# 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?

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## 2 Answers

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 $A$.

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No; if $\lambda$ is not an eigenvalue, $(A-\lambda I)$ is regular, and products of regular matrices are always regular (for example, because they are invertible), so $(A-\lambda I)^k$ is regular as well and $x$ must be $0$ for the equation to hold.

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