topological proof of the Cayley-Hamilton theorem
We begin by showing that the theorem is true if the characteristic polynomial does not have repeated roots, and then prove the general case.
Suppose then that the discriminant of the characteristic polynomial is non-zero, and hence that has distinct eigenvalues once we extend11Technically, this means that we must work with the vector space , where is the algebraic closure of the original field of scalars, and with the extended automorphism with action to the algebraic closure of the ground field. We can therefore choose a basis of eigenvectors, call them , with the corresponding eigenvalues. From the definition of characteristic polynomial we have that
The factors on the right commute, and hence
for all . Since annihilates a basis, it must, in fact, be zero.
To prove the general case, let denote the discriminant of a polynomial , and let us remark that the discriminant mapping
is polynomial on . Hence the set of with distinct eigenvalues is a dense open subset of relative to the Zariski topology. Now the characteristic polynomial map
is a polynomial map on the vector space . Since it vanishes on a dense open subset, it must vanish identically. Q.E.D.
|Title||topological proof of the Cayley-Hamilton theorem|
|Date of creation||2013-03-22 12:33:22|
|Last modified on||2013-03-22 12:33:22|
|Last modified by||rmilson (146)|