Let n≥1. Assume that A is a real n×n matrix which satisfies the equality
A7+A5+A3+A-I=0
Show that det(A)>0
Rewrite as A^7 + A^5 + A^3 + A = I. The determinant of the identity matrix is 1. The determinant of the product of matrices is the product of the matrices' determinants, so if the determinant of A were zero or negative, the given sum would be zero or negative respectively, rather than 1.
added:
there's a flaw in the above argument: it assumes that when two matrices are added together the determinant of the sum has the same sign as the sum of the determinants. I don't have a proof of that.
addendum 2:
That lemma is in fact not in general true, so back to the drawing board.
Edited on August 11, 2023, 11:30 pm
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Posted by Charlie
on 2023-08-11 22:54:03 |
solution
