As usual, let be a linear transformation on a real vector space of dimension . We know that can be put into an almost upper-triangular form
where each block is either a matrix or a matrix with no eigenvalues. Now I assert that
- If is a real number, then exactly of the blocks are the matrix whose single entry is .
- If is a pair of real numbers with , then exactly
of the blocks are matrices with characteristic polynomial . Incidentally, this shows that the dimension of this generalized eigenspace is even.
This statement is parallel to the one about multiplicities of eigenvalues over algebraically closed fields. And we’ll use a similar proof. First, let’s define the polynomial to be if we’re trying to prove the first part, and to be if we’re trying to prove the second part, and let be the degree of . We’ll proceed by induction on the number of blocks along the diagonal.
If then is either one- or two-dimensional. Then the statement reduces to what we worked out by cases earlier. So from here we’ll assume that , and that the statement holds for all matrices with blocks.
Let’s define to be the subspace spanned by the first blocks, and be the subspace spanned by the last block. That is, consists of all but the last one or two rows and columns of the matrix, depending on whether is or . Clearly is invariant under , and the restriction of to has matrix
with blocks. Thus the inductive hypothesis tells us that exactly
of the blocks from to have characteristic polynomial .
We’ll also define to be the linear transformation acting by the block . This is essentially the action of on the quotient space , but we’re viewing as giving representatives in for vectors in the quotient space. This way, if is a vector in this subspace of representatives we can write for some . Further, for some other vector . No matter which form of we’re using, we can see that for some , and further that for some .
Now, either has characteristic polynomial or not. If not, then I say that . This implies that
and thus that both the dimension of this kernel and the number of blocks with characteristic polynomial are the same as for .
So let’s assume and write with and . Then
for some . This implies that , but since is not the characteristic polynomial of , it is invertible on . Thus and .
On the other hand, if the characteristic polynomial of is , then we want to show that
The inclusion-exclusion principle tells us that
We’ll show that , and so its dimension is , and we have the result we want.
So take . Because the characteristic polynomial of is , we know that . Thus . Then
where the last equality holds because the dimension of is , and so the image has stabilized by this point. Thus we can choose so that . And so
which shows that . And thus is in . Since was arbitrary, the whole subspace , which shows that , which completes our proof.
Now, all of that handled we turn to calculate the characteristic polynomial of , only to find that it’s the product of the characteristic polynomials of all the blocks . That is, we will have factors of and factors of . We can thus define this half-dimension to be the multiplicity of the eigenpair. Like the multiplicity of an eigenvalue, it counts both the number of times the corresponding factor shows up in the characteristic polynomial of , and the number of blocks on the diagonal of an almost upper-triangular matrix for that have this characteristic polynomial.