Generalized Eigenvectors of an Eigenpair

Just as we saw when dealing with eigenvalues, eigenvectors alone won’t cut it. We want to consider the kernel not just of one transformation, but of its powers. Specifically, we will say that $v$ is a generalized eigenvector of the eigenpair $(\tau,\delta)$ if for some power $n$ we have

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

The same argument as before tells us that the kernel will stabilize by the time we take $d=\dim(V)$ powers of an operator, so we define the generalized eigenspace of an eigenpair $(\tau,\delta)$ to be

$\displaystyle\mathrm{Ker}\left((T^2-\tau T+\delta I_V)^d\right)$

Let’s look at these subspaces a little more closely, along with the older ones of the form $\mathrm{Ker}\left((T-\lambda I_V)^d\right)$ , just to make sure they’re as well-behaved as our earlier generalized eigenspaces are. First, let $V$ be one-dimensional, so $T$ must be multiplication by $\lambda_0$ . Then the kernel of $T-\lambda I_V$ is all of $V$ if $\lambda=\lambda_0$ , and is trivial otherwise. On the other hand, what happens with an eigenpair $(\tau,\delta)$ ? Well, one application of the operator gives

$\displaystyle(T^2-\tau T+\delta I_V)v=(\lambda_0^2-\tau\lambda_0+\delta)v$

for any nonzero $v$ . But this will always be itself nonzero, since we’re assuming that the polynomial $X^2-\tau X+\delta$ has no roots. Thus the generalized eigenspace of $(\tau,\delta)$ will be trivial.

Next, if $V$ is two-dimensional, either $T$ has an eigenvalue or it doesn’t. If it does, then this gives a one-dimensional invariant subspace. The argument above shows that the generalized eigenspace of any eigenpair $(\tau,\delta)$ is again trivial. But if $T$ has no eigenvalues, then the generalized eigenspace of any eigenvalue $\lambda$ is trivial. On the other hand we’ve seen that the kernel of $T^2-\tau T+\delta I_V$ is either the whole of $V$ or nothing, and the former case happens exactly when $\tau$ is the trace of $T$ and $\delta$ is its determinant.

Now if $V$ is a real vector space of any finite dimension $d$ we know we can find an almost upper-triangular form. This form is highly non-unique, but there are some patterns we can exploit as we move forward.

April 6, 2009 - Posted by John Armstrong | Algebra, Linear Algebra

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