4
$\begingroup$

How does the idea of a "minimal polynomial" for a matrix (i.e. for a matrix $A$, the polynomial, $\mu (x)$, of least degree, such that $\mu (A) =0$) relate the the "minimal polynomial" for some element, $\alpha $ over a field (i.e. the minimal polynomial for $\sqrt {2}$ over $\mathbb{Q} $ is $x^2-2$)? In the previous example, I'm wondering if there some matrix with entries in $\mathbb{Q} $ which has minimal polynomial $x^2-2$ that somehow relates to $\sqrt {2}$?

Thanks!

Improve this question
$\endgroup$
4
  • 4
    $\begingroup$ For any number field $K$ of degree $d$, there is a (non-canonical) embedding of $K$ into $M_d({\bf Q})$. Since this is an embedding of algebras, it preserves minimal polynomials. In general, note that the minimal polynomial of a matrix need not be irreducible. $\endgroup$
    – François Brunault
    Commented Feb 15, 2013 at 14:44
  • 3
    $\begingroup$ In both cases, if c is an element of an algebra A over Q, the minimal polynomial P generates the kernel of the algebra map from Q[t] to A, which sends t to c. Whether c is a matrix or a real number, the subalgebra Q[c] it generates in A is isomorphic to Q[t]/(P). So the idea is the same in both cases. Hence any matrix M over Q with P = t^2-2.Id, generates an algebra isomorphic to Q(sqrt(2)). Francois observes 2x2 such matrices always exist, and I am sure you can write down a 2x2 integer matrix M with M^2 = 2.Id. (As he also observes, P may not be irreducible when A has zero divisors.) $\endgroup$
    – roy smith
    Commented Feb 15, 2013 at 16:13
  • 2
    $\begingroup$ More generally, do you know about "companion matrices", or rational canonical form? For any polynomial P of degree d, the "companion matrix" of P generates a subalgebra of dxd matrices isomorphic to Q[t]/(P). $\endgroup$
    – roy smith
    Commented Feb 15, 2013 at 17:12
  • $\begingroup$ Thanks for the comments! Why not put them as answers? I will have another look at companion matrices and rational canonical form. It's been a while since I looked at either. $\endgroup$
    – Bemao
    Commented Feb 15, 2013 at 19:51

1 Answer 1

Reset to default
3
$\begingroup$

The general notion of minimal polynomial, as Roy Smith indicated in the comments, works whenever you have an algebra $A$ over a field $K$ and an element $a\in A.$ Namely, consider the unique algebra homomorphism $\pi:k[t]\to A$ that sends $a$ to $t.$* Since $k[t]$ is a principal ideal domain, its kernel $\ker\pi$ is $(P),$ where $P$ may be chosen to be a monic polynomial of degree $d\geq 0$, or 0. In the former case, $f$ is the minimal polynomial of $a$ and in the latter case, $a$ generates a polynomial algebra inside $A.$ The canonical object which perhaps deserves the name "minimal polynomial" is the ideal $(P)$. When $A$ is a general commutative ring, same reasoning leads to the ideal $\ker\pi$, but there is no reason to expect that it is principal.

(*) The idea of "prolonging" the element $a$ to the homomorphism $\pi$ is an algebraic version of the "functional calculus". Thus expressions $q(a)$ for various polynomials $q\in K[t]$ make sense: they are simply $\pi(q).$ However, the functional calculus used in analysis and operator algebra theory involves much larger functional spaces than just polynomials.

Another useful interpretation of the minimal polynomial is through the poles of the formal resolvent of $a.$ For any commutative ring $A$ (not necessarily an algebra over a field), consider

$$ (\lambda-a)^{-1}=\sum_{n=0}^{\infty}\lambda^{-n-1}a^n \in A((\lambda^{-1})). $$

A priori, this is just a Laurent formal power series in $\lambda^{-1}$; however,

When $a$ has a minimal polynomial $P(t)$, it is in fact a rational function of $\lambda$ with the denominator (in lowest terms) $P(\lambda),$ and conversely.

Indeed,

$P(a)(\lambda-a)^{-1}=P(\lambda)(\lambda-a)^{-1}$ modulo polynomials in $\lambda.{\ } \square$

See my answer How would you solve this tantalizing Halmos problem? for related ideas.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .