Let V be a n-dimensional vector space and let T be a linear operator on V. Suppose that there exists some positive integer k so that $\ T^k$ = $0$.Prove that $\ T^n$= $0$
I need help, please.
My proof is:
If $\ T^k$ = $0$ then the minimal polynomial divides $\ x^k$, therefore the minimal polynomial must be $\ x^s$ for some s between 1 and n( because the minimal polynomial has degree at most n), but then $\ T^s$= $0$. Therefore we are reduced to the case when $\ T^k$ = $0$ with $k ≤ n$. In that case $\ T^n$ = $\ T^{n-k}$$\ T^k$ = $\ T^{n-k}$$0$= $0$.
Do I have my proof well?
So, what would they do?
Hint: if $T^k x = 0$ but $T^{k-1} x \ne 0$, then $x, Tx, \ldots, T^{k-1} x$ are linearly independent.