An eigenvalue of a linear transformation is the same thing as a root of the characteristic polynomial of . That is, the characteristic polynomial has a factor . We can evaluate this polynomial at to get the linear transformation . Vectors in the kernel of this space are the eigenvalues corresponding to the eigenvector .
Now we want to do the same thing with an eigenpair . This corresponds to an irreducible quadratic factor in the characteristic polynomial of . Evaluating this polynomial at we get the transformation , which I assert has a nontrivial kernel. Specifically, I want to focus in on some two-dimensional invariant subspace on which has no eigenvalues. This corresponds to a block in an almost upper-triangular representation of . So we’ll just assume for the moment that has dimension .
What I assert is this: if is a monic polynomial (with leading coefficient ) of degree two, then either is the characteristic polynomial of or it’s not. If it is, then , and its kernel is the whole of . If not, then the kernel is trivial, and is invertible.
In the first case we can just pick a basis, find a matrix, and crank out the calculation. If the matrix of is
then the characteristic polynomial is . We substitute the matrix into this polynomial to find
On the other hand, if is the characteristic polynomial and is any other monic polynomial of degree two, then , as we just showed. Then we can calculate
for some constants and , at least one of which must be nonzero. If , then is a nonzero multiple of the identity, which is invertible as claimed. On the other hand, if , then
which must be invertible since we assumed that has no eigenvalues.
So for any block, the action of an irreducible quadratic polynomial in either kills off the whole block or has a trivial kernel. This makes it reasonable to define an eigenvector of the eigenpair to be a vector in the kernel , in analogy with the definition of an eigenvector of a given eigenvalue.