TITLE: On two special kind of invertible similar matrices with rational entries
QUESTION [11 upvotes]: Let $A,B \in GL(n, \mathbb Q)$ be two similar matrices i.e. there exists $X \in GL(n, \mathbb Q)$ with $XAX^{-1}=B.$
If there is an integer $s$ such that $A^{s+1}B=BA^s$, then how to prove that $A,B$ are identity matrices?

REPLY [2 votes]: In a first step we show that the eigenvalues of $A$ are roots of unity.
We denote the not necessarily distinct eigenvalues of $A$ by $\{\lambda_1,...,\lambda_N\}$,
$N$ the size of $A$.
As $BA^sB^{-1}=A^{s+1}$, the matrices $A^s$ and $A^{s+1}$ have the same eigenvalues.
This means that there is a permutation $p$ such that $\lambda_i^s=\lambda_{p(i)}^{s+1}$ for all $i$.
It follows inductively that $\lambda_i^{s^j}=\lambda_{p^j(i)}^{(s+1)^j}$ for all $j$.
Now there is a positive integer $d$ such that $p^d=id$. Then we obtain that
$\lambda_i^{(s+1)^d-s^d}=1$ for all $i$ and all eigenvalues are roots of unity.
In the second step we show that $A$ is diagonalisable. Otherwise, let $k+1>1$  denote the
maximal length of a Jordan block of $A$. Now we write $A=S+N$ with a diagonalisable $S$ and a
nilpotent $N$ which commute. We have $N^{k+1}=0$ and $N^{k}\neq0$.
For positive integers $r$, we write
$$A^r=S^r+\binom{r}1\,S^{r-1}N+\cdots+\binom{r}k S^{r-k}N^k.$$
As the eigenvalues of $A$ have modulus 1, there exist positive $a_1,b_1$ such that
$a_1r^k\leq||A^r||\leq b_1r^k$ for all $r$. Here we use any submultiplicative matrix norm. This fact is proved at the end of the solution.
Since $B$ is similar to $A$, there also are positive constants
$a_2,b_2,a_3,b_3$ such that
$a_2r^k\leq||B^r||\leq b_2r^k$ and
$a_3r^k\leq||B^{-r}||\leq b_3r^k$ for all $r$.
Now the equation  $BA^sB^{-1}=A^{s+1}$ implies inductively that
$$\tag{1}B^rA^{s^r}B^{-r}=A^{(s+1)^r}\mbox{ for all }r.$$
We obtain the inequalities
$$a_1 (s+1)^{rk}\leq ||A^{(s+1)^r}||=||B^rA^{s^r}B^{-r}||\leq b_1b_2b_3r^{2k}s^{rk}$$
for all $r$. For sufficiently large $r$, this is false and we have a contradiction.
Therefore $k=0$ and $A$ is diagonalizable.
Consider now the order $m$ of $A$ (and $B$), that is the smallest positive integer such
that $A^m=I$. It exists because the eigenvalues of $A$ are roots of unity and
$A$ is diagonalisable. We claim that $m=1$ and hence $A=B=I$.
Otherwise (1) implies for $r=m$ using $B^m=I$ that $A^{(s+1)^m-s^m}=I$.
Therefore $m$ divides $(s+1)^m-s^m$. Observe that $m$ must be coprime to $s$ and $s+1$ in this case.
Consider now the smallest prime divisor $p$ of $m$ and its multiplicity $e$.
Then by the binomial theorem $(s+1)^p\equiv s^p+1\mod p$ and inductively also
$(s+1)^{p^e}\equiv s^{p^e}+1\mod p$. In particular $(s+1)^{p^e}$ and $s^{p^e}$ are different
modulo $p$. Now the multiplicative group $({\mathbb Z}/p{\mathbb Z})^\times$ has order $p-1$ and
the quotient $u=m/p^e$ is coprime to $p-1$ because $p$ is the smallest prime divisor of $m$.
Therefore also $(s+1)^{up^e}$ and $s^{up^e}$ are different modulo $p$ and hence also modulo
$m$. Contradiction!
Appendix: Consider $A=S+N$ with a diagonalisable $S$ and a nilpotent $N$ which commute. Assume that all eigenvalues of $A$ have modulus 1 and that $k\geq0$ is such that $N^k\neq0$ and $N^{k+1}=0$. Consider any submultiplicative matrix norm $||.||$.
Then there exist positive constants $a,b$ such that for all positive integers $r$ one has $a\,r^k\leq||A^r||\leq b\,r^k$.
For the proof, we use that for positive integers $r$, we can write
$$\tag{2}A^r=S^r+\binom{r}1\,S^{r-1}N+\cdots+\binom{r}k S^{r-k}N^k.$$
Without loss of generality, we can assume that $S$ is diagonal. Now all norms on a finite dimensional vector space are equivalent and
$||S^r||_\infty=1$ for all $r$ and for the matrix norm
$$||(d_{ij})||_\infty=\max_{i}\sum_j|d_{ij}|$$
corresponding to the maximum norm for vectors.
Hence there are positive constants $c,d$ such that for all $r$ we have $d\leq||S^r||\leq c$.
This implies using (2) that
$$||A^r||\leq c+c|||N|r+\cdots\binom rk c||N^k||\leq
     c(1+||N||+\cdots+||N^k||)r^k$$
for all positive integers $k$ as wanted.
We also have for all $r$
$$||A^r||\geq \binom rk d ||N^k||-c-\cdots -\binom r{k-1} c||N^{k-1}||.$$
Now if $r\geq2k$ then $\binom rk\geq \frac {r^k}{2^kk!}$. As above, we can find a positive constant $M$ such that
$$c+\cdots +\binom r{k-1} c||N^{k-1}||\leq Mr^{k-1}$$
for all positive integers $r$. Together, this gives $$||A^r||\geq r^k \left(\frac{d||N^k||}{2^kk!}-\frac Mr\right)$$ for all $r\geq 2k.$
Therefore if $r\geq 2k$ and $r\geq 2M\,2^kk!/(d||N^k||)$ we find that $||A^r||\geq\frac{d||N^k||}{2^{k+1}k!}r^k$.
Finally, if $a$ denotes the minimum of $\frac{d||N^k||}{2^{k+1}k!}$ and $||A^r||/r^k$ for $1\leq r\leq \max(2k,2M\,2^k k!/(d||N^k||)$ then we have
$||A^r||\geq  a r^k$ for all positive $r$ as wanted.