \neq 0$).
We know that all endomorphisms $f$ on $E$ can be written as a linear combination of some projectors of $E$. Otherwise: if $\mathcal P_E$ is the set of all projectors of $E$ then $\text{Span}(\mathcal P_E)= \mathcal L(E)$.


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*Here's a question that attempts to generalize this result:
Let $F$ a vector subspace of $V$ and $\mathcal P_F$ the set of projectors $p$  of $E$ such that $\text{Im} p \subset F$ and $\mathcal V_F=\text{Span}(\mathcal P_F)$.
Determine $\dim(\mathcal V_F)$

*Here's an other question which is not far from this topic
: Determine $\dim (\overline{\mathcal V}_F)$ where $\overline{\mathcal V}_F = \text{Span} (\overline{\mathcal P}_F)$ and $\overline{\mathcal P}_F$ is the set of projectors $p$  of $E$ such that $\text{Im} p = F$ 


I am interested in the second question (the first being more accessible).
 A: For the second question, it seems easiest to first look at the difference between two such projectors. Such a function $f:V\to F$ vanishes on $F$ (both projectors are identity there) and has its image contained in$~F$, so in particular $f^2=0$. Conversely if $f$ has those properties and $p$ is a projector with $\operatorname{Im}p=F$, then $(p+f)^2=p^2+pf+fp+f^2=p+f$ (since $p^2=p$, $pf=f$, and the last two terms vanish), so $p+f$ is a projector, and its image is easily checked to be$~F$. Therefore the set of such differences$~f$ is a vector subspace$~S$ of the space of endomorphisms of$~V$, whose dimension is $\dim(V/F)\dim(F)$. The set$~\overline{\mathcal P}_F$ of projectors is a translate of this subspace$~S$ by some$~p$, which gives an affine space; the span of$~\overline{\mathcal P}_F$ contains $S$ and also$~p$, while $p\notin S$ (unless $\dim F=0$). The presence of $p$ increases the dimension by$~1$. All in all the dimension of the span of$~\overline{\mathcal P}_F$ is $\dim(V/F)\dim(F)+1$ (unless $\dim F=0$, in which case it is$~0$).
