I have a statement: $det(I + A + A^2 + ... + A^{2022}) \ge 0$, where A is a square matrix. The task is to prove it and say, is it possible for it to be equal to $0$?
I'm preparing for the math contest this spring. This problem was last year. I'm very confused with it. As I understand $I$ is identity matrix. It's geometric series of matrices generated by A. I found this formula on the Internet: $$ S_n = \sum_{k=0}^{n-1}A^k = (I-A)^{-1}(I-A^n) $$ So I can rewrite it: $$ S = \sum_{k=0}^{2022}A^k = (I-A)^{-1}(I-A^{2023}) $$ But I don't know what to do next. How to prove that this greater or equal $0$? Maybe I do something wrong or this problem should be solved by using another formula or method? Please explain how to solve it. I appreciate all hints!
Presumably $A$ is meant to be a real matrix, otherwise the statement is not true at all.
Hint: Let $f(x) = 1+x+\ldots+x^{2022}$. If $A$ has eigenvalues $\lambda_1, \ldots, \lambda_n$ (counting multiplicities), then $\det(f(A)) = \prod_j f(\lambda_j)$. What can you say about $f(\lambda_j)$ if $\lambda_j$ is real? What about $f(\lambda_j) f(\overline{\lambda_j})$ if $\lambda_j$ is not real?
Further hint: the fact that $f(x) = (x^{2023}-1)/(x-1)$ may be useful.