If $A_{n\times n}$ is nilpotent matrix then $\beta = (y, Ay,A^2y, \ldots, A^{k-1}y)$ is the basis of vector space $\mathbb R^n$.

Question

$A_{n\times n}$ is nilpotent matrix, $A^k=0$ and $A^{k-1} \neq 0$.

If $A_{nxn}$ is nilpotent matrix and $A^{k-1} \neq 0$ for some $y \in \mathbb R^n$, then prove that $\beta = (y, Ay,A^2y, \ldots, A^{k-1}y)$ is the basis of vector space $\mathbb R^n$.

Answer 1

$A$ is nilpotent in $R^n$, $A^{n-1}(y)\neq 0$. Consider the equation $c_0y+c_1A(y)+...+c_{n-1}A^{n-1}(y)=0$, if you apply $A^{n-1}$ to that equation, you obtain $c_0A^{n-1}(y)=0$, thus $c_0=0$ since $A^{n-1}(y)\neq 0$.

Suppose $c_0,...,c_i=0$ thus $c_{i+1}A^{i+1}(y)+...+c_{n-1}A^{n-1}(y)=0$, if you apply $A^{n-1-i-1}$ you obtain $c_{i+1}A^{n-1}(y)=0$ and $c_{i+1}=0$, recursively you have $c_i=0, i=0,...,n-1$ so $(y,A(y),...,A^{n-1}(y))$ is a base of $R^n$, since it is a free system with $n$ elements.