M_k &= L_k , \\ 
M_i &= M_{i+1}{\mathbin{\Pi}}_{\beta_{i+1}}L_i.\end{aligned}$$ Then, the following hold:

1.  $M_1 = L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}$.

2.  Any eigenvalue of $M_i$ (for $i<k$) may be found as a solution to the equation $$\label{eq:specRecursive}
        \beta_{i+1}\mu^{{\left(i+1\right)}} = \frac{\phi_{L_i}(x)}{\phi_{L_i{^\prime}}(x)}$$ for some $\mu^{{\left(i+1\right)}}\in\text{spec}{\left\{M_{i+1}\right\}}$.

First, we prove statement 1. It can be seen that $$\begin{aligned}
M_{k-1} &= M_k{\mathbin{\Pi}}_{\beta_{k}}L_{k-1} = L_k{\mathbin{\Pi}}_{\beta_{k}}L_{k-1} \nonumber \\ 
&= \frac{1}{\alpha_{k-1}}{\left(\alpha_{k}L_k\otimes D_{k-1} + \alpha_{k-1}{\mathbbm{1}}_{k}\otimes L_{k-1}\right)},\\
M_{k-2} &= M_{k-1}{\mathbin{\Pi}}_{\beta_{k-1}}L_{k-2} \nonumber \\
&= \frac{1}{\alpha_{k-2}}(\alpha_{k}L_k\otimes D_{k-1}\otimes D_{k-2}\ + \nonumber \\
\alpha_{k-1}{\mathbbm{1}}_{k}&\otimes L_{k-1}\otimes D_{k-2} + \alpha_{k-2}{\mathbbm{1}}_{k-1}\otimes {\mathbbm{1}}_{k-2}\otimes L_{k-2}),\end{aligned}$$ and so on, until we have an ${{\vec{\alpha}}}$-weighted sum over all $k$ of the base graphs (with an overall denominator of $\alpha_1=1$), which is precisely $L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}$.

The proof of statement 2 follows as a direct consequence of Lemma \[lm:spec2\], with $K=M_{i+1}, L=L_{i},$ and $\alpha=\beta_{i+1}$.

Theorem \[th:specMain\] provides an algorithm to compute the spectrum of $L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}$, namely:

1.  Compute the relative weight vector ${{\vec{\beta}}}$ from ${{\vec{\alpha}}}$.

2.  Start with $i=k$, where the spectrum of $M_k = L_k$ is known. Decrease $i$ by one.

3.  Compute the spectrum of $M_i$ from the known spectrum of $M_{i+1}$ and Eq. . Decrease $i$ by one.

4.  Perform step 3 repeatedly, halting at $i=0$. Return the spectrum of $M_1 = L^{{\mathbin{\Pi_{\vec{\alpha}}}}k}$.

Therefore, given 