2
$\begingroup$

Let $V$ be a vector space over a field $k$. Let $T$ be a $k$-linear transformation from $V$ to itself. Without using the notion of characteristic polynomial or Cayley-Hamilton theorem from Linear Algebra, how can I show that there exists a unique monic polynomial $p_T(X) \in k[X]$ such that $p_T(T)$ is the zero transformation and that whenever $f(X) \in k[X]$ satisfies the property that $f(T)$ is the zero transformation, then $p_T(X)$ divides $f(X)$ in $k[X]$?

$\endgroup$
  • 1
    $\begingroup$ Nakayama's lemma would seem a good launching point, at least for existence of such a polynomial $\endgroup$ – qbert Jul 24 '18 at 20:56
6
$\begingroup$

Let $n$ be the smallest natural number such that $\operatorname{Id},T,T^2,\ldots,T^n$ are linearly dependent. Then $T^n$ is a linear combination of the other ones. In other words, there's a monic polynomal $p_T(X)$ such that $p_T(T)$ is the null transformation.

Now, if $q(X)$ is a polynomial such that $q(T)=0$, divide $q(X)$ by $p_T(X)$: there are polynomials $q^\star(X),r(X)\in K[X]$ such that $q(X)=p_T(X)q^\star(X)+r(X)$ and that $\deg r(X)<\deg p_T(X)$ or $r(X)=0$. But then$$0=q(T)=p_T(T)q^\star(T)+r(T)=r(T).$$So, by the choice of $n$, $r(X)=0$. In other words, $p_T(X)\mid q(X)$.

$\endgroup$
  • 4
    $\begingroup$ A higher-level way of expressing exactly the same argument: the set of $p \in k[X]$ such that $p(T) = 0$ is an ideal of $k[X]$. Since $k[X]$ is a PID that implies that this "annihilator ideal of $T$" is equal to $\langle p_T \rangle$ for some $p_T \in k[X]$. (In case $V$ is finite-dimensional, that implies that the ideal is nonzero so $p_T \ne 0$.) $\endgroup$ – Daniel Schepler Jul 24 '18 at 21:21
  • $\begingroup$ @DanielSchepler Yes, that's nice. $\endgroup$ – José Carlos Santos Jul 24 '18 at 21:23
3
$\begingroup$

The set of $n\times n$ matrices over $k$ is a so-called algebra over $k$. It is finite dimensional: namely, the dimension is $n^2$. So every element is algebraic over the field. The direct proof is that if $A$ is an $n\times n$ matrix, then the matrices $I, A, A^2, \ldots, A^{n^2}$ are linearly dependent (as there are $n^2+1$ of them, which is bigger than the dimension), so some linear combination of these is zero. That produces a good polynomial: a polynomial $p(x)\in k[x]$ of degree at most $n^2$ such that $p(A)=0$.

It is well-known that algebraic elements in an algebra (over a field) have a minimal polynomial.

$\endgroup$

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.