Can a generalized eigenvectors be generated by a scalar which is not an eigenvalue? I have gone through the definition of generalized eigenvectors. It mentions a scalar value only (not an eigenvalue). Can a scalar other than an eigenvalue generate generalized eigenvectors?
In other words, for the equation $(A - \lambda I)^kx = 0$, for a solution $x$, is it possible that $\lambda$ is not an eigenvalue?

If $(A-\lambda I)^k x = 0$ and $x \neq 0$, this implies that $\det \left((A- \lambda I)^k \right) = 0$, since there exists $x \neq 0\in$ null-space of $(A-\lambda I)^k$. But $$\det \left((A- \lambda I)^k \right) = \left(\det \left(A - \lambda I \right) \right)^k$$ Hence, we get that if $\det \left((A- \lambda I)^k \right) = 0$, then $\left(\det (A- \lambda I) \right)^k=0$, which in-turn implies that $\det (A- \lambda I)=0$. Hence, $\lambda$ is an eigenvalue of $A$.
