- #1

geoffrey159

- 535

- 72

Theorem: Let ##E## be a finite dimensional ##K## vector space, ##K## sub-field of ##\mathbb{C}##, and ##f## be an endomorphism of ##E##. Then, ##f## is diagonalizable if and only if there exists a polynomial ##P## of ##K[X]##, such that ##P(f) = 0##, and ##P## can be written as a product of polynomials of degree 1, and its roots have order of multiplicity 1.

I understand the proof I have for the sufficient condition, but the proof for the necessary condition is hard to follow for me so I tried an alternate way. I would like you to tell me if this is correct please:

##\Leftarrow ## ) Assume that there exists distinct ##\lambda_i##'s for ##i = 1...p##, and a polynomial ##P## in the form ##P = a \prod_{i=1}^p (X - \lambda_i) ## such that ##P(f) = 0##.

I want to show that ##E = \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##, which is a necessary and sufficient condition of diagonalizability.

We have that for any ##x\in E-\{0\}##, ## P(f)(x) = 0##. So there exists ##i \in \{ 1...p \}## such that ##f(x) = \lambda_i x##, and ##x## belongs to the eigenspace ##E_{f,\lambda_i}##. Therefore ##x\in \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##, and ##E \subset \bigoplus_{\lambda \in \text{Sp}(f) } E_{f,\lambda}##. The other inclusion is trivial. So ##f## diagonalizable.