Determine all linear transformations $T:V\rightarrow V$ such that $T=T^2$. Let  $V$ be a vector space. Determine all linear transformations $T:V\rightarrow V$ such that $T=T^2$.
Suppose $x\in V$. Then we can write $x=T(x)+(x-T(x))$. Then $T(x)\in$ Range $T$, and $x-T(x)\in$ kernel of $T$. From this how I to proceed further? 
 A: Given such a $T$, notice that $T(V)$ is a subspace of $V$, and that $T$ acts on $T(V)$ as the identity. Therefore, such linear transformations are a subset of projections from $V$ to subsets of $V$. Is it all of them?
A: There are lots of such linear maps. Given a decomposition of the space $V=R\oplus K$, we have a unique $T:V\to V$ such that $T(V)=R$ and $\ker(T)=K$ (which is the projection determined by the decomposition). 
So what you really want to classify are all decompositions of $V$ as a (ordered) direct sum of subspaces. Given $0\leq k\leq \dim V$, the Grassmannian $Gr_k(V)$ is formed by all subspaces of $V$ with dimension $k$. 
On the other hand, there is a nice way of finding all complements of a fixed subspace $R\subseteq V$, using a fixed complement $R^\perp$ (coming from a inner product, for example): every complement of $R\subseteq V$ is the graphic of a linear map $A:R^\perp\to R$.
In conclusion, there is a bijection between maps $T\in End(V)$ with $T^2=T$ and the set
$$\bigcup_{k=1}^{\dim V}\left(\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)\right).$$
Considering that each $Gr_k(V)$ is a manifold and each $\bigcup_{R\in Gr_k(V)} Hom(R^\perp;R)$ is a bundle over $Gr_k(V)$, I'd say there are lots of solutions.
