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

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The (i,j)-minor of an nxn matrix A, defined as M_{ij}(A)=detA(i|j), is a function that can be viewed as a multilinear map from R^n x ... x R^n to R. The discussion highlights a key point of confusion regarding the independence of the ith component a_i in the calculation of M_{ij}. Specifically, multiplying a_i by a constant c does not yield the expected result of cM_{ij}(a_1,...,a_i,..,a_n) due to the definition of M_{ij} which effectively disregards the ith component. This indicates that the function M_{ij} operates within a domain of n-1 copies of R^n.

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quasar987
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Recall that for an nxn matrix A, the (i,i)-minor of A is defined as M_{ij}(A)=detA(i|j), 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 M_{ij} 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 M_{ij} by a constant c (for instance 0), one gets not cM_{ij}(a_1,...,a_i,..,a_n) as one should, but rather M_{ij}(a_1,...,a_i,..,a_n) because the ith line is not taken into acount altogether! In other words, M_{ij}(a_1,...,a_i,..,a_n) is independent of a_i !

Am I right?
 
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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')?
 
Yes, I suppose this is what the text implied. Ok.
 

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