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The eigenspace of (a square matrix) $A$ corresponding to $\lambda$ is the collection of all vectors $\mathbf{x}$ that satisfy $A\mathbf{x}=\lambda\mathbf{x}$, or equivalently, $(A-\lambda I)\mathbf{x}=\mathbf{0}$.

The generalized eigenspace of $A$ corresponding to $\lambda$ is the collection of all vectors $\mathbf{x}$ for which there exists a positive integer $k$ for which $(A-\lambda I)^k\mathbf{x}=\mathbf{0}$

What I'm looking for is an equivalent point of view as follows: Writing $A\mathbf{x}=\lambda\mathbf{x}$ can be interpreted as the ubiquitous saying that an eigenvector is a vector that only transforms by a factor when $A$ is acting on it. By looking at all vector $\mathbf{x}$ which satisfy $(A-\lambda I)\mathbf{x}=\mathbf{0}$, we obtain exactly those vectors.

But by looking at $(A-\lambda I)^k\mathbf{x}=\mathbf{0}$ I can't see the intuitive idea of defining the generalized Eigenspace like this. Sure, if we want a "simple", deconstructed "look" of what our linear function does (aka maybe an almost diagonal form) it works (if we are over an algebraically closed field), at least the impression I got from my textbook is basically that we are trying to obtain a nearly diagonal matrix and the only way to construct such thing is by looking at generalized Eigenspaces and by quietly doing so, we do obtain what we want.

But that can't be the way this concept was conceived, right?

So to sum the question up: If I want to find all vectors that are only "stretched" with a factor of $\lambda$ by my linear function, I look at the eigenspace $E=\{ \mathbf{x} | (A-\lambda I)\mathbf{x}=\mathbf{0}\}$. Is there a similar thought when looking at the generalized Eigenspace $E^{k}=\{ \mathbf{x} | (A-\lambda I)^k\mathbf{x}=\mathbf{0}\}$? How would one arrive at this definition?

Taleofwoe
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  • Hint: https://en.wikipedia.org/wiki/Jordan_matrix . For some matrices, the Eigenspaces do not fill the complete space. – Gyro Gearloose Nov 19 '20 at 17:40
  • You might find this post to be helpful – Ben Grossmann Nov 19 '20 at 17:43
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    A quick answer: the eigenspace (associated with $\lambda$) is a maximal invariant subspace on which vectors get scaled by $\lambda$, but eigenspaces aren't enough to give us a full decomposition of the space. The associated eigenspace is a maximal invariant subspace over which $A$ has $\lambda$ is its only eigenvalue; weakening the qualifications for an "eigenvector" in this way gives us the complete decomposition that we want – Ben Grossmann Nov 19 '20 at 17:50
  • @Ben Grossman Thank you for your linked post, it was indeed interesting to read as I had a similar thought regarding the nilpotency but couldn't wrap my head around how it connected. Anyway, your quick answer leaves me with more questions than answers though: When you write "The associated eigenspace" at the start of the 2nd sentence, did you mean "generalized" ? And in the same sentence, could you elaborate on why a maximal invariant subspace over which A has only $\lambda$ as its eigenvalues is a weakened condition for eigenvector? Doesn't any vector in an eigenspace have only one unique ev? – Taleofwoe Nov 20 '20 at 00:15
  • Whoops! Yes generalized – Ben Grossmann Nov 20 '20 at 02:04
  • By “weakened” I mean that every eigenvector is automatically a generalized eigenvector but not the other way around – Ben Grossmann Nov 20 '20 at 02:05
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    I'm afraid, I still don't understand the second sentence then. What does "A (...) subspace over which $A$ has $\lambda$ is its only eigenvalue" mean? Could you rephrase it? Do you mean by that, that this subspace contains all vectors which are, if they are scaled by $A$, only scaled by $\lambda$ (but also contains other vectors, which are not scaled)? If yes, then I still don't understand: if the idea is to extend the eigenspace(s) to get a complete decomposition, why are we not just extending by adding some linear independent vectors? Again, why do we arrive at the generalized eigenspace? – Taleofwoe Nov 20 '20 at 09:39
  • A question for your linked post (not enough reputation to comment there): after you define the eigenvector and the generalized v & w respectively, you introduce the nilpotency by telling us to observe $(A-2 I)^2 w=0$ and then you ask what it tells us about A. Then you show that w behaves similar to an eigenvector since Aw=2w+v. But where does the nilpotency play a role? You could have shown that without introducing the nilpotent part by just multiplying Aw. I suspect it has to do with "a transformation can be related to action of a nilpotent over a subspace" but I fail to see the connection. – Taleofwoe Nov 20 '20 at 09:57

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