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