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Mr Davis 97

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- Thread starter Mr Davis 97
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Mr Davis 97

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mfb

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StoneTemplePython

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if you like adjacency matrices, then up to a graph isopmorphism, you can assume your ##K_n## is always in the top left corner of your adjacency matrix for ##K_m##.

In blocked form that top left ##\text{n x n }## submatrix is given by ##\mathbf J_n - \mathbf I_n## indicating that the submatrix (really, subgraph) is a complete graph, because an ##r## dimensional complete graph is specified by an adjacency matrix of ##\mathbf J_r -\mathbf I_r##

(note ##\mathbf J_r = \mathbf 1_r \mathbf 1_r^T## i.e. it indicates the all ones matrix)

- - - -

if you want to have some fun with the graph isomorphism, you can note that permutation matrices are doubly stochastic, and recall that ##\mathbf P^T = \mathbf P^{-1}## since they are orthogonal, so for any arbitrary permutation matrix ##\mathbf P## in ##\mathbb R^{\text{m x m}}## you have

##\mathbf P\big(\mathbf J_m -\mathbf I_m\big) \mathbf P^T = \mathbf P\big(\mathbf J_m\mathbf P^T\big) - \big(\mathbf P\mathbf I_m \mathbf P^T\big) = \mathbf P\big(\mathbf J_m\big) -\big(\mathbf P \mathbf P^T\big) = \mathbf J_m - \mathbf I_m##

In blocked form that top left ##\text{n x n }## submatrix is given by ##\mathbf J_n - \mathbf I_n## indicating that the submatrix (really, subgraph) is a complete graph, because an ##r## dimensional complete graph is specified by an adjacency matrix of ##\mathbf J_r -\mathbf I_r##

(note ##\mathbf J_r = \mathbf 1_r \mathbf 1_r^T## i.e. it indicates the all ones matrix)

- - - -

if you want to have some fun with the graph isomorphism, you can note that permutation matrices are doubly stochastic, and recall that ##\mathbf P^T = \mathbf P^{-1}## since they are orthogonal, so for any arbitrary permutation matrix ##\mathbf P## in ##\mathbb R^{\text{m x m}}## you have

##\mathbf P\big(\mathbf J_m -\mathbf I_m\big) \mathbf P^T = \mathbf P\big(\mathbf J_m\mathbf P^T\big) - \big(\mathbf P\mathbf I_m \mathbf P^T\big) = \mathbf P\big(\mathbf J_m\big) -\big(\mathbf P \mathbf P^T\big) = \mathbf J_m - \mathbf I_m##

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