I've been reading a paper called "Separation of variables for the quantum $Sl(2,R)$ spin chain" in which the author at one point does a power series expansion I do not understand. The problem is this is a crucial point of understanding the whole Bethe ansatz mechanic so I would like to pass this question on to you. He defines the so called transfer matrix
(2.16) $t(u) = tr \, T(u) = A(u) + D (u)$ where
(2.13) $T(u) = \begin{pmatrix} A(u) && B(u) \\ C(u) && D(u) \end{pmatrix} = L_1(u) \cdot\dotsc\cdot L_N(u) = \prod_{n=1}^N L_n(u)$
$T(u)$ is called the monodromy matrix the $L_n$ Operators are Lax matrices and defined through the spin operators via: (2.12) $L_n = \begin{pmatrix} u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3 \end{pmatrix} $ with $I$ being the Identity matrix and $S_a = \frac{1}{2} \sigma_a , S_\pm = S_1 \pm i S_2 $ . The $L$ matrices act nontrivially in different spaces $L_1$ acts in $V_0 \otimes V_1 \otimes \mathrm{I}_2 \otimes \cdots \otimes \mathrm{I}_N$ nontrivial only in $V_0$ and $V_1$ and so on, $L$ acts on $V_0 \otimes V_1 \otimes \dots \otimes V_N $ thus $T$ acts nontrivially on $V_0 \otimes V_1 \otimes \dots \otimes V_N $ The matrix $L$ and $T$ itself act on our auxiliary space $V_0$ while their entries act on our physical space $V_1 \otimes \cdots \otimes V_N$
Now to the main question, in the paper he states: "Substituting (2.13) into (2.16) and taking into account the explicit form of the Lax operator (2.12) one finds that $t(u)$ is a polynomial of degree $N$ in the spectral parameter $u$ with the operator valued coefficients: $t(u) = 2u^N + q_2u^{N-2} + \dots + q_N $"
I don't see how exactly he arrives at that form? I substitue 2.13 into 2.16 and arrive at: $t(u) = A(u) + D (u) = tr_0(L_1(u) \dots L_N(u))$ now if I take 2.12 into account I get: $t(u) = tr_0(L_1(u) \dots L_N(u)) = tr_0(\begin{pmatrix} u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3 \end{pmatrix}_1 \dots \begin{pmatrix} u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3 \end{pmatrix}_N)$ using $tr_{a,b}(A \otimes B ) = tr(A) tr(B) $
I arrive at the form: $t(u) = (u \mathrm{I} + i S_3 + u \mathrm{I} - iS_3)^N = (2u\mathrm{I})^N$.
I would like to know now how does he obtain the solution: $t(u) = 2u^N + q_2u^{N-2} + \dots + q_N $
Is there something I overlooked?
EDIT: I screwed up taking the right trace, question can be deleted.
Let $L$ denote the matrix $$ L = \pmatrix{u \mathrm{I} + i S_3 && i S_- \\ i S_+ && u \mathrm{I} - i S_3} $$ My best guess is that whatever the author is getting at has something to do with the fact that $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_N $$ or, if $N$ is odd, $$ \operatorname{tr}(L^N) = 2I\,u^N + q_{2}u^{N-2} + \cdots + q_Nu $$ I could come up with a proof that this will generally happen (i.e. there are no terms $u^{N - k}$ for odd $k$) if that's something you're interested in (this is purely a consequence of how matrix multiplication works). In the mean time, note that $$ \operatorname{tr}(L^2) = 2u^2 - 2S_3^2 - S_+\,S_- - S_-\,S_+\\ \operatorname{tr}(L^3) = 2u^3 - (6S_3^2 + 3 S_+S_- + 3S_-S_+)u $$