Let $A\in\mathbb C^{n\times n}$ be nilpotent. A Jordan basis of $A$ is a basis of $\mathbb C^n$ with respect to which $A$ has Jordan normal form. Assume that we do not know the Jordan structure of $A$. Given a basis of the kernel of $A$, is there a criterion to decide on whether this basis can be extended to a Jordan basis of $A$ (maybe in terms of powers of $A^*$ or whatever)?


This should always be possible.

Given a basis $\mathcal{B}$ of $\ker A$ you can check whether or not it already is a basis of $\mathbb{C}^n$. If it is not a basis you proceed as in the proof of the existence of the Jordan normal form and add vectors from $\ker A^m \backslash \ker A^{m-1}$ to $\mathcal{B}$ until you get a basis of $\mathbb{C}^n$. As you are working over $\mathbb{C}$ we know that a Jordan basis exists, so this procedure yields a basis of $\mathbb{C}^n$ after finitely many steps.

As you can start this procedure with any basis of $\ker A$, this means that any basis of $\ker A$ can be extended to a Jordan basis of $A$.

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    $\begingroup$ I am sorry, but you are wrong. For example, assume we are in $\mathbb C^3$ and let $\{u,w,v\}$ be a Jordan basis of $A$, that is, $\{u,v\}$ is a basis of the kernel and $Aw=u$. Now, try to extend the basis $\mathcal B := \{u+v,u-v\}$ of the kernel to a Jordan basis. You won't succeed because none of the vectors is contained in the image of $A$. Moreover, I am not interested in algorithms, but rather in direct conditions. $\endgroup$ – Friedrich Philipp Mar 24 '17 at 15:47
  • $\begingroup$ Ah, yes, you are right. $\endgroup$ – Elvorfirilmathredia Mar 24 '17 at 16:11

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