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

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In summary, the (i,i)-minor of an nxn matrix A is defined as M_{ij}(A)=detA(i|j), where A(j|i) is obtained by removing the ith line and jth column. This function can also be viewed as a map from R^n x ... x R^n to R, taking n vectors and calculating the determinant of the resulting matrix. Similarly, M_{ij} can be seen as a multilinear map from R^n x ... x R^n to R. However, there may be some ambiguity in its definition, as it is not clear whether its domain includes all n or only 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 [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?
 
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  • #2
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')?
 
  • #3
Yes, I suppose this is what the text implied. Ok.
 

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

1. What is the (i,j)-minor of a matrix?

The (i,j)-minor of a matrix refers to the determinant of the submatrix obtained by deleting the i-th row and j-th column of the original matrix. It is denoted by Mi,j.

2. How is the (i,j)-minor related to the concept of multilinear maps?

The (i,j)-minor can be seen as a multilinear map from the i-th row and j-th column vectors to a scalar value. In other words, it represents the contribution of these two vectors to the overall determinant of the matrix.

3. Why is the (i,j)-minor important in linear algebra?

The (i,j)-minor is important because it allows us to calculate the determinant of a matrix by breaking it down into smaller submatrices. This makes it easier to compute and understand the properties of a matrix, such as whether it is invertible or singular.

4. Can the (i,j)-minor be used to find the rank of a matrix?

Yes, the (i,j)-minor can be used to find the rank of a matrix. The rank of a matrix is equal to the maximum number of linearly independent rows or columns, which is equivalent to the highest number of non-zero (i,j)-minors.

5. How does changing the (i,j)-minor affect the determinant of a matrix?

Changing the (i,j)-minor of a matrix can affect the determinant in different ways, depending on the properties of the matrix. For example, if the (i,j)-minor is multiplied by a constant, the determinant will also be multiplied by that constant. However, if the (i,j)-minor is replaced with a different value, the overall determinant may change significantly.

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