3
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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 $\forall i, \ker(\varphi-\lambda_i) = \ker(\varphi-\lambda_i)^{\mu_i}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.