The Unapologetic Mathematician

Mathematics for the interested outsider

Determining Generalized Eigenvectors

Our definition of a generalized eigenvector looks a lot like the one for an eigenvector. But finding them may not be as straightforward as our method for finding eigenvectors. In particular, we’re asking that the vector be in the kernel of not one transformation, but some transformation in a whole infinite list of them. We can’t just go ahead and apply each transformation, hoping to find one that sends our vector to zero.

Maybe the form of this list can help us. We’re really just taking the one transformation T-\lambda1_V and applying it over and over again. So we could start with v and calculate (T-\lambda1_V)v, and then (T-\lambda1_V)^2v, and so on, until we end up with (T-\lambda1_V)^nv=0. That’s all well and good if v is a generalized eigenvector, but what if it isn’t? At what point do we stop and say we’re never going to get to zero?

The first thing we have to notice is that as we go along the list of transformations, the kernel never shrinks. That is, if (T-\lambda1_V)^iv=0 then

\displaystyle(T-\lambda1_V)^{i+1}v=(T-\lambda1_v)(T-\lambda1_V)^iv=(T-\lambda1_V)0=0

Thus, we have an increasing sequence of subspaces

\displaystyle 0=\mathrm{Ker}\left((T-\lambda1_V)^0\right)\subseteq\mathrm{Ker}\left((T-\lambda1_V)^1\right)\subseteq\mathrm{Ker}\left((T-\lambda1_V)^2\right)\subseteq...

Next we have to recognize that this sequence is strictly increasing until it levels out. That is, if \mathrm{Ker}\left((T-\lambda1_V)^{i-1}\right)=\mathrm{Ker}\left((T-\lambda1_V)^i\right) then \mathrm{Ker}\left((T-\lambda1_V)^i\right)=\mathrm{Ker}\left((T-\lambda1_V)^{i+1}\right). Then, of course, we can use an inductive argument to see that all the kernels from that point on are the same. But why does this happen? Well, let’s say that (T-\lambda1_V)^iv=0 implies that (T-\lambda1_V)^{i-1}v=0 (the other implication we’ve already taken care of above). Then we can see that (T-\lambda1_V)^{i+1}v=0 implies that (T-\lambda1_V)^iv=0 by rewriting them:

\displaystyle\begin{aligned}(T-\lambda1_V)^{i+1}v&=0\\\Rightarrow(T-\lambda1_V)^i(T-\lambda1_V)v&=0\\\Rightarrow(T-\lambda1_V)^{i-1}(T-\lambda1_V)v&=0\\\Rightarrow(T-\lambda1_V)^iv&=0\end{aligned}

where we have used the assumed implication between the second and third lines.

So once this sequence stops growing at one step, it never grows again. That is, if the kernels ever stabilize the sequence looks like

\displaystyle\begin{aligned}...\subsetneq\mathrm{Ker}\left((T-\lambda1_V)^{n-1}\right)\subsetneq\mathrm{Ker}\left((T-\lambda1_V)^n\right)\\=\mathrm{Ker}\left((T-\lambda1_V)^{n+1}\right)=...\end{aligned}

and \mathrm{Ker}\left((T-\lambda1_V)^n\right) is as large as it ever gets.

So does the sequence top out? Of course, it has to! Indeed, each step before it stops growing raises the dimension by at least one, so if it didn’t stop by step d=\dim(V) it would get bigger than the space V itself, which is absurd because these are all subspaces of V. So \mathrm{Ker}\left((T-\lambda1_V)^d\right) is the largest of these kernels.

Where does this leave us? We’ve established that if v is in the kernel of any power of T-\lambda1_V it will be in \mathrm{Ker}\left((T-\lambda1_V)^d\right). Thus the space of generalized eigenvectors with eigenvalue \lambda is exactly this kernel. Now finding generalized eigenvectors is just as easy as finding eigenvectors.

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February 17, 2009 - Posted by | Algebra, Linear Algebra

5 Comments »

  1. [...] points, it’s going to be useful to have a sort of dual to the increasing chain of subspaces we found yesterday. Instead of kernels, we’ll deal with [...]

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  2. [...] turns out that generalized eigenspaces do capture this notion, and we have a way of calculating them to boot! That is, I’m asserting that the multiplicity of an eigenvalue is both the number of [...]

    Pingback by The Multiplicity of an Eigenvalue « The Unapologetic Mathematician | February 19, 2009 | Reply

  3. [...] a single transformation whose only eigenvalue is . That is, for sufficiently large powers . We’ve seen that is sufficiently large a power to take to check if is nilpotent. We’ve also seen that [...]

    Pingback by Nilpotent Transformations I « The Unapologetic Mathematician | February 26, 2009 | Reply

  4. [...] polynomial of , whose roots are the eigenvalues of . For each eigenvalue , we can define the generalized eigenspace as the kernel , since if some power of kills a vector then the th power [...]

    Pingback by Jordan Normal Form « The Unapologetic Mathematician | March 4, 2009 | Reply

  5. [...] same argument as before tells us that the kernel will stabilize by the time we take powers of an operator, so we [...]

    Pingback by Generalized Eigenvectors of an Eigenpair « The Unapologetic Mathematician | April 6, 2009 | Reply


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