For a positive integer , the chromatic polynomial is defined to be the number of proper vertex colourings of using a fixed palette of colours.
We prove that this counting function is a polynomial by induction on the number of edges. If has vertices and no edges, every assignment of colours is proper, soOtherwise choose an edge . Every proper colouring of either gives different colours, in which case it is a colouring of , or gives them the same colour, in which case it corresponds to a proper colouring of the contracted graph . Hence the deletion-contraction recurrence for the chromatic polynomial isBoth terms on the right are polynomials by induction, so is a polynomial.
Factor the proposed polynomial:In particular,By the two-colourability criterion for bipartite graphs, every finite bipartite graph has at least one proper two-colouring, so its chromatic polynomial is positive at . Therefore cannot be the chromatic polynomial of a bipartite graph.
It is nevertheless a chromatic polynomial. Start with the complete graph , whose vertices may be coloured inways, and attach one leaf to any vertex. After the triangle is coloured, the leaf has available colours. The chromatic polynomial after attaching a leaf therefore gives
Let and supposeIf some coefficient is nonzero, let be the largest index with . Choose a pathwhose endpoints realize the graph diameter. Every initial segment is a shortest path, since a shorter route from to could be followed by the remaining segment to shorten the path from to . Thus
By the walk count from powers of an adjacency matrix,Taking the entry of the assumed relation and using maximality of givesa contradiction. Every coefficient is therefore zero, proving the linear independence of adjacency powers up to the diameter:
Take the infinite familyEach complete graph is connected and has graph diameter one. If is its adjacency matrix, thenand direct multiplication gives the adjacency-matrix quadratic relation for a complete graphConsequentlyso , which are precisely the powers through , are linearly dependent for every .
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