For some technical points, it’s going to be useful to have a sort of dual to the increasing chain of subspaces we found yesterday. Instead of kernels, we’ll deal with images.
Specifically, if then I say . Indeed, the first statement asserts that there is some so that . But then , and so it’s the image of under as well. So we have a decreasing sequence
Just like last time, these stabilize by the time we get to the th power, where . Instead of repeating everything, let’s just use the rank-nullity theorem, which says for each power that . Now if then we calculate
where in the second line we used the stability of the sequence of kernels from yesterday. This tells us that for all these higher powers of .