Nilpotent Operator Minimal Polynomial

Question

Given a Nilpotent Operator $F: V\rightarrow V$ with index $k$, that being the smallest $k$ such that $F^k = 0$, show the minimal polynomial is $M(t) = t^k$. It's obvious that $M(F) = 0$, but how do I know there is no Polynomial of $deg < k$ that also satisfies the condition?

Answer 1

The essential point to use is that any annihilating polynomial of the operator $F$ must be a (polynomial) multiple of its minimal polynomial, or equivalently the minimal polynomial must divide any given annihilating polynomial. Since by hypothesis $X^k$ is an annihilating polynomial, the only candidates for a minimal polynomial are its monic divisors, which are $1,X,X^2,\ldots,X^k$. But by what is given, all of them except the last one are not annihilating polynomials. It remains as only possibility that $X^k$ must be the minimal polynomial.