topological proof of the Cayley-Hamilton theorem

We begin by showing that the theorem is true if the characteristic polynomialMathworldPlanetmathPlanetmath does not have repeated roots, and then prove the general case.

Suppose then that the discriminantMathworldPlanetmathPlanetmathPlanetmathPlanetmath of the characteristic polynomial is non-zero, and hence that T:VV has n=dimV distinct eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath once we extend11Technically, this means that we must work with the vector spaceMathworldPlanetmath V¯=Vk¯, where k¯ is the algebraic closureMathworldPlanetmath of the original field of scalars, and with T¯:V¯V¯ the extended automorphism with action T¯(va)T(V)a,vV,ak¯. to the algebraic closure of the ground field. We can therefore choose a basis of eigenvectorsMathworldPlanetmathPlanetmathPlanetmath, call them 𝐯1,,𝐯n, with λ1,,λn the corresponding eigenvalues. From the definition of characteristic polynomial we have that


The factors on the right commute, and hence


for all i=1,,n. Since cT(T) annihilates a basis, it must, in fact, be zero.

To prove the general case, let δ(p) denote the discriminant of a polynomialPlanetmathPlanetmath p, and let us remark that the discriminant mapping


is polynomial on End(V). Hence the set of T with distinct eigenvalues is a dense open subset of End(V) relative to the Zariski topologyMathworldPlanetmath. Now the characteristic polynomial map


is a polynomial map on the vector space End(V). Since it vanishes on a dense open subset, it must vanish identically. Q.E.D.

Title topological proof of the Cayley-Hamilton theoremMathworldPlanetmath
Canonical name TopologicalProofOfTheCayleyHamiltonTheorem
Date of creation 2013-03-22 12:33:22
Last modified on 2013-03-22 12:33:22
Owner rmilson (146)
Last modified by rmilson (146)
Numerical id 7
Author rmilson (146)
Entry type Proof
Classification msc 15-00