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)?
Can basis of kernel be extended to a Jordan basis?
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I understand that you are not interested in the algorithm, but an illustration of the algorithm will help characterize the kernel's basis.
The following figure is an illustration of a case $G$ is the generalized eigenspace such that $6$ is the size of the largest block. In this example, there are $2$ Jordan blocks of size $6$, and $1$ block each for size $4$, $3$, and $2$. The vertices in the diagram are Jordan basis elements. The vertices on the left of the line are the basis of the kernel. Each segment going to the right or bottom represents taking preimages under $N$.
As you already know, not every basis of the kernel can extend to a Jordan basis. I propose the following criterion:
Extending the basis of the kernel to Jordan basis
- Let $u_1, \ldots, u_r$ be a basis of the kernel of $N$. For each $i\leq r$, let $m_i\geq 0$ be the largest integer such that $u_i\in N^{m_i}G$.
- Find $s=\sum_{i\leq r} (m_i+1)$. This gives the total length of the preimage chains.
- If $s<\mathrm{dim} G$, then $u_1, \ldots, u_r$ cannot be extended to Jordan basis. If $s=\mathrm{dim} G$, then such an extension is possible.
Application to your example
If $u,v,w$ is Jordan basis with $u, v$ is a basis of the kernel and $Nw=u$, then the basis $u,v$ has $s=2+1=3$ with $2$ is from $u$, $1$ is from $v$. A basis $u+v, u-v$ of the kernel has $s=1+1=2<3$ with both $1$ are from $u+v$ and $u-v$.

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 $A$.