## Two facts about the kernel of a linear transformation on an F[x]-module

Let $V$, $T$, $B$, $A$, $E$, and $\varphi$ be as described in the previous exercise.

1. Show that $x \xi_j = \omega_j + f_j$, where $f_j \in \sum_i F\xi_i$ is in the $F$-vector space spanned by the $\xi_i$.
2. Show that $\mathsf{span}_{F[x]}(\xi_1,\ldots,\xi_n) = \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n) + \mathsf{span}_F(\xi_1,\ldots,\xi_n)$.

Recall that $\omega_j = x \xi_j - \sum a_{i,j} \xi_i$ by definition, so that $x \xi_j = \omega_j + \sum a_{i,j} \xi_i = \omega_j + f_j$ as desired.

We claim that in fact $x^t \cdot \xi_j \in \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n) + \mathsf{span}_F(\xi_1,\ldots,\xi_n)$ has this form for any $t$, and prove this by induction. The base case is already shown; for the inductive step, suppose the conclusion holds for $x^t \xi_j$. Now $x^{t+1} \xi_j = x^t(\omega_j + f_j)$; Now $x^t \omega \in \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n)$ and $f_j = \sum a_{i,j} \xi_i$, so that $x^t f_j$ is also in the desired span. Thus $\mathsf{span}_{F[x]}(\xi_1,\ldots,\xi_n) \subseteq \mathsf{span}_{F[x]}(\omega_1,\ldots,\omega_n) + \mathsf{span}_F(\xi_1,\ldots,\xi_n)$. The reverse inclusion is immediate, and so these subspaces are equal.