For graphs \(G_0\) and \(G_1\) show that \(G_0\) is isomorphic to \(G_1\) if and only if there exists permutation matrix \(P\) such that \(P^{T}A_{G_0}P = A_{G_1}\) where \(A_{G_0}\) and \(A_{G_1}\) are the adjacency matricies for \(G_0\) and \(G_1\) respectively.

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For graphs \(G_0\) and \(G_1\) show that \(G_0\) is isomorphic to \(G_1\) if and only if there exists permutation matrix \(P\) such that \(P^{T}A_{G_0}P = A_{G_1}\) where \(A_{G_0}\) and \(A_{G_1}\) are the adjacency matricies for \(G_0\) and \(G_1\) respectively.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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@geerky42 I cant msg you back... you are not a fan of me
But yes I have had way harder things answered on this site, and this is actually linear algebra which many people here love. The question is posted on math stack exchange as well but the answer is not rigorous and they don't like when you repost questions even if the other thread is dead...

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