 Suppose p is a permutation of the rows of the identity matrix where pn vp is equal to a. We say that a and a' are orthogonally equivalent. Earlier we proved that isomorphic graphs have orthogonally equivalent adjacency matrices and conversely if the adjacency matrices are orthogonally equivalent then the graphs are isomorphic. And this gives us a way to easily determine if two graphs are isomorphic, provided we can easily find the right permutation matrix. But this is a hard problem. However, orthogonal equivalence is a special case of similar matrices. Suppose pn vp is equal to a, then we say that a and a' are similar matrices. The main difference is that p might not be a permutation matrix. But here's a key fact about similar matrices. Similar matrices have the same eigenvalues with the same multiplicities. Could this be useful? Let's find out. So before we start, we note that the adjacency matrix for a graph is square, symmetric, with real entries. And these are the required conditions for a powerful theorem about the eigenvalues of a. If a is a real symmetric matrix, all of the eigenvalues are real, eigenvectors for distinct eigenvalues are orthogonal, and a has an eigen decomposition that looks suspiciously like our formula for the orthogonal equivalence. So we might find the adjacency matrix for the graph and then find its eigenvalues. So our adjacency matrix will be, and we can find its eigenvalues using a complicated method of going online and using Wolfram Alpha. And this gives us our eigenvalues. What's useful here is that because all of the eigenvalues are real and there's a complete set of eigenvectors, then there are easy numerical methods for finding the eigenvalues. What if we switch the vertices around? So let's try a relabeling like, how about this one? Our new adjacency matrix is, and we find the eigenvalues to be, which are the same as they were before. Since isomorphic graphs have similar adjacency matrices, this gives us the following result. If two graphs are isomorphic, their adjacency matrices have the same eigenvalues with the same multiplicities. The contra positive then tells us that if there is a difference in the eigenvalues or the multiplicities, then the graphs cannot be isomorphic. For example, we'll find the adjacency matrices and eigenvalues for the two graphs shown. The graph on the left has adjacency matrix, and the eigenvalues are. The graph on the right has adjacency matrix, and the eigenvalues are. And since the eigenvalues are different, the graphs are not isomorphic. Or we can do the same thing for these graphs. The graph on the left has adjacency matrix, and eigenvalues. The graph on the right has adjacency matrix, and eigenvalues. And since the graphs have the same eigenvalues with the same multiplicities, they are isomorphic. Or are they? The graphs can't be isomorphic since one has two components while the other is connected. And so this leads to two problems. Suppose two graphs have the same number of vertices, the same number of edges, the same degree sequence, and the same eigenvalues. These are necessary conditions for the two graphs to be isomorphic. But even then they might not be isomorphic. We can't prove they're isomorphic until we find an isomorphism. So let's take a look at that next.