The kernel and image of $T^n$ I need help with this question:
Let $V$ be a finite vector space where $ \dim V = n $, over the complex numbers
and let $ T: V\to V $ be a linear transformation.
Prove that $ V = \ker(T^n) \oplus Im(T^n) $

Note that in general, although kernel and image of an endomorphism have complementary dimensions, they need not form a direct sum. What you must show here is that for the case of $T^n$ (with $n$ the dimension of the space), kernel and image do always form a direct sum. 
This can be shown by showing that the intersection of these subspaces is reduced to $\{0\}$. Suppose for a contradiction there is a nonzero vector $w\in\ker(T^n)\cap\operatorname{im}(T^n)$. Then $w=T^n(v)$ for some $v$, with obviously $v\neq0$. The condition on $w$ gives $T^{2n}(v)=0$, so $m=\min\{\,i\in\Bbb N\mid T^i(v)=0\,\}$ is well defined and $n<m\leq2n$ (the first inequality is the one that will be useful). Because the dimension of the space is only $n$, the $m$ nonzero vectors $v,T(v),T^2(m),\ldots,T^{m-1}(v)$ must be linearly dependent. Then one can write one of those vectors, say $T^i(v)$ as linear combination of the vectors after it in the list. Now apply the linear operator $T^{m-1-i}$ to this relation; this maps $T^i(v)$ to $T^{m-1}(v)\neq0$, but it kills all the vectors occurring in the linear combination giving $T^i(v)$; this is a contradiction.
