What can I say about the following matrix $A$?

Question

Given a $n \times n$ complex matrix $A$ such that $A^k = I_n$, where $k > 1$. Also assume $1$ is not an eigenvalue of $A$. Then which of these follow?

(a) $A$ is diagonalizable.

(b) $A + A^2 + \cdots + A^{k-1} = O_n$

(c) $\mbox{tr}(A) + \mbox{tr}(A^2) + \cdots + \mbox{tr}(A^{k-1}) = -n$

(d) $A^{-1} + A^{-2} + \cdots + A^{-(k-1)} = -I_n$

Now I have the following observations and answers:

(b) is false as multiplying the equation by $A$ we get

$$A^2 + A^3 + \cdots + A^{(k-1)} + I_n = O_n$$

So, subtracting from the original equation, we get $A-I_n = O_n \implies A =I_n$ which is not true as $k>1$. Also,

$$A^{-1} = A^{(k-1)} \implies (A^{(k-1)})^{-1} = A^{-(k-1)} = A$$

So how can I conclude about other parts? Any hints would be nice. Thank you.

Answer 1

Here $$A=(a_{ij})_{n \times n\qquad \forall\quad a_{ij} \in \mathbb{C}}$$

where $A$ satisfies $$A^k=I_n\implies A^k-I_n=0$$

So minimal polynomial of $A$ is $m_A(x)=x^k-1$

Now $m_A(x)=0\implies x^k-1=0 \qquad \text{for}\quad k\gt 0,\quad k \in \mathbb{Z}$

$\implies x^k=1\implies x=1^{\frac{1}{k}}=1, w, w^2, . . . , w^{k-1}$

Therefore $$m_A(x)=x^k-1=(x-1)(x-w)(x-w^2)...(x-w^{k-1})$$

So the minimal polynomial has distinct linear factors.

Hence $A$ is diagonalizable. $\quad$(Option $(a)$ is Correct)

Now from the $k^{th}$ root os unity, $$1+w+w^2+...+w^{k-1}=0\implies w+w^2+...+w^{k-1}=-1 $$ By Cayley's theorem $$A+A^2+A^3+...+A^{k-1}=-I$$ $\quad$(Option $(b)$ is not Correct)

Again $$\text{trace}(A+A^2+A^3+...+A^{k-1})=\text{trace}(-I)=-n$$ $$\implies \text{trace}(A)+\text{trace}(A^2)+...+\text{trace}(A^{k-1})=-n$$ $\quad$(Option $(c)$ is Correct)

Again $$A^k=I\implies A^{-1}=A^{k-1},\quad A^{-2}=A^{k-2},\quad A^{-3}=A^{k-3},\quad...,\quad A^{-(k-2)}=A^{2},\quad A^{-(k-1)}=A$$

Also $$A+A^2+A^3+\quad...\quad+A^{k-2}+A^{k-1}=-I$$ $$\implies A^{-(k-1)}+A^{-(k-2)}+\quad . . . \quad +A^{-2}+A^{-1}=-I$$ $$\implies A^{-1}+A^{-2}+\quad . .. \quad+A^{-(k-1)}=-I$$ $\quad$(Option $(d)$ is Correct)

Answer 2

The following statements are equivalent:

1. $A^k = I $
2. $A^k - I = 0$
3. $(A - I)(I + A + \cdots + A^{k-1}) = 0$
4. $(A - \lambda_1 I)(A - \lambda_2 I) \cdots (A - \lambda_k I) = 0$

where $\lambda_1,\dots,\lambda_k$ are the $k$ complex solutions to $\lambda^k = 1$.

By statement 4, we have $p(A) = 0$ where $p(x) = (x-\lambda_1)\cdots (x - \lambda_k)$. Thus, the minimal polynomial of $A$ divides $p$. Thus, the minimal polynomial of $A$ has no repeated factors. Thus, $A$ is diagonalizable.

As you noted, b is false. Multiplying statement 3 by $(A - I)^{-1}$ on both sides confirms your result (since $1$ is not an eigenvalue of $A$, $(A-I)$ is invertible).

To confirm that c is correct, it suffices to note that we have $$ I_n + A + A^2 + \cdots + A^{k-1} = 0 \implies A + A^2 + \cdots + A^{k-1} = -I_n $$ and then take the trace of both sides of the above equation.

To confirm that d is correct, it suffices to note that $B = A^{-1}$ also satisfies $B^k = I_n$, which means that $B$ satisfies $B + B^2 + \cdots + B^{k-1} = -I_n $.