Let $T, N :V \to V$, N is nilpotent, T is invertible and $N*T = T*N$. Prove $N+T$ invertible.

Question

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?

Answer 1

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!

Answer 2

You can not prove this. Take $T=0$.

Answer 3

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 $$(T^{k-1} + T^{k-2}(-N) + \ldots + T(-N)^{k-2} + (-N)^{k-1}) T^{-k}$$