# finding two conditions so that for two polynomials, there exists exactly one matrix

Let $K$ be a field, and $f, g \in K[t]$, the ring of polynomials over $K$.

I want to find a necessary and a sufficient condition for $f$ and $g$, so that there exists an (except for similar matrices) unique matrix $A$ with entries of $K$, so that it's minimal polynomial is equal to $f$ and it's characteristic polynomial is equal to $g$.

My attempt: Considering that the minimal polynomial always divides the characteristic polynomial, $g|f$ must apparently be a necessary condition for $f$ and $g$. On the other hand, it's true that $f(A) = g(A) = 0$ if $f$ is the characteristic and $g$ is the minimal polynomial of a matrix $A$. But how can I now find conditions of the said form, so that there's except for similar matrices one and only one $A$ that satisfies this?

To answer your question you need the so called invariant factors e.g. https://www2.bc.edu/~reederma/LinAlgAdv.pdf but there are lot of material on line.

The invariant factors of $A \in \mathbb{K}^{n,n}$are monic polynomials $f_1,f_2,\cdots,f_n$ such that $f_{i} | f_{i+1}$, $f_n$ is the minimal polynomial of $A$ and the product $f_1 \cdots f_n = \pm \chi_A$ is the characteristic polynomial of $A$. Each conjugation class of matrices in $\mathbb{K}^{n,n}$ is determined by the sequence $f_1,\cdots,f_n$ and reciprocally, any sequence of monic polynomials $f_1,\cdots,f_n$ such that $f_{i} | f_{i+1}$ and the product $f_1 \cdots f_n$ has degree $n$ gives a conjugation class.

From now assume that $K$ is an algebraically closed field and assume that $f$ and $g$ are as in the OP. Split $f = \prod_i (x-\lambda_i)^{m_i}$ where $\lambda_i \neq \lambda_j$ for $i \neq j$. Then $g = \prod_i (x-\lambda_i)^{c_i}$ with $1 \leq m_i \leq c_i$ and $\sum_j c_j = n$. If we want that $f$ and $g$ determine the conjugation class then the numbers $m_i, c_i$ $i=1,\cdots,k$ must determine the sequence $f_1,\cdots,f_n$ of invariant factor. To be so for each $i=1,\cdots,k$ the two numbers $m_i,c_i$ must determine the exponent $0 \leq r_{ij}$ of $(x-\lambda_i)$ of the invariant factor $f_j$. By the divisibility property we know that $r_{ij} \leq r_{i (j+1)}$. So at this point we have a combinatorial problem. Namely, we have a $k \times n$ matrix $F$ with non negative integers $r_{ij}$ such that

1)the sequence is incresing on each row i.e. $r_{ij} \leq r_{i (j+1)}$,

2)we know the last column of $F$ (i.e. we know the decomposition of the minimal polynomial hence $r_{i n} = m_i$),

3)we know the sum $c_i$ of each row of $F$ (since the product $f_1 \cdots f_n$ is $g$ ).

The question is which are the conditions on the last column and the $c_i$ to determine univocally the whole $k \times n$ matrix ?

If I am not wrong the condition is that for each $i$ one of the following holds:

(i) $m_i = 1$

(ii) $m_i = c_i$

(iii) $m_i + 1 = c_i$

So the two polynomials $f,g$ are the minimal and the characteristic polynomial of a (unique up to conjugation) matrix $A \in \mathbb{K}^{n,n}$ if and only if :

1) $f = \prod_i (x-\lambda_i)^{m_i}$ where $\lambda_i \neq \lambda_j$ for $i \neq j$ and $g = \prod_i (x-\lambda_i)^{c_i}$ with $1 \leq m_i \leq c_i$ and for each $i$ the numbers $m_i,c_i$ satisfies one of the three conditions (i), (ii) or (iii) above.

If $\mathbb{K}$ is not algebraically closed then you can replace the polynomials $(x - \lambda_i)$ by monic irreducible polynomials $p_i \in \mathbb{K}[x]$ and obtain a similar result. Namely, $f = \prod_i p_i^{m_i}$ where $p_i \neq p_j$ for $i \neq j$ and $g = \prod_i p_i^{c_i}$, now you need that $\sum_i c_i \deg(p_i) = n$ but the conditions on $m_i, c_i$ are the same as for $\mathbb{K}$ algebraically closed.

• For a non algebraically closed field, can't you replace "root" by "irreducible factor", and still have the same claim ? – Sylvain L. Jun 17 '15 at 16:31
• Actually, my claim is not correct so I am going to edit my answer. I will try to give more details for $K$ algebraically closed. I will think about the non algebraically closed case and let you know. – Holonomia Jun 17 '15 at 17:13
• At first sight sight, i would expect something like : "$f|g$, any root of $g$ is a root of $f$ and $f/g$ has simple roots". No ? (and for a general field : $f|g$, every irreducible factor of $g$ is an irreducible factor of $f$ and $f/g$ is without squared factor.) – Sylvain L. Jun 17 '15 at 17:27
• It is "almost" like that. There is one case in which $f/g$ can contain squares e.g. the obvious example is $f(x)=x$ and $g(x)=-x^3$ in dimension 3. – Holonomia Jun 17 '15 at 20:03
• Hmm, i'm not sure to understand your example. If $f(x)=x$ is the minimal polynomial of $A$, then is'n't $A=0$ and then uniquely determined ? – Sylvain L. Jun 17 '15 at 20:34