factorisations overs matrices

Question

Can anyone explain to me this passage Given to matix $A$ and $B$ taht commutes and $q$ a positive integer when you consider the polynomial $X^{2q+1}+1$ it has as roots $\{-1,\alpha_1,......,\alpha_q,\beta_1,......,\beta_q\}$ such that the $\beta_i$'s are the conjugates of the $\alpha_i$'s than $A^{2q+1}+B^{2q+1}=(A+B)\prod_{k=1}^{q} (A+\alpha_k B)(A+\beta_k B)$ could anyone help me to understand this please ?

Answer 1

From the factorisation (over $\mathbf C$): $$X^{2q+1}+1=(X+1)\prod_{k=1}^q(X-\alpha_k)(X-\beta_k)$$ (note there's a sign error in your formula) you can deduce by homogeneisation the factorisation (set $X=Y/Z$ and multiply both sides by $Z^{2q+1}$): $$Y^{2q+1}+Z^{2q+1}=(Y+Z)\prod_{k=1}^q(Y-\alpha_kZ)(Y-\beta_kZ)$$ Now in this polynomial $\in\mathbf C[Y,Z]$, you can substitute $A$ and $B$ to $Y$ and $Z$ respectively, since $A$ and $B$ commute.