How to prove that equality? Let $E$ be a vector space of finite dimension and $f:E\mapsto E$ be a linear map (that is $f$ is an endomorphism) such that $(f\circ f \circ f) (E)=f(E)$. I want to prove that $E=f(E)\bigoplus Ker(f)$. I do not know how to prove that. Thanks
 A: I assume that $E$ is a finite dimensional space:
Firstly, let $y\in f(E)\cap Ker(f)$ then there's $x\in E$ such that $y=f(x)$ so
$$0=f(y)=f^2(x)$$
hence $x\in Ker f^2$. Moreover since $f^2(f(E))=f(E)$ then the restriction of $f^2$ on $f(E)$ is surjective and then injective (in finite dimensional space) then $Ker f^2=\{0\}$ so $x=0$ and then $y=0$ hence
$$f(E)\cap Ker(f)=\{0\}\tag1$$
Secondly, by the rank-nullity theorem we have 
$$\dim E=\dim f(E)+\dim Ker f\tag 2$$
so by $(1)$ and $(2)$ we conclude the desired result.
A: Another approach, using matrices. 
Assume that $B=\{b_1,\ldots,b_n\}$ is a basis of $E$ and $f$ is realized by a matrix $A=(a_{ij})\in\mathbb F^{n\times n}$, i.e.,
$$
f(e_i)=\sum_{j=1}^n a_{ij}e_j, \quad i=1,\ldots,n.
$$
Now, $f\circ f\circ f=f$ implies that $A^3=A$, which means that $A$ is diagonalizable (as it is annihilated by a polynomial of simple roots) and its eigenvalues are among $\{0,1,-1\}$.
Also
$$
E=X_0\oplus X_1 \oplus X_{-1},
$$
where $X_0$, $X_1$ and $X_{-1}$ are the corresponding eigenspaces, and in particular $X_0=\ker f$. Clearly
$$
AX_0=f(X_0)=\{0\},\,AX_0=f(X_1)=X_1,\,AX_{-1}=f(X_{-1})=X_{-1},
$$
and hence
$$
f(E)=X_1\oplus X_{-1},
$$
and thus
$$
E=f(E)\oplus \ker f.
$$
