Solution (#1027) Let A be the adjacency matrix of a bipartite graph with vertices v1, ,vn As the graph is bipartite we can partition the vertex set into disjoint
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[PDF] Solution (#1027) Let A be the adjacency matrix of a bipartite graph
Solution (#1027) Let A be the adjacency matrix of a bipartite graph with vertices v1, ,vn As the graph is bipartite we can partition the vertex set into disjoint
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Solution(#1027) LetAbe the adjacency matrix of a bipartite graph with verticesv1,...,vnAs the graph is bipartite
we can partition the vertex set into disjoint subsetsV1andV2such that every edge of the graph connects a vertex
inV1to one inV2.Say thatvi1,...,vimare the vertices inV1and letPbe any permutation matrices whose firstm
columns areeT i1,eT i2,...,eT im.Denote the remaining columns ofPaseT
im+1,eT im+2,...,eT in.We have for any1kmthat
APeT k=AeT ik=n r=m+1b rkeT ir=n r=m+1b rkPeT r for somebrias the graph is bipartite. So P -1APeT k=n r=m+1b rkeT r.In a similar fashion we have form+ 1knthat
P -1APeT k=m r=1b rkeT r. Finally asPis orthogonal andAis symmetric thenP-1AP=PTAPis symmetric and we have P -1AP=0mmB B T0nn for somem×(n-m)matrixB.The eigenvalues ofAare the same as those ofP-1AP. Ifλ= 0then there is nothing to prove. Ifλis a non-zero
eigenvalue ofP-1APthis means that there is a vector v=v1 v 2 (wherev1is inRmandv2is inRn-m) such that P -1APv1 v 2 =0mmB B T0nn v1 v 2 =Bv2 BTv1 =λv1