"Let M be the given matrix of order n and its Jordan Canonical Form be J . Prove that the minimal and characteristic polynomial of M are same, if and only if, distinct eigenvalues of M are associated with distinct jordan blocks in J ." (Source: https://www.quora.com/When-are-minimal-and-characteristic-polynomials-of-a-matrix-the-same )

Is that correct? And if its, then...why? How to prove that?

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    $\begingroup$ What is meant by “distinct eigenvalues are in different Jordan blocks”? Isn’t this always the case? $\endgroup$ – paulinho Jun 15 at 14:35
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    $\begingroup$ @paulinho This is supposed to mean that there is only one Jordan block associated with each eigenvalue $\endgroup$ – Ben Grossmann Jun 15 at 14:36

The statement is poorly phrased, but the idea that I think was intended is correct. In particular, we have the following:

Claim: A matrix $M$ has equal characteristic and minimal polynomials if and only if for each of its eigenvalues, $M$ has only one Jordan block.

For more information on matrices like these, see Horn and Johnson's Matrix Analysis. In that context, such matrices are referred to as "non-derogatory".

To see that the statement holds, it suffices to understand how the minimal and characteristic polynomials relate to the Jordan form of $M$. In particular, suppose that the minimal polynomial of $M$ is given by $$ p(x) = (x-\lambda_1)^{m_1} \cdots (x - \lambda_k)^{m_k} $$ where $\lambda_1,\dots,\lambda_m$ are distinct. For each $j = 1,\dots,k,$ $m_j$ is the size of the largest Jordan block associated with $\lambda_j$.

On the other hand, the characteristic polynomial is given by $$ \chi(x) = (x - \lambda_1)^{d_1} \cdots (x - \lambda_k)^{d_k}. $$ In general, because $p(x) \mid \chi(x)$, it must be that $d_j \geq m_j$ for all $j = 1,\dots,k$. Note that for each $j$, $d_j$ is the sum of the sizes of all Jordan blocks associated with $\lambda_j$.

With these characterizations, it is clear that if $\lambda_j$ has more than one Jordan block in the Jordan form, then it must hold that $d_j > m_j$. Equivalently, if $d_j = m_j$ for all $j$ (so that $p = \xi$), then each $\lambda_j$ has only one Jordan block in the Jordan form.

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  • $\begingroup$ Thank you :) Unfortunately I don't get that proof, but I've searched the term "non-derogatory matrix" and I've found this site: solitaryroad.com/c154.html . In the Theorem 1 it comes to "where g(λ) is the greatest common divisor of all minors of order n-1 of the determinant |λI - A| ." Could we just show that in this case g(λ) equals 1? If so, how could we do that? $\endgroup$ – Hououin Kyouma Jun 15 at 16:42
  • $\begingroup$ @HououinKyouma What part do you not get? The proof that I've hinted at is very simple, and I've never heard of this characterization via size $n-1$ minors $\endgroup$ – Ben Grossmann Jun 15 at 18:11
  • $\begingroup$ By $m_k$ do you mean $m_j$ and by $λ_k$ do you mean $λ_j$ ? $\endgroup$ – Hououin Kyouma Jun 15 at 21:27
  • $\begingroup$ How does that make it equal the characteristic polynomial? And how do we know that it's the neccessary condition for that equality? $\endgroup$ – Hououin Kyouma Jun 15 at 22:12
  • $\begingroup$ Yes, that's what I meant. See my latest edit regarding your second comment. $\endgroup$ – Ben Grossmann Jun 16 at 5:12

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