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The (i,j)-minor of a matrix: multilinear map?

  1. Jun 10, 2008 #1

    quasar987

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    Recall that for an nxn matrix A, the (i,i)-minor of A is defined as [itex]M_{ij}(A)=detA(i|j)[/itex], where A(j|i) stands for the matrix (n-1)x(n-1) obtained from A by removing the ith line and jth column.

    Also note that we can view det as a map from R^n x ... x R^n to R taking n vectors from R^n, staking them as the lines of a nxn matrix and taking the determinant of that. And it is a well known fact from linear algebra that this map is n-linear.

    In the same way, we can view [itex]M_{ij}[/itex] as a function from R^n x ... x R^n to R, and it is in this context that I ask if the (i,j)-minor of a matrix is a multilinear map.

    My book says that it is, but I find it odd that if one multiplies the ith component a_i of [itex]M_{ij}[/itex] by a constant c (for instance 0), one gets not [itex]cM_{ij}(a_1,...,a_i,..,a_n)[/itex] as one should, but rather [itex]M_{ij}(a_1,...,a_i,..,a_n)[/itex] because the ith line is not taken into acount altogether! In other words, [itex]M_{ij}(a_1,...,a_i,..,a_n)[/itex] is independant of a_i !

    Am I right?
     
    Last edited: Jun 10, 2008
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  3. Jun 10, 2008 #2

    morphism

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    I guess it depends on how exactly you're defining the function M_ij, i.e. does its domain consist of n or n-1 copies of R^n (where in the latter case we're discarding the 'ith component')?
     
  4. Jun 11, 2008 #3

    quasar987

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    Yes, I suppose this is what the text implied. Ok.
     
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