## Generalized Eigenspaces are Still Invariant and Disjoint

When working over an algebraically closed field we found that generalized eigenspaces are invariant and disjoint from each other. The same holds now that we’re allowing eigenpairs for transformations on real vector spaces.

First off, the generalized eigenspace of an eigenpair is the kernel of a polynomial in . Just like before, this kernel is automatically invariant under , just like the generalized eigenspace is.

Generalized eigenspaces of distinct eigenvalues are disjoint, as before. But let be an eigenvalue, be an eigenpair, and be a vector in both generalized eigenspaces. The invariance of under shows that if is a generalized eigenvector of , then so is . Just like we did before, we can keep hitting with until one step before it vanishes (which it eventually must, since it’s a generalized eigenvector of ). So without loss of generality we can assume that is an actual eigenvector of and a generalized eigenvector of .

Now we can use the generalized eigenvector property to write

but since is an eigenvector with eigenvalue , this says

If is nonzero, this can only be true if is a root of , which we assumed not to be the case.

Finally we consider two distinct eigenpairs and , and a generalized eigenvector of both, . Another argument like that above shows that without loss of generality we can assume is an actual eigenvector of . This eigenspace is the kernel , which is thus invariant, and *another* argument like before lets us assume that is an actual eigenvector of both eigenpairs. Thus we have

Subtracting, we find

If , this makes an eigenvector with eigenvalue . On the other hand, if then , and we conclude that .

At the end of the day, no nonzero vector can be a generalized eigenvector of more than one eigenvalue *or* eigenpair.

[…] know that these subspaces are mutually disjoint. We also know that the dimension of is equal to the multiplicity of , which is the number of […]

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