1
$\begingroup$

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)

$\endgroup$
2
$\begingroup$

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 $p_v=q_A=p_A$.

$\endgroup$
  • $\begingroup$ I'm still having problems parsing through (b) => (c). I'm not sure how to establish a relationship between the polynomials of $A$ and the compaion matrix $C$. $\endgroup$ – emka Mar 31 '15 at 6:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.