Polynomial module associated with a linear operator

Let $V$ be an $F$ -vector space and $T:V\to V$ an $F$ -linear map. The polynomial module associated with $T$ is the $F[X]$ -module on $V$ whose action is

\left(\sum_{j=0}^m a_jX^j\right)\cdot v =\sum_{j=0}^m a_jT^j(v),\qquad T^0=\operatorname{id}_V.

This is the unique action of the polynomial ring $F[X]$ for which constants act by the original scalar multiplication and $X\cdot v=T(v)$ . Finite dimensionality is not required.

Why this satisfies the module axioms

Polynomial evaluation gives $(f+g)(T)=f(T)+g(T)$ , $(fg)(T)=f(T)g(T)$ , and $1(T)=\operatorname{id}_V$ . The multiplicative identity follows by expanding products and using $T^iT^j=T^{i+j}$ . Each $f(T)$ is linear, so the two distributive module axioms follow as well.

The ring of all linear endomorphisms of $V$ may be noncommutative; the scalar operators and the powers of this single $T$ commute, which is all the construction needs.

Morphisms and invariant subspaces

For operators $T$ on $V$ and $S$ on $W$ , a linear map $u:V\to W$ is an $F[X]$ -module homomorphism exactly when $uT=Su$ . A subspace is an $F[X]$ -submodule exactly when it is $T$ -invariant.

For finite-dimensional $V$ , the minimal polynomial generates the annihilator of this module. The construction is the bridge to rational canonical form from the module structure theorem.

References

The polynomial identities above verify existence directly; the images of constants and $X$ prove uniqueness. The intertwining condition extends from $X$ to every polynomial by induction and linearity.