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I just solved some exercises on minimal polynomials and i remember that there is a relation between the minimal polynomial and the jordan normal form.

But my question is the following : knowing the minimal polynomial of a given matrix $A$ what information can we get about the jordan normal form (without computing it).

Thanks in advance!

Rodrigo de Azevedo
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sigmatau
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  • http://math.stackexchange.com/questions/82607/why-does-the-largest-jordan-block-determine-the-degree-for-that-factor-in-the-mi?rq=1 has some information. I'm not sure there's much more you can do. – Evan Aug 12 '13 at 16:08

2 Answers2

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The multiplicity of an eigenvalue in the minimal polynomial is the size of that eigenvalue's largest Jordan block. See the "Complex matrices" topic in Wikipedia article "Jordan normal form."

user0
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Suppose a matrix $A\in M_n(K)$ has the characteristic polynomial \begin{align*} p(\lambda)\,=\, \big|A-\lambda I_n\big|\,&=\,(-1)^n\big(\lambda -\lambda_1\big)^{m_1}\cdots\big(\lambda -\lambda_p\big)^{m_p}, \end{align*} then the Jordan normal form of $A$ is given by $$ J\,=\,\widetilde J_{m_1}(\lambda_1)\oplus\cdots\oplus\widetilde J_{m_p}(\lambda_p) $$ where $\widetilde J_{m_i}(\lambda_i)$ is the Jordan matrix associated with the eigenvalue $\lambda_i$. If \begin{align*} m(\lambda)\,=\,(\lambda-\lambda_1)^{k_1}\cdots(\lambda-\lambda_p)^{k_p}. \end{align*} is the minimal polynomial of $A$, and $\dim E(\lambda_i)=r_i$ for each $i$, then the Jordan matrices are given by \begin{align} \widetilde J_{m_i}(\lambda_i) \,=\, J_{d_{i1}}(\lambda_i)\oplus\cdots\oplus J_{d_{ir_i}}(\lambda_i) \end{align} where $J_{d_{ij}}(\lambda_i) $ is a Jordan block of order $d_{ij}$, and the largest block is of size $k_i$. Usually, when $n\leq4$, you can immediately deduce $r_i$ (and hence, $d_{ij}$) after getting the minimal polynomial $m(\lambda)$ without computing $\dim E(\lambda_i)$.

PermQi
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