If $B$ and $C$ are $n * n$ complex matrices such that $BC=qCB$ for some $q$ in $ℂ$, then we can apply the following argument to show that $q^m = 1$ for some $m ≥ 1$.
Let's assume that $q$ is an eigenvalue of $BC$ with corresponding eigenvector $v$. Then we have:
$BCv = qCBv = qC(Bv)$
Since v is nonzero, we can divide both sides by Cv to obtain:
$Bv = qv$
Therefore, $v$ is also an eigenvector of $B$ with eigenvalue $q$. Applying the same argument to $C$, we see that $v$ is also an eigenvector of $C$ with eigenvalue $q$.
Now, let's consider the matrix product $BC...CB$, where there are $m$ factors of $BC$ and $m-1$ factors of $CB$. By repeatedly applying the above argument, we see that each factor of $BC$ or $CB$ contributes a factor of $q$ to the eigenvalue of the product. Therefore, the eigenvalues of $BC...CB$ are $q^m$ times the eigenvalues of the identity matrix, which are all equal to $1$. Thus, we have:
$det(BC...CB) = q^{m*n}$
On the other hand, we can also write $BC...CB as (BC)^{m-1}CB$, so we have:
$det(BC...CB) = det((BC)^{m-1}) * det(CB)$
Since $BC=qCB$, we have $det(BC) = q^n det(CB)$, so $det((BC)^{m-1}) = q^{m*n} det(CB)^{m-1}$. Substituting this into the above equation, we get:
$q^{m*n} det(CB)^{m-1} det(CB) = q^{m*n} det(CB)^m$
Since $det(CB) ≠ 0$ (otherwise $B$ or $C$ would be singular), we can divide both sides by $det(CB)^{m-1} det(CB)$ to obtain:
$q^{m*n} = det(CB)$
Therefore, we have shown that $q^m = det(CB)^{1/n}$ for some $m ≥ 1$. Since $det(CB)$ is a complex number of magnitude $1$ (since $B$ and $C$ are both invertible and have the same size), $q^m$ must also be a complex number of magnitude $1$, so we must have $q^m = 1$ for some integer $m ≥ 1$. Please cheak this or give another solution
The statement is false. Consider e.g. $C=0$ and $|q|\ne1$.
For a less obvious counterexample, consider $B=\pmatrix{2&0\\ 0&1},\,C=\pmatrix{0&1\\ 0&0}$ and $q=2$.
The statement in question is true if $BC$ is not nilpotent, i.e., if $BC$ has at least one nonzero eigenvalue, but your proof is valid only when both $B$ and $C$ are invertible.
Suppose $BC$ has at least one nonzero eigenvalue. Then $q$ must be nonzero, otherwise $BC=qCB=0$ has only zero eigenvalues. Let $\lambda$ be any nonzero eigenvalue of $BC$. Since $CB=\frac{1}{q}BC$, $\frac{1}{q}\lambda$ is an eigenvalue of $CB$. However, by Sylvester’s secular theorem, $BC$ and $CB$ have the same spectrum. Hence $\frac{\lambda}{q}$ is also an eigenvalue of $BC$. So, inductively, every number in the list $\{\lambda,\frac{\lambda}{q},\frac{\lambda}{q^2},\frac{\lambda}{q^3}\ldots\}$ is an eigenvalue of $CB$. Since $CB$ has only finitely many eigenvalues, this list must contain repeated values. Hence $\frac{\lambda}{q^r}=\frac{\lambda}{q^s}$ for some $r<s$. In turn, $q^{r-s}=1$.