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Let $A$ be a square matrix, so $A$ has some Jordan Normal form. Then $A$ has a minimal polynomial, say $m(X)=\prod_{i=1}^k (t-\lambda_i)^{m_i}$.

Wikipedia says

The factors of the minimal polynomial $m$ are the elementary divisors of the largest degree corresponding to distinct eigenvalues.

So $m_i$ is the size of the largest Jordan block of $\lambda_i$. Why is this exactly?

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    You know the minimal polynomial of a single Jordan block, no? See what happens if you try to find the minimal polynomial of a block-diagonal matrix with just one eigenvalue, but with Jordan blocks of various sizes...2011-11-16

2 Answers 2