When is a linear operator nilpotent? My textbook has the following question:

Suppose $T \in \mathcal{L}(V)$ and there exists a positive integer $n$ such that $T^n = 0$.

When can a linear operator raised to a power equal $0$?
 A: The key word you are looking for is nilpotent operator (or matrix, if you want to.)
An elementary (and very important!) result is that any linear operator $T:V\to V$ on a finite dimensional vector space can be decomposed uniquely into a sum $D+N$ where $D$ is a diagonalizable (more generally, semi-simple) operator and $N$ is nilpotent and, moreover, $DN=ND$. This is usually called a Jordan–Chevalley decomposition.
If you are expressing $T$ as a matrix, then this decomposition is quite straightforward: first, find a basis in which your matrix is upper triangular (this can always be done over the complex numbers). Then the decomposition is given by taking $D$ the diagonal part of your matrix and $N$ the strictly upper triangular remaining part.
An essential example is given by an $n\times n$ matrix $J$ that is zero everywhere except in the first upper diagonal where it has a line of $1$s, usually called a Jordan block. You can check then that $J^n=0$ but $J^{n-1}\neq 0$. One of the most important results in elementary linear algebra is that this is example suffices to pin down any nilpotent matrix and, by way of the Jordan--Chevalley decomposition, any matrix at all, as a sum of a diagonal and a nilpotent matrix.
A: Assume $T(a,b,c)=(0,a,b)$ therefore $T^3=0.$
