Given $T: V \to V$ and $T^2 = T$, how is $V = \ker(T) \oplus \ker(T-I)$? 
$V$ is a finite-dimensional vector space over $\mathbb{F}$, and $T$ is a linear operator on $V$ such that $T^2 = T$. Show that $V = \ker(T) \oplus \ker(T-I)$

It is easy to see that $\ker(T) \cap \ker(T-I) = \{0\}$, so that is sorted.
Since $T^2 = T$, we know that $\forall v \in V$, $T(T(v)) = T(v)$. $\ker(T)$ is essentially the set of vectors $\in V$ for which $T(v) = 0$. As a result, $T^2(v) = T(T(v)) = 0$ but that doesn't tell us anything new. I found a related question here but that is the converse of the result I wish to obtain.
Any hints/answers would be great (preferably something to point me in the right direction so I can work it out on my own)
 A: Hint. For any $v \in V$,
$$v = (v - T(v)) + T(v).$$
A: $v=(v-Tv)+Tv$ Note that $T((v-Tv))=Tv-T^{2}v=0$ and $(I-T) (Tv)=Tv-T^{2}v=0$.
A: Take any $v\in V $ and let $u= T(v)$, then $$(T-I)u = (T^2-T)(v)=0$$ so $u\in ker (T-I)$. Now let $w= v-u$ then $$T(w) = T(v-T(v))=(T-T^2)(v)=0$$
which means that $w\in ker(T)$ and $v=w+u$ so we are done.
A: On a more abstract level, we can work with the minimal polynomial: from $T^2=T$ it follows that $T^2-T=0$, meaning that the minimal polynomial divides the polynomial $X^2-X$. That leaves us with only three possibilities. The minimal polynomial can be any of these: $X,~X-1,~X(X-1)$. In all cases, we can read out the nontrivial (i.e. dimension greater than zero) generalized eigenspaces: they are $\ker T$ for the first one, $\ker T-I$ for the second one, and both for the last one. We also have that any finite dimensional vector space is the direct sum of its generalized eigenspaces, that is, $V\in\{\ker T,\ker(T-I),\ker T\oplus\ker(T-I)\}$. But also, in the cases where $V$ is just one of the kernels, the other kernel will be trivial, and the trivial vector space is essentialy the identity element of direct sums, so even in those two cases we can write $V=\ker T\oplus\ker(T-I)$.
