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 |
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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 |