I was curious whether the direct sum of generalized eigenspaces in a finite dimensional vector space is the largest decomposition of this space invector spaces that are invariant under the endomorphism or are there counterexamples available?

  • $\begingroup$ What would it mean that one decomposition is "larger" than another one? $\endgroup$ – Martin Argerami May 30 '13 at 22:56
  • $\begingroup$ the question is whether all invariant subspaces are generalized eigenspaces? I see that we have generalized eigenspace => invariant subpace, but is the converse also true? $\endgroup$ – user66906 May 30 '13 at 22:57

No. If you take the sum of two generalized eigenspaces, it will still be an invariant subspace, since generalized eigenspaces correspond to the blocks in the Jordan decomposition.

Even in finite dimension, the number of invariant subspaces can be infinite. To see the most basic example, consider the identity endomorphism: it obviously has infinitely many invariant subspaces. Or consider $$ V=\begin{bmatrix}1&0&0\\0&0&0\\0&0&0\end{bmatrix} $$ as an endomorphism of $\mathbb R^3$. Then, for each $t\in[0,1]$, the subspace $$ X_t=\{(0,ct,c(1-t)):\ c\in\mathbb R\} $$ is invariant for $V$.

  • $\begingroup$ and by also taking those into account? or even better: how do I find all invariant subspaces for a given endomorphism? $\endgroup$ – user66906 May 30 '13 at 23:19
  • $\begingroup$ The number of invariant subspaces is often infinite (even uncountable). See the edit. $\endgroup$ – Martin Argerami May 30 '13 at 23:45
  • $\begingroup$ so is it in generally impossible to determine all invariant subspaces right? $\endgroup$ – user66906 May 31 '13 at 8:11
  • $\begingroup$ Looks like it, yes. $\endgroup$ – Martin Argerami May 31 '13 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy