Theorem on characteristic polynomials and minimal polynomials.

Question

Let $A$ be an $n\times n$ matrix. Let $q_A(t)$ and $p_A(t)$ represent the minimal and characteristic polynomial respectively. Then, the following are equivalent:

(a) $\deg(q_A(t))=\deg(p_A(t))$

(b) $p_A(t)=q_A(t)$

(c) $A$ is similar the companion matrix $C$ of $p_A(t)$.

(a)$\Rightarrow$(b) If $q_A(t)$ is the minimal polynomial of $A$ and $p_A(t)$ is the characteristic polynomial of $A$ then both $q_A(A)=0=p_A(t)$. Since $q_A(t)$ is the minimal polynomial, $q_A(t)|p_A(t)$. Because $q_A(t)$ is unique, this forces $q_A(t)=p_A(t)$.

(b)$\Rightarrow$(c) Suppose that the characteristic and minimal polynomial of $A$ are the same. Let $C$ be the companion matrix of $p_A(t)$. Then $q_C(t)=p_C(t)$. Because in both matrices we have that characteristic polynomials equal to minimal polynomials we know that each eigenvalue has one block in $J_A$ and $J_C$ (Jordan canonical forms). I'm stuck trying to put facts together to show similarity.

Also, I'd like to verify my (a)$\Rightarrow$(b)

Answer 1

What you write for (a)$\Rightarrow$(b) is not quite correctly formulated (you never use the hypothesis (a) for instance; also "because ... is unique" is never in itself an argument; the characteristic polynomial is unique too, but that does not make it equal to the minimal polynomial).

From $p_A[A]=0$ (Cayley-Hamilton) one gets $q_A\mid p_A$, unconditionally. From (a) and the fact that $q_A,p_A$ are both by definition monic polynomials one then gets (b) (a monic polynomial divides no other monic polynomial of the same degree).

The implication (b)$\Rightarrow$(c) is harder. The main intermediate result to obtain is that there exists a vector$~v$ such that $v,Av,A^2v,\ldots,A^{n-1}v\,$ are linearly independent (and therefore a basis); such a vector is called a cyclic vector for$~A$, and transformed to this basis the matrix of $A$ will be a companion matrix (and then it must be the one for the polynomial $q_A=p_A$), which will finish the implication.

You can do this using the Jordan Canonical Form (JCF), but (since this is assuming an algebraically closed field) there is a simpler (mostly because it avoids the fairly difficult proof that JCF exists) argument using just that the field is infinite. For the argument using the JCF, take a basis$~\mathcal B$ transformed to which $A$ has a JCF. Then take for $v$ the sum of all the final basis vectors of all the Jordan blocks. For a polynomial$~P$ to have $P[A](v)=0$, the components of $P[A](v)$ in each Jordan block must be zero, and one can show that this requires $P$ to be divisible by the minimal polynomial$~q_A$. Then if $q_A=p_A$ and if $P$ is to be nonzero, one must have $\deg P\geq \deg P_A=n$, from which it follws that $v$ is a cyclic vector for$~A$.

Here's a proof using just that the base filed is infinite. For each vector $v$ there is a minimal degree monic polynomial $P_v$ such that $P_v[A](v)=0$, which divides all other such polynomials. Then $v$ will be cyclic if and only if its polynomial $P_v=p_A$. Clearly $q_A=p_A$ is a necessary condition to have any cyclic vectors (because one has $P_v\mid q_A$, since $q_A[A](v)=0(v)=0$) but it is in fact also a sufficient condition. Like any polynomial, $q_A$ has but a finite number of monic divisors, and for any such proper divisor$~d$ one knows that $d[A]\neq0$ by minimality of the minimal polynomial; stated differently $\ker(d[A])\neq V$ (the whole space). But then the union of $\ker(d[A])$ as $d$ runs over the finitely many proper divisors of $q_A$ cannot fill up $V$ either (over an infinite field no finite union of proper subspaces can fill up the whole vector space). It suffices to take $v$ outside this union to ensure that $p_v=q_A=p_A$.