I am given that for some $A_{n\times n},B_{n\times m}$, $z\in\mathbb{R}^n$ if $zA^kB=0\forall k=0,1,\dots n-1$ then this implyes $z=0$
could anyone help me to realize its linear algebraic significance in terms of eigenvalue eigenvector?
$zB=0,zAB=0,zA^2B=0,\dots, zA^{n-1}B=0$
Further I thought: If I take any polynomial $\phi(A)=I+\alpha_1A+\alpha_2A^2+\dots+\alpha_{n-1}A^{n-1}$ Then clearly $ z\phi(A)B=0$