= \alpha K\otimes D_L + {\mathbbm{1}}_K \otimes L.$$ If $U_K$ is a unitary that diagonalizes $K$, we conjugate the above equation with the unitary $U_K\otimes{\mathbbm{1}}_L$, and look at the resulting block matrix. Each block corresponds to an eigenvalue of $K$, and thus the $j$-th block is given by $\alpha\kappa_jD_L + L$. The full spectrum may then be expressed as a disjoint union of the block spectra, $$\text{spec}{\left(K{\mathbin{\Pi_\alpha}}L\right)} = \bigsqcup_{j=1}^{|K|}\text{spec}{\left(\alpha\kappa_jD_L + L\right)}.$$ Now, we apply \[eq:thm310\] to $K{\mathbin{\Pi_\alpha}}L \equiv {\left(\alpha K\right)}{\mathbin{\Pi}}L$ and use the fact that $\phi_{\alpha K}(x) = \det{\left[x{\mathbbm{1}}- \alpha K\right]} = \alpha^{n_K}\det{\left[\frac{x}{\alpha}{\mathbbm{1}}- K\right]}\equiv \alpha^{n_K}\phi_{K}{\left(\frac{x}{\alpha}\right)}$. This yields Eq. , as desired.

Now we show that if the eigenvalues of $K$ and the polynomials $\phi_L$ and $\phi_{L{^\prime}}$ are known, then there is a straightforward procedure to compute the eigenvalues of $K{\mathbin{\Pi_\alpha}}L$.

\[lm:spec2\] Let $K$ and $L$ be graph Laplacians, as before. Each eigenvalue of the product characteristic polynomial $\phi_{{\mathbin{\Pi}}}$ can be found as a solution of the equation $$\label{eq:recursive}
  \alpha\kappa_i = \frac{\phi_{L}{\left(x\right)}}{\phi_{L{^\prime}}{\left(x\right)}}$$ for some $K$-eigenvalue $\kappa_i$.

Any eigenvalue of the product graph must be a zero of the left-hand side of Eq. (\[eq:HPchar\]) and, by equality, a zero of the right-hand side. Now, the degree of polynomial $\phi_K$ is $n_K$, which 