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Fedor Petrov
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Well, if $V=\{1,\ldots,n\}$, $(a_{ij})_{1\leqslant i,j\leqslant n}$ is the adjacency matrix, and it is non-singular, then its determinant (considered as a sum over permutations) has a non-zero term, which $\prod_i a_{i,\sigma(i)}$. It corresponds to a disjoint collection of cycles (these are cycles of the permutation $\sigma$ on $V$) which cover all vertices.

Well, if the adjacency matrix is non-singular, its determinant has a non-zero term, which corresponds to a disjoint collection of cycles which cover all vertices.

Well, if $V=\{1,\ldots,n\}$, $(a_{ij})_{1\leqslant i,j\leqslant n}$ is the adjacency matrix, and it is non-singular, then its determinant (considered as a sum over permutations) has a non-zero term $\prod_i a_{i,\sigma(i)}$. It corresponds to a disjoint collection of cycles (these are cycles of the permutation $\sigma$ on $V$) which cover all vertices.

Source Link
Fedor Petrov
  • 108.9k
  • 9
  • 264
  • 459

Well, if the adjacency matrix is non-singular, its determinant has a non-zero term, which corresponds to a disjoint collection of cycles which cover all vertices.