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
(In reply to
A Theorem by Brian Smith)
With help from Wolfram Alpha, The roots of P(x)=x^7+x^5+x^3+x-1 are 0.6245, -0.8285+/-0.6543*i, -0.1536+/-1.0938*i, and 0.6699+/-0.8539*i.
The determinant of A is the product of its eigenvalues, and by the theorem those eigenvalues are limited to the seven roots of P(x). A given to be a real matrix, so the complex eigenvalues must come in conjugate pairs.
(-0.8285+0.6543*i) * (-0.8285-0.6543*i) = 1.1145
(-0.1536+1.0938*i) * ( -0.1536-1.0938*i) = 1.2200
(0.6699+0.8539*i) * (0.6699-0.8539*i) = 1.1779
Then the determinant of A is some product of powers of 0.6245, 1.1145, 1.2200, and 1.1779. These are all positive, so any possible determinant of A must be positive.
re: A Theorem - Solving the given problem
