# Matrix on complex field.

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.

• For $2)$, consider $A = \left(\begin{array}{rr} -1 & 0 \\ 0 & -1\end{array}\right)$. – Cameron Williams Jun 23 '15 at 1:16
• @cameron this is already a diagonal maatrix such that $A^2=I_2$ and eigen values are -1,-1, its characteristic poly is $(x+1)^2$ $tr(A)=-2$ – aryan Jun 23 '15 at 1:25
• @CameronWilliams you just buldozen my answer :( – aryan Jun 23 '15 at 1:28
• @CameronWilliams according the your hint ans are 1),3),4). can we reach to general result by a simple example? – aryan Jun 23 '15 at 1:34
• @aryan : No, you can not generalize the result but you can discard the options, like by the given matrix option 2) is not right. – Chiranjeev_Kumar Jun 23 '15 at 2:09

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 $$A + A^2 + \cdots + A^{k-1} = -I$$
4) is true. Note that $A$ is necessarily invertible, and that $$A + A^2 + \cdots + A^{k-1} = -I \implies\\ A^{-k}(A + A^2 + \cdots + A^{k-1}) = -A^{-k} \implies\\ A^{-1} + A^{-2} + \cdots + A^{-k+1} = -A^{-k} = -[A^{k}]^{-1} = -I^{-1} = -I$$\