Identity linear transformation? For a linear transformation $A$ it is known that $A^3=A$. $A$ is the linear transformation over finite dimensional vector space $V$. Now I am supposed to prove that 
$$\ker (A-I) + \operatorname{im} (A-I) = V$$
where $\ker$ means a set of vectors which are transformed to zero vector
$\operatorname{im}$ is image of $A-I.$
And $+$ is the direct sum
I identity matrix
What I want to know is, is $A=I $? If it is, is there a way to prove it? And if it's not, what else could it be?

Well, $A=I$ is certainly one solution for $A^3=A$, but not the only solution, e.g. $0$ or any reflection as a linear transformation is a solution, in particular $A=-I$.
Suppose $v\in{\rm im}(A-I)\cap\ker(A-I)$, i.e. $v=(A-I)u$ for some $u$, and $Av=v$. 
We want to prove $v=0$. For that,
$$0=(A^3-A)u=A(A+I)(A-I)u=A(A+I)v=A^2v+Av\\
A^2v=-Av$$
But we also have $Av=v$ and thus $A^2v=A(Av)=Av=v$, the above line thus gives $v=-v$.
Now, with the dimension theorem for $A-I$ (having $\dim({\rm im}(A-I))+\dim(\ker(A-I))=\dim V$), the proof completes.
