Which of the following are true about a square matrix of order $n$?

Question

An $n\times n$ complex matrix $A$ satisfies $A^k=I_n$, the $n\times n$ identity matrix, where $k$ is a positive integer $\gt1$. Suppose $1$ is not eigenvalue of $A$. Then which of the following are necessarily true?

1. $A$ is diagonalizable.
2. $A+A^2+A^3+\dots+A^{k-1}=0$
3. $t(A)+tr(A^2)+\dots+tr(A^{k-1})= -n$
4. $A^{-1}+A^{-2}+...+A^{-(k-1)}=-I_n$

I think option $1$ is true. But not able to decide about rest of options.

Answer 1

With the given hypotheses, you can see that $X^k-1$ annihilates $A$, and since the polynomial has no multiple roots this implies that $A$ is diagonalisable (with eigenvalues contained in the set of $k$-th roots of unity). But $A-1$ is invertible, from which it follows that the quotient polynomial $(X^k-1)/(X-1) =X^{k-1}+ X^{k-2}+\cdots+X+1$ already annihilates $A$. It follows that the expression in point 2 is definitely not $0$, but rather $-I$. Taking traces, we see that point 3 does hold.

Obviously $A$ is invertible and $A^{k-1}=A^{-1}$, so the relation $A^{k-1}+ A^{k-2}+\cdots+A=-I_n$ can be read as $A^{-1}+ A^{-2}+\cdots+A^{-(k-1)}=-I_n$, which is point 4.

Answer 2

Let $\mu$ denote the minimal polynomial of $A$ over $\mathbb C$ and $\chi$ its characteristic polynomial.

Lemma: $\mu$ and $\chi$ have the same set of roots (see here).

1. True. $X^k-1$ is a polynomial with simple roots that annihilate $A$, therefore $A$ is diagonalizable. Furthermore, if you let $\mu$ denote the minimal polynomial of $A$ over $\mathbb C$, then $\mu$ divides $X^k-1$. By the lemma, the eigenvalues of $A$ are among the $k$-th roots of unity (except $1$ by assumption).

2. True. Since $1$ is not an eigenvalue of $A$, it is not a root of $\chi$, hence not a root of $\mu$. So $\gcd(X-1, \mu)=1$ This combines with the fact that $\mu$ divides $X^k-1=(X-1)(X^{k-1}+\ldots+1)$ implies that $\mu$ divides $X^{k-1}+\ldots+1$. Hence $X^{k-1}+\ldots+1$ annihilates $A$.

3. True. Let $\alpha_1,\ldots,\alpha_n$ denote the eigenvalues of $A$ counted with their multiplicity(they need not be distinct). Since $A$ is diagonalizable, $tr(A)=\sum_{i=1}^n \alpha_i$ and $tr(A^{2})=\sum_{i=1}^n \alpha^{2}_i$, and it goes on until $ tr(A^{k-1})=\sum_{i=1}^n \alpha^{k-1}_i$.

So $tr(A)+tr(A^2)+\dots+tr(A^{k-1})= \sum_{j=1}^{k-1}\sum _{i=1}^n \alpha^{j}_i=\sum _{i=1}^n \left( \sum_{j=1}^{k}\alpha^{j}_i - 1\right) = -n $ since the sum of the $k$-th roots of unity is $0$.

4. True. The eigenvalues of $A$ are nonzero, therefore $A$ is invertible and $A^{-1}$ is diagonalizable and its eigenvalues are the inverses of the eigenvalues of $A$ (with the same multiplicities). But, if $\alpha$ is a $k$-th root of unity, $\frac{1}{\alpha}=\bar{\alpha}$, which is still a $k$-th root of unity. Therefore (ask for details if needed) , $A$ annihilates $\frac{X^k-1}{X-1}$