The Unapologetic Mathematician

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.