Just as we saw when dealing with eigenvalues, eigenvectors alone won’t cut it. We want to consider the kernel not just of one transformation, but of its powers. Specifically, we will say that is a generalized eigenvector of the eigenpair if for some power we have
The same argument as before tells us that the kernel will stabilize by the time we take powers of an operator, so we define the generalized eigenspace of an eigenpair to be
Let’s look at these subspaces a little more closely, along with the older ones of the form , just to make sure they’re as well-behaved as our earlier generalized eigenspaces are. First, let be one-dimensional, so must be multiplication by . Then the kernel of is all of if , and is trivial otherwise. On the other hand, what happens with an eigenpair ? Well, one application of the operator gives
for any nonzero . But this will always be itself nonzero, since we’re assuming that the polynomial has no roots. Thus the generalized eigenspace of will be trivial.
Next, if is two-dimensional, either has an eigenvalue or it doesn’t. If it does, then this gives a one-dimensional invariant subspace. The argument above shows that the generalized eigenspace of any eigenpair is again trivial. But if has no eigenvalues, then the generalized eigenspace of any eigenvalue is trivial. On the other hand we’ve seen that the kernel of is either the whole of or nothing, and the former case happens exactly when is the trace of and is its determinant.
Now if is a real vector space of any finite dimension we know we can find an almost upper-triangular form. This form is highly non-unique, but there are some patterns we can exploit as we move forward.