3
$\begingroup$

Let $A$ be an $n × n$ matrix with rational entries such that the minimal polynomial of $A$ is $x^3 + 2x+2$.

Prove that $3$ divides $n$.

I think there is no rational root of this polynomial but how I use this. Please give me hint.

$\endgroup$
5
$\begingroup$

That is a monic polynomial with integer coefficients, so any rational roots must be integer. Also none of the divisors $-2,-1,1,2$ of the constant term $2$ are roots, so there are no rational roots at all. Being of degree$~3$ this means the polynomial is irreducible over$~\Bbb Q$.

Now use the theorem (kind of converse to Cayley-Hamilton) that the characteristic polynomial$~\chi$ divides a power of the minimal polynomial. (This is so because the characteristic polynomial of $A$ is the product of the invariant factors associated to $IX-A$, all of which divide to minimal polynomial.) The minimal polynomial being irreducible this means that $\chi$ itself is a power of the minimal polynomial, whence its degree$~n$ is a multiple of the degree $~3$ of the minimal polynomial.

$\endgroup$
2
$\begingroup$

The argument I gave in my other answer can be used in situation where the minimal polynomial has only one irreducible factor (but it might have a multiplicity); as soon as there are two different factors, not much can be achieved along those lines. But for the even more special situation that the minimal polynomial is itself irreducible, as is the case here, there is another type of argument that can be applied.

If $A$ is any $n\times n$ matrix with rational entries, the vector space $\def\Q{\Bbb Q}\Q^n$ can be made into a module over the polynomial ring $\Q[X]$ by having $X$ act as multiplication by$~A$ (and constant polynomials by scalar multiplication). If $P\in\Q[X]$ is any polynomial annihilating$~A$ (that is $P[A]=0$) then the action passes to the quotient to define a $\Q[X]/(P)$-module structure on $\Q^n$. Now here one can take in particular $P=X^3+2X+3$ which is irreducible over$~\Q$, so that $K=\Q[X]/(P)$ is a field, over which $\Q^n$ becomes a vector space. One then has $$n=\dim_\Q(\Q^n)=[K:\Q]\dim_K(\Q^n)=3\dim_K(\Q^n),$$ which proves that $3\mid n$.

$\endgroup$

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.