Does a Jordan chain always start with an eigenvector?

Question

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 $(A-\lambda I)v_{i} = v_{i-1}$?

Answer 1

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 $\lambda$.