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If p(x) minimal polynomial and f(x) characteristic polynomial are given, and f(x) is divisible by p(x). can I always find a matrix that has f(x) as characteristic polynomial and p(x) as minimal polynomial? (p(x) and f(x) also have same roots) And how can I prove this?

Rodrigo de Azevedo
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Armin
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    Welcome to MSE. Please read this text about how to ask a good question. – José Carlos Santos Jan 03 '22 at 09:34
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    The minimal polynomial will always be a factor of the characteristic polynomial so you are no better off knowing the minimal polynomial in addition to the characteristic polynomial. So your question boils down to being a duplicate of one already tackled here; https://math.stackexchange.com/questions/1658106/determining-a-matrix-from-its-characteristic-polynomial/1658112 I found some questions and answers at the following link that may help with this topic; https://math.okstate.edu/people/binegar/4063-5023/4063-5023-s07.pdf – Martin Hansen Jan 03 '22 at 10:23
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    I don't think I follow your reasoning @MartinHansen. Yes the minimal polynomial will be a factor of the characteristic polynomial but knowing them both gives you more information than just knowing the characteristic polynomial. You can't figure out the minimal polynomial from the characteristic polynomial.

    Though I do agree there's still not enough information and the reasoning provided in your links basically explain why.

     – Rioghasarig Jan 03 '22 at 13:53
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    @Armin It's not entirely clear what you are asking. Martin's comment seems to assume that you are asking whether it's possible to recover the matrix of a linear transformation (up to similarity) using its minimal and characteristic polynomials, which has already been addressed. I suspect that what you are actually asking is whether there exists a matrix that has any given characteristic and minimal polynomial, which I do not believe has been addressed. It would be helpful if you could edit your question to make it clearer which of these you mean. – Ben Grossmann Jan 03 '22 at 14:50

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I assume that when you say $p$ and $f$ have the same roots, you mean that they have the same roots over a field extension such that the polynomial $f$ splits (or that you are taking roots from an algebraically closed field like $\Bbb C$). For example, if we take $p(x) = (x^2 + 1)$ and $f(x) = (x^2 + 1)(x^2 + 4)$ to be polynomials with coefficients in $\Bbb R$, then $p$ and $f$ technically satisfy your requirements (if "roots" refer merely to zeros in $\Bbb R$), but there is no matrix with minimal polynomial $p$ and characteristic polynomial $f$.

Yes, you can always find such a matrix. One nice way to do so is to produce a matrix that is almost in Frobenius normal form. Denote $q(x) = f(x)/p(x)$. Express $q$ as a product $$ q(x) = q_1(x) q_2(x)\cdots q_k(x) $$ such that each polynomial $q_k$ divides $p$ (note that this must be possible because $p$ and $q$ have "the same roots"). It follows that the block-diagonal matrix $$ M = \pmatrix{C_{q_1}\\ & \ddots \\ && C_{q_k}\\ &&& C_{p}} $$ has the desired minimal and characteristic polynomial.

To prove that this matrix works, it suffices to note that for block-diagonal matrices $$ M \pmatrix{A_1\\ & A_2\\ && \ddots \\ &&& A_k}, $$ the minimal polynomial of $M$ is the least common multiple of the minimal polynomials of $A_1,\dots,A_k$, and the characteristic polynomial of $M$ is the product of the characteristic polynomials of $A_1,\dots,A_k$.

Note: The matrix I have given is not in Frobenius normal form; for that we would have the further divisibility requirement that $q_1 \mid \cdots \mid q_k$.

Ben Grossmann
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