For a matrix $A$, Let $a_{i,i+1} = c_i$ for $i=1,...n$, and the other matrix elements are zero.
Prove that the determinant of $I + A + A^2 + ... + A^n = (1-c)^{n-1}$
where $c = c_1...c_n$
The problem is from http://staff.math.su.se/mleites/books/prasolov-1994-problems.pdf
Here is its original wording and solution. Can someone explain it for me? still confused even after reading it.
As John Hughes and I have indicated in the comments, the indices are meant to be taken modulo $n$. So for instance in the case of $n = 4$, the matrix $A$ we are talking about is $$ A = \pmatrix{&c_1\\&&c_2\\&&&c_3\\c_4} $$ with the blank entries equal to zero. With that established, the steps in the solution make sense; we need only verify that $\det(I - A) = 1-c$. That is, we need to compute the determinant of $$ I - A = \pmatrix{1&-c_1\\&1&-c_2\\&&\ddots&\ddots\\ &&&&-c_{n-1}\\-c_n&&&&1}. $$ If we use the Laplace expansion along the final row, we have $$ \det(I - A) = (-1)^{n-1}\cdot (-c_n) \det(M_1) + 1 \cdot \det(M_2), $$ where $$ M_1 = \pmatrix{-c_1\\1&-c_2\\&\ddots&\ddots\\ &&&-c_{n-2}\\ &&&1&-c_{n-1}}, \\ M_2 = \pmatrix{1&-c_1 \\ & 1&-c_2\\ &&\ddots&\ddots\\&&&&-c_{n-2} \\ &&&&1}. $$ Because $M_1$ and $M_2$ are lower and upper triangular respectively, we quickly find that $\det(M_1) = (-1)^{n-1} c_1 \dots c_{n-1}$ and $\det(M_2) = 1$. The conclusion follows.