- #1
robforsub
- 16
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If given a n*n matrix with all rows and columns sum to 0, how do I argue that all its (n-1)*(n-1) minor have the same determinant up to a sign?
Since all rows and columns all sum to 0, then I know that any column is a linear combination of all others, so that the determinant of this n*n matrix must be zero, then since the determinant is calculated using minors, it seems to imply that all (n-1)*(n-1) minors must have the same determinant up to a sign, but how do I rigorously prove that?
Since all rows and columns all sum to 0, then I know that any column is a linear combination of all others, so that the determinant of this n*n matrix must be zero, then since the determinant is calculated using minors, it seems to imply that all (n-1)*(n-1) minors must have the same determinant up to a sign, but how do I rigorously prove that?