The Unapologetic Mathematician

Mathematics for the interested outsider

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 (\tau,\delta) is the kernel \mathrm{Ker}\left((T^2-\tau T+\delta I_V)^d\right) of a polynomial in T. Just like before, this kernel is automatically invariant under T, just like the generalized eigenspace \mathrm{Ker}\left((T-\lambda I_V)^d\right) is.

Generalized eigenspaces of distinct eigenvalues are disjoint, as before. But let \lambda be an eigenvalue, (\tau,\delta) be an eigenpair, and v be a vector in both generalized eigenspaces. The invariance of \mathrm{Ker}\left((T^2-\tau T+\delta I_V)^d\right) under T shows that if u is a generalized eigenvector of (\tau,\delta), then so is Tu-\lambda u. Just like we did before, we can keep hitting v with T-\lambda I_V until one step before it vanishes (which it eventually must, since it’s a generalized eigenvector of \lambda). So without loss of generality we can assume that v is an actual eigenvector of \lambda and a generalized eigenvector of (\tau,\delta).

Now we can use the generalized eigenvector property to write

\displaystyle(T^2-\tau T+\delta I_V)^dv=0

but since v is an eigenvector with eigenvalue \lambda, this says

\displaystyle(\lambda^2-\tau\lambda+\delta)^dv=0

If v is nonzero, this can only be true if \lambda is a root of X^2-\tau X+\delta, which we assumed not to be the case.

Finally we consider two distinct eigenpairs (\tau_1,\delta_1) and (\tau_2,\delta_2), and a generalized eigenvector of both, v. Another argument like that above shows that without loss of generality we can assume v is an actual eigenvector of (\tau_2,\delta_2). This eigenspace is the kernel \mathrm{Ker}(T^2-\tau_2T+\delta_2I_V), which is thus invariant, and another argument like before lets us assume that v is an actual eigenvector of both eigenpairs. Thus we have

\displaystyle\begin{aligned}(T^2-\tau_1T+\delta_1I_V)v&=0\\(T^2-\tau_1T+\delta_1I_V)v&=0\end{aligned}

Subtracting, we find

\displaystyle -(\tau_1-\tau_2)Tv+(\delta_1-\delta_2)v=0

If \tau_1-\tau_2\neq0, this makes v an eigenvector with eigenvalue \frac{\delta_1-\delta_2}{\tau_1-\tau_2}. On the other hand, if \tau_1=\tau_2 then \delta_1\neq\delta_2, and we conclude that v=0.

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

About these ads

April 7, 2009 - Posted by | Algebra, Linear Algebra

1 Comment »

  1. [...] 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 [...]

    Pingback by Decomposing Real Linear Transformations « The Unapologetic Mathematician | April 9, 2009 | Reply


Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 392 other followers

%d bloggers like this: