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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 independant of a_i !
Am I right?
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 independant of a_i !
Am I right?
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