Index of a Nilpotent matrix I was wondering why there can't be a  nilpotent matrix of index greater than its no.  of rows.
Like why there does not exist a nilpotent matrix of index 3 in $M_{2×2}(F)$
When I look up on the internet it says that it is related to ring theory and an element being nilpotent is $x^n = 0$.
I have a very basic understanding of rings but I know groups so I thought for $x^n=0$ n must be less than or equal to the order of group  and thus this thing. Am I right?
 A: The characteristic polynomial of any $n\times n$ matrix is of degree at most $n$, and the matrix is a root of that polynomial.
Since the minimal polynomial of a nilpotent must divide $x^N$ for some $N$, and it also divides the characteristic polynomial, you have that the minimal polynomial is of the form $x^k$ for some $0\leq k\leq n$.

Geometrically, another way to look at it is that, viewing a nilpotent matrix $T$ as a linear transformation of $V=F^n$, $V\supseteq T(V)\supseteq T^2(V)\supseteq\cdots\supseteq \{0\}$ is a descending chain of subspaces of $V$.
Now, it cannot be the case at any point that $T^k(V)=T^{k+1}(V)$, because if that were the case, the chain would remain stable and would never reach zero no matter how high $k$ goes.
So the chain is a strictly descending chain of subspaces of the $n$ dimensional space $V$. Then at each link, you must decrease by at  least one dimension.  But no chain in $V$ is deeper than $n$ links, so you are guaranteed it will take no more than $n$ applications of $T$ to reach zero.
