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Determinants, multiplication

  1. Jan 30, 2010 #1
    1. The problem statement, all variables and given/known data

    A is an nxn matrix. Suppose A has the form ([tex]^{U}_{W}[/tex][tex]^{V}_{X}[/tex]) 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


    2. Relevant equations

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

    3. The attempt at a solution

    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.
     
  2. jcsd
  3. Jan 30, 2010 #2
    I would try showing this using cofactor expansion, specifically along a row/column with a lot of zeros (so you are only dealing with one term).
     
  4. Jan 31, 2010 #3
    Is this true?

    Let [tex]\rho[/tex] act on (1 2 ... n1) and [tex]\sigma[/tex] act on (n1+1 ... n1 + n2)

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

    detA = [tex]\sum[/tex] [tex]_{\rho\sigma\in sym(n)}[/tex] sign ([tex]\rho\sigma[/tex])[tex]\Pi[/tex] ai i[tex]\rho\sigma[/tex]
    = [tex]\sum[/tex] [tex]_{\rho\in sym(n)}[/tex] sign ([tex]\rho[/tex])[tex]\Pi[/tex] ai i[tex]\rho[/tex] x[tex]\sum[/tex] [tex]_{\sigma\in sym(n)}[/tex] sign ([tex]\sigma[/tex])[tex]\Pi[/tex] ai i[tex]\sigma[/tex]
    =detUdetX
     
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