proof of necessary and sufficient condition for diagonalizability
First, suppose that is diagonalizable. Then has a basis whose elements are eigenvectors of associated with the eigenvalues respectively. For each , as is an eigenvector, its annihilator polynomial is . As these vectors form a basis of , we have that the minimal polynomial (http://planetmath.org/MinimalPolynomialEndomorphism) of is which is trivially a product of linear factors.
Now, suppose that for some . Let . Consider the - cyclic subspace generated by , . Let be the restriction of to . Of course, is a cyclic vector of , and then . This is really easy to see: the dimension of is , and it’s also the degree of . But as divides (because ), and divides (Cayley-Hamilton theorem), we have that divides . As these are two monic polynomials of degree and one divides the other, they are equal. And then by the same reasoning . But as divides , then as , we have that divides , and then has no multiple roots and they all lie in . But then so does . Suppose that these roots are . Then , where is the eigenspace associated to . Then is a sum of eigenvectors. QED.
|Title||proof of necessary and sufficient condition for diagonalizability|
|Date of creation||2013-03-22 14:15:45|
|Last modified on||2013-03-22 14:15:45|
|Last modified by||rspuzio (6075)|