A surprising result about the product of Blaschke matrices

Upon setting $\lambda_{n+1}=0$, we see that $$S(x) = \frac{1}{x}\begin{bmatrix} 0 & 1 \end{bmatrix}\left[\mathbf M_{n+1}(x)\mathbf M_n(x)\mathbf M_{n-1}(x)\dots\mathbf M_{2}(x)\mathbf M_1(x)\right]\begin{bmatrix} 0 \\ 1\end{bmatrix}.$$ Thus (working with $n$ instead of $n+1$), we see that we have to control the lower right entry of $$\mathbf M(x)=\mathbf M_n(x)\mathbf M_{n-1}(x)\dots\mathbf M_{2}(x)\mathbf M_1(x).$$

By induction, one verifies that $g_i(\lambda_i)-g_i(x)$ is a scalar multiple (factor depending on $i$) of $\frac{x-\lambda_i}{1-x\lambda_{i-1}}$ for $i\ge 2$. Set $$\Gamma(x)=\prod_{i=1}^n\frac{x-\lambda_i}{1-x\lambda_i}\text{ and }g(x)=\prod_{i=1}^n(g_i(\lambda_i)-g_i(x)).$$ Thus $$ g(x)=c\Gamma(x)(1-x\lambda_n)\tag{1} $$ for a constant $c$.

In the following, we drop the argument $x$ if the meaning is clear. From the recursion for $g_i$ we obtain $$\mathbf M_i\begin{bmatrix} 1 \\ g_i\end{bmatrix}=(g_i(\lambda_i)-g_i(x))\begin{bmatrix} 1 \\ g_{i+1}\end{bmatrix}.$$ Iterated application yields $$\mathbf M\begin{bmatrix} 1 \\ g_1\end{bmatrix}=g\begin{bmatrix} 1 \\ g_{n+1}\end{bmatrix}.\tag{2}$$

For a term $A$ in $x$ define $\tilde A$ by $\tilde A(x)=A(1/x)$. Then $$\tilde{\mathbf M}_i^t\mathbf M_i=(g_i(\lambda_i)^2+1)\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix},$$ hence $$\tilde{\mathbf M}^t\mathbf M=\gamma\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix},$$ where $$\gamma=\prod_{i=1}^n(g_i(\lambda_i)^2+1).$$ Now we apply the tilde transformation to equation (2): $$\tilde{\mathbf M}\begin{bmatrix} 1 \\ \tilde g_1\end{bmatrix}=\tilde g\begin{bmatrix} 1 \\ \tilde g_{n+1}\end{bmatrix}.$$ Transposing and multiplying from the right with $M$ yields $$\gamma\begin{bmatrix} 1 & \tilde g_1 \end{bmatrix}=\begin{bmatrix} 1 & \tilde g_1 \end{bmatrix}\tilde{\mathbf M}^t\mathbf M=\tilde g\begin{bmatrix} 1 & \tilde g_{n+1} \end{bmatrix}\mathbf M.$$ Together with the previous relation (2) for $\mathbf M$ we obtain four linear equations for the four entries of $\mathbf M=\begin{bmatrix} m_1 & m_2\\ m_3 & m_4 \end{bmatrix}$ in terms of $\gamma,g_1,\tilde g_1,g_{n+1},\tilde g_{n+1},g,\tilde g$, namely \begin{align*} m_1+m_2g_1&=g\\ m_3+m_4g_1&=gg_{n+1}\\ m_1\tilde g+m_3\tilde g_{n+1}\tilde gt&=\gamma\\ m_2\tilde g+m_4\tilde g_{n+1}\tilde g&=\tilde g_1\gamma. \end{align*} The rank of the system is $3$. Use the first, second and fourth equation to express $m_1,m_2,m_3$ in terms of $m_4$. As expected, the third equation does not allow to solve for $m_4$, rather it collapses to $$(g_{n+1}\tilde g_{n+1} + 1)g\tilde g=(g_1\tilde g_1+1)\gamma,\tag{3}$$ a relation we will need in a moment.

We can determine $m_4$ from $$m_1m_4-m_2m_3=\det\mathbf M=\gamma\Gamma(x)$$ and obtain $$ m_4(x)=\frac{(g\tilde g_1g_{n+1} + \tilde g\Gamma)\gamma}{(g_{n+1}\tilde g_{n+1} + 1)g\tilde g}.$$

Together with (3) this gives $$m_4(x)=\frac{g\tilde g_1g_{n+1} + \tilde g\Gamma}{g_1\tilde g_1 + 1}.$$

For $1\le i\le n-1$ we have $g(\lambda_i)=0$, and $\lambda_i$ is not a pole of $\tilde g_1$, nor of $g_{n+1}$.

However, $\lambda_i$ is a root of $\Gamma$, and potentially a pole of $\tilde g$. So in order to evaluate $\tilde g\Gamma$ in $\lambda_i$ we apply the tilde operation to equation (1) and obtain $$\tilde g=c\tilde\Gamma\left(1-\frac{\lambda_n}{x}\right).$$

It follows from the definition of $\Gamma$ that $\tilde\Gamma\Gamma=1$, hence $$\tilde g\Gamma=c\frac{x-\lambda_n}{x}.$$

Now we can evaluate in $\lambda_i$ and $\lambda_j$, and obtain $$ \frac{m_4(\lambda_i)}{m_4(\lambda_j)} =\frac{\lambda_j\;(g_1(\lambda_j)g_1(1/\lambda_j) + 1)\;(\lambda_n - \lambda_i)}{\lambda_i\;(g_1(\lambda_i)g_1(1/\lambda_i) + 1)\;(\lambda_n - \lambda_j)}.$$ Thus, as explained in the beginning, with $\lambda_n=0$ we get the case $n-1$ of the question.