Let $T, N :V \to V$, N is nilpotent, T is invertible and $N*T = T*N$. Prove $N+T$ invertible. Lets say we have $T,N:V \to V$ . N is nilpotent, T is invertible and $T*N=N*T$.
Prove that $N+T$ is invertible.
Any suggestion on how I can prove this? we know that $\exists k$ $s.t$ $N^k=0$.
I was thinking of $(N+T)^k$ as a start but got stuck. Any Ideas?
 A: Note that $N, T$ commute, so we may see the following is true for polynomials in $t, n$.
$$t+n \mid t^{k} - n^{k}$$
for all $k \in \mathbb{N}$. But there exists a $k$ such that $N^k = 0$. Pick this $k$. Then the inverse is 
$$(T^{k-1} + T^{k-2}(-N) + \ldots + T(-N)^{k-2} + (-N)^{k-1}) T^{-k}$$
A: Let
$m \in \Bbb N \tag 1$
be the smallest positive integer such that
$N^m = 0; \tag 2$
if we assume $N \ne 0$ (in the contrary case there is nothing to prove), then
$m > 1; \tag 3$
we write
$T + N = T(I + T^{-1}N); \tag 4$
now,
$TN = NT \tag 5$
implies 
$T^{-1}N = NT^{-1}, \tag 6$
whence
$(-T^{-1}N)^m = (-1)^m T^{-m} N^m = 0.  \tag 7$
We recall the polynomial identity
$x^m - 1 = (x - 1) \displaystyle \sum_0^{m - 1} x^{m - 1 - i}, \tag 8$
which we may also write as
$1 - x^m = (1 - x)  \displaystyle \sum_0^{m - 1} x^{m - 1 - i}; \tag 9$
we apply this identity taking
$x = -T^{-1}N, \tag{10}$
and find
$I = I - (-T^{-1}N)^m = (I + T^{-1}N) \displaystyle \sum_0^{m - 1} (-T^{-1}N)^{m - 1- i}, \tag{11}$
which shows that $I + T^{-1}N$ has an inverse
$(I + T^{-1}N)^{-1} =  \displaystyle \sum_0^{m - 1} (-T^{-1}N)^{m - 1- i}. \tag{12}$
Now since both $T$ and $(I + T^{-1}N)$ are invertible, so is their product
$T + N = T(I + T^{-1}N), \tag{13}$
and we are done!
A: You can not prove this. Take  $T=0$.
