Is this Jordan form possible?
$$J=\begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$$
Motivation
I was trying to find how can I know the Jordan form of a 7x7 matrix $A$ with one eigenvalue of multiplicity 7.
Suppose that $\dim(\ker(A-\lambda\mathbb{I}))=3$, which means that there will be 3 Jordan blocks. And $\dim(\ker(A-\lambda\mathbb{I})^3)=\dim(\ker(A-\lambda\mathbb{I})^4)$, i.e, the biggest Jordan block is 3x3.
This yields to possibilities:
$J_1 =\begin{pmatrix} \lambda & 1 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$
and
$J_2 =\begin{pmatrix} \lambda & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & \lambda & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & \lambda & 1 & 0 & 0 & 0\\ 0 & 0 & 0 &\lambda & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &\lambda & 1 & 0\\ 0 & 0 & 0 & 0 &0& \lambda & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & \lambda\\ \end{pmatrix}$
What information do I need in order to distinguish both cases? I think that $J_1$ is not possible, so the only possibility is $J_2$, but I can't prove it.
Should I look the dimension of $\dim(\ker(A-\lambda\mathbb{I})^2)$?
Both options are possible. Just notice that if both $A = J_1$ and $A = J_2$ satisfy all the assumptions you impose.
To distinguish the two cases, you would need to be able to say something more about the sizes of Jordan blocks. For easier notation, let $B := A - \lambda I$. It would help if you could say something about ranks (codimensions of kernels) of powers of $B$. You already know that $rank \ B = 4$, $rank \ B^3 = 0$. For case $J_1$ you have $ rank \ B^2 = 1$; for case $J_2$ you have $ rank \ B^2 = 2$. So, if you can figure out $rank B^2$, you have the solution.