Let , , , , , and be as described in the previous exercise.
- Show that , where is in the -vector space spanned by the .
- Show that .
Recall that by definition, so that as desired.
We claim that in fact has this form for any , and prove this by induction. The base case is already shown; for the inductive step, suppose the conclusion holds for . Now ; Now and , so that is also in the desired span. Thus . The reverse inclusion is immediate, and so these subspaces are equal.