Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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
1  
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
show 1 more comment

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.

share|improve this answer
    
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
1  
@Amitabha: Yes this works regardless of commutativity, and works for any size minor. –  Stopple Jun 9 '11 at 14:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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