Given a matrix $A$ with eigenvalue $\lambda$ and a corresponding eigenvector $v_1$:
- Is it possible that the Jordan chain $v_1$ generates does not span the full generalised eigenspace of $A$ w.r.t $\lambda$? ie. There exists some vector $v$ that is not in the span of the vectors in that Jordan chain?
Here's an example I came up with, but I'm not sure if I made any mistake:
Let $A=\text{zero matrix}$, then $(1,0,...,0)$ is an eigenvector of eigenvalue $0$, but it only generates a Jordan chain of length 1 (only contains itself), since the range of $(A-\lambda I)$ only contains $0$.
- If 1. is true, then how could we find all vectors in the full generalised eigenspace of $A$ with respect to $\lambda$?
For example, it's possible that for some matrix A and eigenvalue $\lambda$, there is some $v_1$ such that $Av_1=\lambda v_1$ but $v_1$ only generates a Jordan chain of say length 3, while $(A-\lambda I)$ has nullity 4, then how could we find a 4th generalised eigenvector?
here full generalised eigenspace = set of all generalised eigenvectors corresponding to $\lambda$ together with the zero vector.