The Euclid's algorithm for computing the greatest common divisor of two polynomials $r_i(z)$ and $r_0(z)$ can be done iteratively as follows, where we assume $\deg r_ 1 > \deg r_0$. $$r_{-1}(z)=q_1(z)r_0(z)+r_1(z),\deg r_{-1}=\deg q_1+\deg r_0,\deg r_0>\deg r_1,$$ $$r_{0}(z)=q_2(z)r_1(z)+r_2(z),\deg r_{0}=\deg q_2+\deg r_1,\deg r_1 > \deg r_2,$$ $$r_{1}(z)=q_3(z)r_2(z)+r_3(z),\deg r_{1}=\deg q_3+\deg r_2,\deg r_2 > \deg r_3,$$ $$\vdots$$
This iteration can be rewritten in the matrix form as follows: $$ \begin{pmatrix} q_i(z) & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} r_{i-1}(z) \\ r_{i}(z) \end{pmatrix} = \begin{pmatrix} r_{i-2}(z) \\ r_{i-1}(z) \end{pmatrix} $$ We define $U_i(z),V_i(z)$ as $\begin{pmatrix} U_i(z) & U_{i-1}(z) \\ V_i(z) & V_{i-1}(z) \end{pmatrix}= \prod_{k=1}^{i} \begin{pmatrix} q_k(z) & 1 \\ 1 & 0 \end{pmatrix} $
According to equation $11$ of this article on page 91, $$\deg U_i = \deg r_{-1} - \deg r_{i-1}$$ will be satisfied. Is this obvious?
Author say this is equivalent to $\deg r_{i-1}=\deg r_{-1} - \sum_{j=1}^i \deg q_{i}$.
How to prove this ?
