How to prove that the square matrix $A_{n}$ matrix is nilpotent such that $A^{(n-1)}=0$

Question

The matrix A looks like this: $$A=\begin{bmatrix} 0 & 1 & 0 & 0 & .&.&. &0\\ 0 & 0 & 2 & 0 & .&.&. &0\\ 0 & 0 & 0 & 3 & .&.&. &0 \\ .\\.\\.\\ \\ 0 & 0 & 0 & .&.&. & &n-1 \\ 0 & 0 & 0 & 0 & .&.&. &0 \end{bmatrix}$$

I know it's nilpotent (all the eigenvalues are zero), but can't prove it's nilpotent after the $(n-1)$ index. I understand that the diagonal "moves up" every multiplication but is there a simpler way to formalize this?

Answer 1

If we denote by $e_1,\dotsc,e_n$ the standard basis of $K^n$ (assuming $A\in {\rm GL}_n(K)$), then we have $Ae_1=0$ and $Ae_{i+1}=i\cdot e_i$ for $i=1,\dotsc,n-1$. Hence $A^{i}e_{i}=0$ for $i=1,\dotsc,n$. Then $A^n e_i = 0$ for all $i=1,\dotsc,n$, i. e. $A^n=0$ since $e_1,\dotsc,e_n$ is a basis.

Note, that we have $A^{n-1}e_n = (n-1)!\cdot e_1$, which is non-zero if ${\rm char}(K)\ge n$ or ${\rm char}(K) =0$. Hence $A^{n-1}\neq 0$.

Answer 2

This results from Hamilton-Cayley, since the characteristic polynomial is $x^n$.

Furthermore, you can prove by induction that, denoting $ A^k=\bigl(a^{(k)}_{ij}\bigr)_{1\le i,j\le n}$, all coefficients are $0$ except $$a^{(k)}_{i\,i+k}=i(i+1)\dotsm(i+k-1),\quad i=1,\dots,n-k.$$