So I saw this problem: Is there an upper triangular matrix $A$ such that $A^n\neq 0$ but $A^{n+1}=0$? Prove or disprove.
I said no, and my reasoning was that the matrix must have a zero diagonal since $(A^k)_{ii}=a_{ii}^k$. Then the matrix must be srtictly upper diagonal, and when we multiply it the diagonal of zeros starts "moving up" and eventually the matrix is zero and it follows that it must be in at most $n$ steps.
Is there a neater way to do this? Like looking at characteristic polynomials or something? I was thinking that the minimal polynomial should be of the form $t^k$ and thus the characteristic polynomial ought to be $t^n$ as the minimal divides the characteristic, and by Cayley Hamilton we ought to have $A^n=0$, but I cannot see why the minimal polynomial must be $t^k$. If I was working over a complex (or alg closed field then yes since it could only have zero eigenvalues). Any thoughts?
Thanks,