Context: In linear algebra, a nilpotent matrix is a square matrix $N$ such that
$$
N^{k}=0\,
$$
for some positive integer $k$. (The smallest such $k$ is sometimes called the index of $N$.)
Question: How can one show that a matrix is nilpotent or not? Direct matrix multiplication for checking the definition is an obvious choice but seems not efficient. Is there any other method to do it?
Best Answer
A matrix is nilpotent iff its only eigen value (in the algebraic closure of the base field) is $0$. This follows easily from Caley - Hamilton Theorem. See https://en.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem
Thanks to Kezer for pointing out the taking the algebraic closure is needed.