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Let $\mathsf{F}$ be any field. Let $A$ be an $n \times n$ matrix over $\mathsf{F},$ whose rank is $r \le n.$ Let $\mu \in \mathsf{F}[x]$ be the minimal polynomial of $A.$

What does $\deg(\mu)$ tell about $A$? Is it related to the rank of $A$?

Edit: As noted by Qiaochu, $\deg(\mu)$ is not equal to rank. Are there known cases (or conditions) where $\deg(\mu)$ is actually equal to the rank of $A$?

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In general there is no relationship between this degree and the rank (consider diagonal matrices). The degree of the minimal polynomial tells you something about how many eigenvalues $A$ can have, and it also tells you something about how large the corresponding Jordan blocks can be; see . – Qiaochu Yuan Aug 17 '12 at 2:10
@QiaochuYuan Thanks. Could you please convert it to an answer. And also, (in the answer) are there certain known cases where this degree is equal to the rank? – user2468 Aug 17 '12 at 2:16
As long as you don't have an eigenvalue that belongs to two different Jordan blocks... – J. M. Aug 17 '12 at 2:34
Also $deg(\mu) \leq r+1$ : Link – Q-rious Aug 12 '14 at 17:20

1 Answer 1

If $A$ has distinct $n$ non-zero eigen values in an extension field of $F$, $\mu$ is the characteristic polynomial of $A$. Hence deg $\mu = n =$ rank $A$.

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