# Prove that if the minimum polynomial is the product of distinct factors then the transformation is diagonalizble.

I know that If $$f$$ is diagonalisable then its minimal polynomial is the product of distinct linear factors.

Now, how to prove the converse. That is:

Let $$A:E\to E$$ be a linear transformation from a finite dimensional vector space. If the minimum polynomial is the product of distinct factors as $$m(\lambda)=(\lambda-\lambda_1)...(\lambda-\lambda_k)$$ then $$A$$ is diagonalizble.

I believe that the characteristic polynomial must be like

$$p(\lambda)=(\lambda-\lambda_1)^{r_1}...(\lambda-\lambda_k)^{r_k}$$ right?

If so, then with the basis which is formed by the eigenvectors of $$A$$, $$A$$ is diagonalizble.

This argument is true?

I see two ways of proving this. The first is a very quick proof, if you know the following result about the relationship between the minimal polynomial and the Jordan canonical form (this is a useful result to know, but perhaps not so obvious to prove):

Theorem $$1$$: Let $$\lambda_1, \dots, \lambda_k$$ be all the distinct eigenvalues of the linear transformation $$A: E \to E$$. For each $$i$$, let $$n_i$$ be the maximum size of the Jordan blocks corresponding the eigenvalue $$\lambda_i$$. Then, the minimal polynomial of $$A$$ is $$$$m_A(t) = \prod_{i=1}^k (t-\lambda_i)^{n_i}.$$$$ (FYI: the number $$n_i$$ is also the nilpotency index of $$(A-\lambda_iI)$$ when restricted to the generalised eigenspace corresponding to $$\lambda_i$$.)

In your question, all the $$n_i$$'s are $$1$$ by assumption, so by Theorem $$1$$, the maximum size of the Jordan block corresponding to $$\lambda_i$$ is $$1$$. It follows that all the Jordan blocks are $$1 \times 1$$. This is equivalent to $$A$$ being diagonalizable.

The second proof uses the following lemma about polynomials of operators and the relationship between their kernels:

Lemma: Suppose $$E$$ is a vector space over a field $$\mathbb{F}$$, and let $$f_1(t), \dots, f_k(t)$$ be pairwise coprime polynomials with coefficients in $$\mathbb{F}$$. Then, for any linear map $$A: E \to E$$, we have that $$$$\text{ker}\left( f_1(A) \circ \dots \circ f_k(A)\right) = \bigoplus_{i=1}^k \text{ker}f_i(A),$$$$ i.e $$$$\text{ker} \left( \prod_{i=1}^k f_i(A) \right) = \bigoplus_{i=1}^k \text{ker}f_i(A)$$$$

The proof of the lemma is by induction on $$k$$, which you should definitely attempt. To apply this, we take $$f_i(t) = t-\lambda_i$$. Then the lemma says $$$$\text{ker}\left( m_A(A) \right) = \bigoplus_{i=1}^k \text{ker}(A-\lambda_iI).$$$$ Since $$m_A(t)$$ is the minimal polynomial of $$A$$, we have $$m_A(A) = 0$$; i.e the kernel is all of $$E$$. Also, note that $$\text{ker}(A-\lambda_iI)$$ is precisely the eigenspace $$E_{\lambda_i}$$ of $$A$$ correpsonding to $$\lambda_i$$. Hence, we have shown that

$$$$E = \bigoplus_{i=1}^k E_{\lambda_i}.$$$$

Recall that $$A$$ is diagonalizable if and only if we have such a direct sum decomposition; hence this completes the proof.

$$E=\oplus N(A-\lambda_i I)$$. The restriction of $$A-\lambda_i I$$ to $$E_{\lambda_i}=N(A-\lambda_iI)=0$$ implies that the restriction of $$A$$ to $$E_{\lambda_i}$$ is $$\lambda_i I$$.

• Sorry I edited your answer with my notations :)! Your answer seems correct and of course sweet. Let me think a bit more. – Majid May 23 at 13:54
• Just one question! The eigenvectors do not need to be orthogonal? I mean we always can have the spac as the dirrect some of eigenspaces an dit does not matter that the eigenvectors give an orthonormal basis or not? – Majid May 23 at 14:15