Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 4 down vote accepted

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 $A$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.