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 \begin{equation} m_A(t) = \prod_{i=1}^k (t-\lambda_i)^{n_i}. \end{equation} (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 \begin{equation} \text{ker}\left( f_1(A) \circ \dots \circ f_k(A)\right) = \bigoplus_{i=1}^k \text{ker}f_i(A), \end{equation} i.e \begin{equation} \text{ker} \left( \prod_{i=1}^k f_i(A) \right) = \bigoplus_{i=1}^k \text{ker}f_i(A) \end{equation}

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 \begin{equation} \text{ker}\left( m_A(A) \right) = \bigoplus_{i=1}^k \text{ker}(A-\lambda_iI). \end{equation} 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

\begin{equation} E = \bigoplus_{i=1}^k E_{\lambda_i}. \end{equation}

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 $\lambda_i I$.

  • $\begingroup$ Sorry I edited your answer with my notations :)! Your answer seems correct and of course sweet. Let me think a bit more. $\endgroup$ – Majid May 23 '19 at 13:54
  • $\begingroup$ 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? $\endgroup$ – Majid May 23 '19 at 14:15

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