An $n\times n$ complex matrix $A$ satisfies $A^k=I_n$, where $I_n$ is $n\times n$ identity matrix and $k$ is positive integer $\gt1$. Suppose that 1 is not an eigen value of $A$. Then which of the fallowing is necessarily true?
1) $A$ is diagonalizble
2) $A+A^2+...+A^{k-1}=0$
3) $tr(A)+tr(A^2)+...+tr(A^{k-1})=-n$
4) $A^{-1}+A^{-2}+...+A^{-(k-1)}=-I_n$
I think 1) and 2) are ture.
If $A^k = I$, then we have $A^k - I = 0$. Factoring, we have $$ (A - I)(I + A + A^2 + \cdots + A^{k-1}) = 0 $$ Since $1$ is not an eigenvalue, $A-I$ is invertible so that $$ I + A + A^2 + \cdots + A^{k-1} = 0 $$ In fact, $A$ satisfies the hypothesis of your question if and only if the above equality holds. Let $p(x)$ denote the polynomial $$ p(x) = 1 + x + x^2 + \cdots + x^{k-1} $$
1) necessarily holds. Note that $p(x)$ factors as $$ p(x) = \prod_{j=1}^{k-1} (x - e^{2 \pi ij/k}) $$ since the minimal polynomial of $A$ divides $p$, we may conclude that $A$ is diagonalizable.
2) Never holds. In particular, since $p(A) = 0$, we have $$ A + A^2 + \cdots + A^{k-1} = -I $$
3) holds by the above equality
4) is true. Note that $A$ is necessarily invertible, and that $$ A + A^2 + \cdots + A^{k-1} = -I \implies\\ A^{-k}(A + A^2 + \cdots + A^{k-1}) = -A^{-k} \implies\\ A^{-1} + A^{-2} + \cdots + A^{-k+1} = -A^{-k} = -[A^{k}]^{-1} = -I^{-1} = -I $$\
