Universal Property for Cayley-Hamilton theorem

Question

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 $k[X]$?

Answer 1

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, $p_T(X)\mid q(X)$.

Answer 2

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.