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    A is an nxn matrix. Suppose A has the form (^{U}_{W}^{V}_{X}) in which U, V, W, X are n1xn1, n1x n2, n2xn1 and n2xn2 matrices respectively, such that n1 + n2 = n. If V=0, show that detA = detUdetX

    detA := \sum _{\rho\in sym(n)} sign (\rho)\Pi ai i\rho

    I don't really know how to go about this. If I expand along the 1st row I will get each of the u entries of the first row multiplied by their minor and sign summed together, and the v coefficients will all be zero. I don't know how to write this using the correct notation or where I would go from here.
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    What I would do is think about what has to be true about the permutation in order for the product to not include any term 'from V'.

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    Think about partitioning \rho into two permutations, one on (1 2 ... n1) and one on (n1+1 ... n1+n2)
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    So could I say

    detA = \sum _{\rho\in sym(n)} sign (\rho)\Pi ai i\rho + \sum _{\sigma\in sym(n)} sign (\sigma)\Pi ai i\sigma

    where \rho acts on (1 2 ... n1) and \sigma acts on (n1+1 ... n1 + n2)

    V = 0 so \sum _{\sigma\in sym(n)} sign (\sigma)\Pi ai i\sigma = 0

    so det A = \sum _{\rho\in sym(n)} sign (\rho)\Pi ai i\rho

    I'm not sure that I've completely got my head around the link between permutations and determinants so this may all be wrong. I'm not sure where to go from here.
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    No, that's not right. You seem to have said the (non-zero terms of the) sum can be split into permutations that act only on \rho acting on (1 2 ... n1) and permutations \rho acting only on (n1+1 ... n1 + n2).

    But in fact any permutation that can be partitioned into a permutation on (1 2 ... n1) and (n1 +1 ... n1+n2) will give a non-zero term. (Where by non-zero we mean a term not involving an element of V).

    More explicitly, the non-zero terms are those generated by all products of the form \rho \sigma.

    Given that information, can you rewrite your sum for the determinant and rearrange it to look like the product of two determinant sums?
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    Is this true?

    Let \rho act on (1 2 ... n1) and \sigma act on (n1+1 ... n1 + n2)

    So detU = \sum _{\rho\in sym(n)} sign (\rho)\Pi ai i\rho
    and detX \sum _{\sigma\in sym(n)} sign (\sigma)\Pi ai i\sigma

    detA = \sum _{\rho\sigma\in sym(n)} sign (\rho\sigma)\Pi ai i\rho\sigma
    = \sum _{\rho\in sym(n)} sign (\rho)\Pi ai i\rho x\sum _{\sigma\in sym(n)} sign (\sigma)\Pi ai i\sigma
    =detUdetX
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    Yes. (Not an awful lot of justification, but what you've posted is true, assuming I'm not misreading the fairly grotesque LaTeX).
 
 
 
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