Let $\phi$ be an endomorphism on an $n$-dimensional $F$-vector space $V$ and let $$P_\phi(x) = \pm \prod_i (x-\lambda_i)^{\mu_i}$$ be the characteristic polynomial of $\phi$ where $\lambda_1, \dots, \lambda_r \in F$ are the pairwise distinct eigenvalues of $\phi$. Suppose that $\dim\ker(\phi-\lambda_i\operatorname{Id})^{\mu_i-1} = \dim\ker(\phi-\lambda_i \operatorname{Id})^{\mu_i}$ for $i=1\dots,r$.

Question: Is $\phi$ diagonalizable?

My thoughts: I assume that the Statement is true, but I have not found a good argument yet. By the condition we know that $\ker(\phi-\lambda_i \operatorname{Id})^{\mu_i-1} = \ker(\phi-\lambda_i \operatorname{Id})^{\mu_i}$ for $i=1\dots,r$, as $\ker(\phi-\lambda_i \operatorname{Id})^{\mu_i-1}$ is a subspace of $\ker(\phi-\lambda_i \operatorname{Id})^{\mu_i}$ with same dimension. If I could find a way to reduce that successively to $\ker(\phi-\lambda_i \operatorname{Id}) = \ker(\phi-\lambda_i \operatorname{Id})^{\mu_i}$ which means that algebraic and geometric multiplicity are the same, so $\phi$ would be diagonalizable.

Could you please help me completing my argument? Thank you!


It's false. Indeed, if we consider the matrix $$A=\begin{bmatrix} 2 & 1 & & & & \\ & 2 & & & & \\ & & 2 & & & \\ & & & 3 & 1 & \\ & & & & 3 & \\ & & & & & 3\end{bmatrix}$$ then its characteristic polynomial is $$ (x-2)^3(x-3)^3$$ and if we denote by $(e_1,\dots,e_6)$ the vectors of the canonical basis of $F^6$, then $$\ker(A-2I)^2 = \ker(A-2I)^3 = \langle e_1, e_2, e_3\rangle$$ and $$\ker(A-3I)^2 = \ker(A-3I)^3 = \langle e_4, e_5, e_6\rangle$$ but $A$ isn't diagonalizable (for instance, $\ker(A-2I)=\langle e_1,e_3\rangle \ne \ker(A-2I)^2$).

You really need to have $\forall i, \ker(\varphi-\lambda_i) = \ker(\varphi-\lambda_i)^{\mu_i}$.


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