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.

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