# Can a regular graph have the same Laplacian spectrum with a non-regular one?

So the Laplacian matrix of an undirected graph $$G$$ is $$L(G)=D(G)-A(G)$$, where $$D(G)$$ is the diagonal degree matrix and $$A(G)$$ is the adjacency matrix, as usual.

I can easily prove the case when the other graph $$H$$ is regular but with a different order, using the trace. But I'm struggling with the non-regular case.

Let the degrees of the graph $$G$$ be $$d_1, d_2, \dots, d_n$$.
Then the trace of $$L(G)$$ gives us the sum $$\sum_{i=1}^n d_i$$. On the other hand, the trace of $$L(G)$$ is the sum of the eigenvalues; therefore $$\sum_{i=1}^n d_i$$ is determined by the Laplacian spectrum.
In $$L(G)^2$$, the diagonal entries are $$d_1^2 - d_i, \dots, d_n^2 - d_n$$, so the trace of $$L(G)^2 + L(G)$$ gives us the sum $$\sum_{i=1}^n d_i^2$$. On the other hand, this trace can also be computed from the eigenvalues: it is $$\sum_{i=1}^n (\lambda_i^2 + \lambda_i)$$. Therefore $$\sum_{i=1}^n d_i^2$$ is also determined by the Laplacian spectrum.
Therefore the spectrum of $$G$$ lets us compute $$n \sum_{i=1}^n d_i^2 - \left(\sum_{i=1}^n d_i\right)^2 = \sum_{i \ne j} (d_i - d_j)^2$$ which is $$0$$ if and only if the graph is regular. (This is essentially the equality case of the Cauchy-Schwarz inequality.) As a result, it's also determined by the Laplacian spectrum of $$G$$ whether or not $$G$$ is a regular graph; a regular graph cannot be cospectral with a non-regular one.