Let $k$ be an odd integer. Let $A$ be a matrix such that $A^k = A + I$. Prove that $\det(A) > 0$.

Question

Problem

Let $k$ be an odd integer. Let $A$ be a matrix such that $A^{k} = A+ I$. Prove that $\det(A) > 0$.

My attempt:

We know that $\det(A) = \lambda_1\dots\lambda_n$ where $\lambda_1,\dots,\lambda_n$ are the complex eigenvalues of $A$. We know that $\lambda_1,\dots,\lambda_n$ are roots of $X^{2k+1} - X -1$. I can see graphically that the only real root of $X^{2k+1} -X -1$ is positive by plotting some graphs, and since the product of two conjugate complex numbers is positive, I think it has something to do with it.

The issue is that if $\lambda$ is an eigenvalue of $A$, I'm not sure that $\bar{\lambda}$ is an eigenvalue as well since $A$ possibly has complex coefficients (it's not specified in the statement).

Answer 1

Let $n$ be such that $A$ is an $n×n$ matrix. Let us also assume that $k$ is positive.

Then $$\det(A) =\prod_i \lambda_i^{c_i},$$ where the $\lambda_i$s are the algebraic eigenvalues of $A$ and $c_i$, for each such integer $i$, is the algebraic multiplicity of $\lambda_i$ as an algebraic eigenvalue of $A$. Then each $\lambda_i$ is the root of the polynomial $X^{2k+1}-X-1$. The $c_i$s as above are positive integers that sum to $n$.

The challenge of the problem, however, is this: We do not know a priori precisely which subset of the roots of the equation $X^{2k+1}-X-1$, are the algebraic eignevalues of $A$, and furthermore, with which multiplicity does it appear as an algebraic eigenvalue of $A$. For example, the real root of the polynomial $X^{2k+1}-X-1$ may be the one algebraic eignvalue of $A$, with multiplicity $n$.

However, if $A$ is real, then one can look at the polynomial $\det(A-\lambda I)$ [a real polynomial in $\lambda$] and note the following:

1. If $\lambda_i$ is a complex non real algebraic eigenvalue of $A$ with multiplicity $c_i$ for some positive intrger $c_i$, then $\bar{\lambda}_i$ is also an algebraic eigenvalue of $A$, with the same multiplicity $c_i$. However, for each such $c_i$ note that $\lambda_i^{c_i}\bar{\lambda}_i^{c_i}$ is $|\lambda_i|^{2c_i}$ which is necessarily positive.

2. So from 1. above it follows that, assuming $A$ is real, that $\det(A)$ must be of an integral power of the real root $y$ of the polynomial $X^{2k+1}-X-1$, times a positive number [make sure you see why]. More precisely, $\det(A)$ must be of the form

$$\det(A) = y^c \prod_i \lambda_i^{c_i}\bar{\lambda}_i^{c_i}$$ $$= \ y^c \prod_i |\lambda_i|^{2c_i}.$$ But as already observed in the OP, $y$ is positive. So from this it follows that $\det(A)$ must also be positive, if $A$ is real.

If $A$ is allowed to not be real then the claimed result does not always hold. Indeed, let $A$ be a $1×1$ matrix where the one entry of $A$ is a complex non-real root of $X^{2k+1}-X-1$. Then $\det(A)$ is simply that nonreal root, which is not positive.

Answer 2

COMMENT.-The problem have correct sens only when the matrices are with entries belonging to an ordered field; we choose real numbers. It is difficult to have examples illustrating the problem. We examine the cas $k=3$ with matrices $2\times 2$ so we have $$A^3=A+I$$ which is equivalent to $$A(A^2-I)=I$$ In other words, $A$ should be invertible and having as inverse $A^2-I$.Because $$\det(A)\det(A^2-I)=1$$ $\det(A)$ and $\det(A^2-I)$ must have equal sign (could it be negative?).

If $A=\pmatrix{a&b\\c&d}$ then $$\begin{cases}\det(A^2-I)=a^2d^2+b^2c^2+1-(a^2+2abcd+2bc+d^2)\\\det(A)=ad-bc\end{cases}$$ I stop here. Anyone could try to find out an example or counterexample with this.

Answer 3

Assuming that $A$ is intended to be a real matrix, say $A \in \mathrm{Mat}_n(\mathbb R)$, then its eigenvalues (say $\lambda_1,\ldots,\lambda_n\}$ where the $\lambda_i$ are not necessarily distinct) consist of pairs $(w,\bar{w})$ with $w \neq \bar{w}$ and the set of real roots of $f(t) = t^{2k+1}-t-1$.

But $f'(t) = (2k+1)t^{2k}-1$, hence if we let $\eta_k = (\frac{1}{2k+1})^{1/2k}$, we see that $f'(t)$ is positive outside of the interval $[-\eta_k,\eta_k]$ and negative in the interior of that interval (and vanishes at the end points). Thus $f$ is increasing on $(-\infty,-\eta_k)$, decreasing $(-\eta_k,\eta_k)$ and increasing again on $(\eta_k,\infty)$. But it is clear that $f$ is negative on $[-\eta_k,\eta_k]$, hence $f$ has exactly one real root, which must be greater that $\eta_k$, and hence is positive.

It follows that $\det(A)$ is a product of the moduli of its non-real eigenvalues times its unique real, and necessarily positive, eigenvalue, and hence $\det(A)>0$ as required.