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.


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.

  • $\begingroup$ What a great answer! Thank you very much for this elegant and very instructive answer. There is only one thing I didn't get it: it is not clear to me why $\det(M-\lambda I_6)=\det(\lambda^2 I-BE^T)$. Can you elaborate a little more? $\endgroup$ – jobe Feb 9 '18 at 12:38
  • $\begingroup$ Since $B_x$ and $\lambda I_3$ commute, $\det(M-\lambda I)=\det(\lambda ^2I -E_xB_x)=\det(\lambda^2 I-BE^T)$. $\endgroup$ – loup blanc Feb 9 '18 at 13:05
  • $\begingroup$ I still don't get it. In the first determinant, $\det(M-\lambda I_6)$, we have $6\times 6$ matrices, while in the second one, $\det(\lambda^2 I_3-E_\times B_\times)$, we have $3\times 3$ matrices. Why $\lambda^2$? I think you are using some property I don't know. $\endgroup$ – jobe Feb 9 '18 at 15:18
  • $\begingroup$ If $CD=DC$, then $\det(\begin{pmatrix}A&B\\C&D\end{pmatrix})=\det(AD-BC)$. $\endgroup$ – loup blanc Feb 10 '18 at 11:22
  • $\begingroup$ Thank you. I was ignorant of that property. $\endgroup$ – jobe Feb 11 '18 at 15:58

Very short: You can easily get examples of all ranks less than $n$, when the matrix is $n \times n$. A very simple example, take a matrix with zeros on the diagonal, zeros below, and ones above the diagonal. It is nilpotent, that is, there is some positive integer $k$ such that $A^k=0$.

An extended hint: When all eigenvalues are zero, then, if $x$ is an eigenvector, it will certainly belong to the kernel! So, if there is a basis of eigenvectors then the matrix must be zero.

  • $\begingroup$ Thank you for your answer. Perhaps I was not very clear in my first formulation of this question, so I have reformulated it. The characteristic polynomial is $\lambda^6$ and the kernel has dimension 2, these two facts can be determined through software or brute force. Determination of generalized eigenvectors through brute force for this matrix is not so easy and so I began wondering if there is a more elegant and efficient method to obtain these results. That is what I am looking for as an answer. $\endgroup$ – jobe Feb 8 '18 at 12:44

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