Make an $n\times n$ matrix $A$ such that $A,A^2,...A^{n-1}$ are not zero, but $A^n=0$

Question

Make an $n\times n$ matrix $A$ such that $A,A^2,\dots,A^{n-1}$ are not zero, but $A^n=0$.

By trial and error I just stumbled upon the matrix form which would contain all zero elements other than at $(i,i+1)$. How do I prove this?

Answer 1

Let $A=[0,e_1,e_2,\dots,e_{n-1}]$ where $e_1,\dots,e_n$ is the canonical base of $\mathbb{R}^n$. Then $$A^2=A\cdot A=[0,e_1,e_2,\dots,e_{n-1}]\cdot [0,e_1,e_2,\dots,e_{n-1}]=[0,0,e_1,e_2,\dots,e_{n-2}]$$ and, recursively, we have that for $k=1,\dots,n-1$, $$A^k=[\underbrace{0,\dots, 0}_{k},e_1,e_2,\dots,e_{n-k}]\not=0.$$ Finally $$A^n= A\cdot A^{n-1}= [0,e_1,e_2,\dots,e_{n-1}]\cdot [\underbrace{0,\dots, 0}_{n-1},e_1]=[\underbrace{0,\dots, 0}_{n}]=0.$$

Answer 2

Suppose that $n=4$. Take$$A=\begin{pmatrix}0&1&0&0\\0&0&1&0\\0&0&0&1\\0&0&0&0\end{pmatrix}.$$Then$$A^2=\begin{pmatrix}0&0&1&0\\0&0&0&1\\0&0&0&0\\0&0&0&0\end{pmatrix}\text{, }A^3=\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix},$$and $A^4=0$. The general case is similar.

Answer 3

The matrix having $\alpha_i\ne0$ at position $(i,i+1)$ and zero elsewhere describes a linear map $f\colon V\to V$ ($V$ any $n$-dimensional vector space) such that: $$ f(e_1)=\alpha_1e_2,\quad f(e_2)=\alpha_2e_3,\quad \dots,\quad f(e_{n-1})=\alpha_{n-1}e_n,\quad f(e_n)=0 $$ where $\{e_1,\dots,e_n\}$ is a basis of $V$.

Clearly, $f^k(e_1)\ne0$ for $k<n$, since $\alpha_k\ne0$ for $k=1,2,\dots,n-1$. However, $f^n(e_k)=0$ for every $k=1,2,\dots,n$.

By the way, given a matrix $A$ with that property, choose $v$ such that $A^{n-1}v\ne0$. Then $\{v,Av,A^2v,\dots,A^{n-1}v\}$ is a basis of $\mathbb{R}^n$ with respect to which the matrix has the form above, with $\alpha_k=1$ for $k=1,2,\dots,n-1$.