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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.

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In your special case, do you want minus signs on $E^T$, $F^T$ and $J^T$ as well? – Neil Strickland Jun 8 '11 at 17:04
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. – Amitabha Lahiri Jun 8 '11 at 17:28
Just out of curiousity, is there any motivation behind this question? I am not being negative, really just curious. – Vladimir Dotsenko Jun 8 '11 at 17:30
Yes, I found this problem while trying to count the degrees of freedom in a particular system. – Amitabha Lahiri Jun 8 '11 at 17:41
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 – Gerhard Paseman Jun 8 '11 at 17:58

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's_Expansion_Theorem

This might help.

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Thanks. I'll have a look. – Amitabha Lahiri Jun 9 '11 at 1:59
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? – Amitabha Lahiri Jun 9 '11 at 6:52
@Amitabha: Yes this works regardless of commutativity, and works for any size minor. – Stopple Jun 9 '11 at 14:19

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