proof of necessary and sufficient condition for diagonalizability

First, suppose that T is diagonalizablePlanetmathPlanetmath. Then V has a basis whose elements {v1,,vn} are eigenvectorsMathworldPlanetmathPlanetmathPlanetmath of T associated with the eigenvaluesMathworldPlanetmathPlanetmathPlanetmathPlanetmath {λ1,,λn} respectively. For each i=1,,n, as vi is an eigenvector, its annihilator polynomial is mvi=X-λi. As these vectors form a basis of V, we have that the minimal polynomialPlanetmathPlanetmath ( of T is mT=lcm(X-λ1,,X-λn) which is trivially a product of linear factors.

Now, suppose that mT=(X-λ1)(X-λp) for some p. Let vV. Consider the T - cyclic subspace generated by v, Z(v,T)=v,Tv,,Trv. Let Tv be the restriction of T to Z(v,T). Of course, v is a cyclic vector of Z(v,Tv), and then mv=mTv=χT. This is really easy to see: the dimensionPlanetmathPlanetmathPlanetmath of Z(v,T) is r+1, and it’s also the degree of mv. But as mv divides mTv (because mTvv=0), and mT divides χTv (Cayley-Hamilton theoremMathworldPlanetmath), we have that mv divides χTv. As these are two monic polynomials of degree r+1 and one divides the other, they are equal. And then by the same reasoning mv=mTv=χT. But as mv divides mT, then as mv=mTv, we have that mTv divides mT, and then mTv has no multiple roots and they all lie in k. But then so does χTv. Suppose that these roots are λ1,,λr+1. Then Z(v,T)=λiEλi, where Eλi is the eigenspaceMathworldPlanetmath associated to λi. Then v is a sum of eigenvectors. QED.

Title proof of necessary and sufficient condition for diagonalizability
Canonical name ProofOfNecessaryAndSufficientConditionForDiagonalizability
Date of creation 2013-03-22 14:15:45
Last modified on 2013-03-22 14:15:45
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 13
Author rspuzio (6075)
Entry type Proof
Classification msc 15A04