linear algebra - Efficient method for computing the properties of a block anti-diagonal matrix

EDIT: the title and content of this question were reformulated for the sake of clarity and to avoid diverting attention from the main issue.

I was trying to answer a question about a system of linear differential equations and it was necessary to find the eigenvalues, eigenvectors and generalized eigenvectors for the following matrix:
$$
M=\begin{bmatrix}0&E_{\times}\\B_{\times}&0\end{bmatrix},
$$
where $E=[E_1,E_2,E_3]^T,B=[B_1,B_2,B_3]$, $E^TB=0$ and $E_\times,B_\times$ are the representation of vectors through antisymmetric matrices:
$$
E_{\times}= \begin{pmatrix} 0 & -E_{3} & E_{2}\\ E_{3}& 0 & -E_{1}\\ -E_{2} & E_{1} & 0 \\ \end{pmatrix},\qquad B_{\times}= \begin{pmatrix} 0 & -B_{3} & B_{2}\\ B_{3}& 0 & -B_{1}\\ -B_{2} & B_{1} & 0 \\ \end{pmatrix}.
$$

With WolframAlpha it is possible to see that all eigenvalues are zero and the kernel has dimension 2. I have tried to find some method to obtain these results without resort to software or brute force (solve a linear system step by step), but I could not.

Then my question is the following: is it possible to determine the null space, eigenvalues and generalized eigenvectors for matrix $M$ efficiently without resort to software or brute force?

It is always possible to compute them the hard way: computing the characteristic polynomial $\det(M-\lambda 1)$, finding its roots, substituting in $(M-\lambda 1)v=0$ to find eigenvectors $v$ and using them to determine the generalized eigenvectors. What I am looking for is a answer that computes the null space, eigenvalues and (generalized) eigenvectors making use of the properties of matrix $M$, not brute force.

Answer

We assume that the system $\{E,B\}$ is linearly independent.

Note that (double cross product) $E_xB_xX=BE^TX-X(E^TB)=BE^TX$, that is, $E_xB_x=BE^T$ and, in the same way, $B_xE_x=EB^T$.

$M[X,Y]^T=[E\times Y,B\times X]^T$ and a basis of $\ker(M)$ is $\{[B,0]^T,[0,E]^T\}$.

Moreover $\det(M-\lambda I_6)=\det(\lambda^2I-BE^T)$; $BE^T$ has rank $1$ and trace $0$; then it is nilpotent and $M$ also.

About the generalized eigenvectors, $M^2=diag(BE^T,EB^T)$ and $M^2[X,Y]^T=0$ can be written $BE^TX=0,EB^TY=0$, that is, $X\in E^{\perp},Y\in B^{\perp}$, that implies that $dim(\ker(M^2))=4$.

It is not difficult to see that $M^3=0$ and we are done.