2025-11-13

This following is a very brief summary of what happened in class on 2025-11-13.

Beginning with:

a field, $F$
a finite-dimensional $F$ -vector space, $V$
an $F$ -linear endomorphism, $T : V \to V$
we put an $F [x]$ -module structure on $V$ by letting $x$ act via $T$ . We then noted that, as an $F [x]$ -module, $V$ is still finitely generated. Since the ring $F [x]$ is a PID, our fundamental theorem guarantees an $F [x]$ -module isomorphism of the form

V ≃ F [x] / ⟨ a_{1} (x) ⟩ \oplus \dots \oplus F [x] / ⟨ a_{m} (x) ⟩ \oplus (F [x])^{n}

for some nonzero nonunits $a_{1} (x), \dots, a_{m} (x) \in F [x]$ with $a_{1} (x) ∣ a_{2} (x) ∣ \dots ∣ a_{m} (x)$ in $F [x]$ . We noted that the free rank, $n$ , had to be 0 (since $V$ is finite-dimensional as an $F$ -vector space), and also noted that we can make the invariant factors $a_{i} (x)$ unique by insisting they are monic, i.e., each is of the form $a (x) = b_{0} + b_{1} x + \dots + b_{k - 1} x^{k - 1} + x^{k}$ .

We then proceeded to analyze the direct sum decomposition on the right. See our notes for full details, but in short we noted that for each summand $F [x] / ⟨ a (x) ⟩$ , the set ${\overset{―}{1}, \overset{―}{x}, \dots, {\overset{―}{x}}^{k - 1}}$ is an $F$ -vector space basis and with respect to this basis the matrix for $T$ (which acts as multiplication by $x$ ) is

[\begin{matrix} 0 & 0 & \dots & \dots & - b_{0} \\ 1 & 0 & \dots & \dots & - b_{1} \\ 0 & 1 & \dots & \dots & - b_{2} \\ ⋮ & ⋮ & ⋱ & \dots & ⋮ \\ 0 & 0 & \dots & 1 & - b_{k - 1} \end{matrix}]

This ultimately led to the rational canonical form for the linear transformation $T$ .

We gave one example of how this all looks for a $6$ -dimensional $Q$ -vector space with given invariant factors.

Next class we will see how to compute the invariant factors (and rational canonical form) of a given linear transformation.

Concepts

References

Dummit & Foote, Abstract Algebra: Section 12.2