Can you find a matrix by its minimal and characteristic polynomials?

Question

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?

Answer 1

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 $q_1 \mid \cdots \mid q_k$.