Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Does a Jordan chain always start with an eigenvector? If so when computing a Jordan chain for a particular matrix do you have to start with an eigenvector, $v_0$ for particular eigenvalue then just go up the chain with $(A-\lambda I)v_{i} = v_{i-1}$?

share|cite|improve this question
up vote 1 down vote accepted

Yes, a Jordan chain necessarily starts with an eigenvector, because $(A-\lambda I)v_0$ should be zero (otherwise, you could "extend" the chain further down).

When you have a Jordan canonical basis, the initial vectors of the chains corresponding to $\lambda$ will form a basis of the eigenspace corresponding to $\lambda$.

share|cite|improve this answer

Your Answer


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