I have a 4n$\times$4n matrix, which can be written as \begin{pmatrix} 0 & A &B &C \cr D& 0& E & F \cr G& H & 0 & J \cr K& L& M& 0 \end{pmatrix}

each entry being an n$\times$n matrix with vanishing determinant. Is there a rule for checking if the full matrix has zero determinant? How about the special case \begin{pmatrix} 0 & A &B &C \cr -A^T & 0& E & F \cr -B^T & E^T & 0 & J \cr -C^T & F^T & J^T & 0 \end{pmatrix}

still with vanishing determinants for each n$\times$n matrix?

(The n is the dimension of an SU group -- I can probably work out the SU(2) or n=3 case by brute force, but I would like to know if there is some method that does not require explicit calculation.)

Many thanks in advance for any help or suggestion.

  • 1
    $\begingroup$ In your special case, do you want minus signs on $E^T$, $F^T$ and $J^T$ as well? $\endgroup$ Jun 8, 2011 at 17:04
  • $\begingroup$ No, actually E,F,J are antisymmetric, so $E^T = -E$ etc (for n=3, which makes the determinant vanish). A,B,C are not antisymmetric, they only have vanishing determinants (one row vanishes). For higher n I am not absolutely certain what I will get in the special case. $\endgroup$ Jun 8, 2011 at 17:28
  • 2
    $\begingroup$ Just out of curiousity, is there any motivation behind this question? I am not being negative, really just curious. $\endgroup$ Jun 8, 2011 at 17:30
  • $\begingroup$ Yes, I found this problem while trying to count the degrees of freedom in a particular system. $\endgroup$ Jun 8, 2011 at 17:41
  • $\begingroup$ If the matrices commuted (perhaps most of the pairs instead of all of them), then you could reduce the problem to the determinant of a nxn matrix product. Or if e.g. A B and C were simultaneously diagonalizable, you could then check if say the first n rows had full rank. Apart from that, I can only suggest the standard methods without shortcuts. Gerhard "Ask Me About System Design" Paseman, 2011.06.08 $\endgroup$ Jun 8, 2011 at 17:58

1 Answer 1


It would be nice if the rule for determinants for $2\times2$ matrices generalized to the case of $2n\times 2n$ matrices:

$\det \begin{pmatrix} A & B \cr C & D \end{pmatrix} =\det A \det D - \det B\det C$,

but this is sadly not true.

Nonetheless, the familiar Laplace expansion theorem for minors of order $n-1$ does have a generalization to minors of any order, including, in this case, minors of order $2n$ of a $4n \times 4n$ matrix, see http://www.proofwiki.org/wiki/Laplace's_Expansion_Theorem

This might help.

  • $\begingroup$ If I can work with 3n$\times$3n minors, whose determinants are be the cofactors for the $n\times n$ matrices along the top $n$ rows, that would be good. The proof does not mention anything about the commutativity of the submatrices. I assume it works even when none of the submatrices commute? $\endgroup$ Jun 9, 2011 at 6:52
  • 1
    $\begingroup$ @Amitabha: Yes this works regardless of commutativity, and works for any size minor. $\endgroup$
    – Stopple
    Jun 9, 2011 at 14:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.