2

Reading the Wikipedia article on the topic, when you have a defective matrix that lacks a full rank eigenspace, you can form 'generalised eigenvectors' by looking at the equation $(A - \lambda I)v = v_\lambda$, where $v_\lambda$ denotes a true eigenvector and $v$ a generalised one (corresponding to the same eigenvalue).

Opening up the expression, we're looking for a vector that maps to a multiple of itself plus the true eigenvector, ie. $Av = \lambda v + v_\lambda$, which makes some sense when considering an action like a shear transformation, but it's a little hazy.

My questions are: why is this a good definition for these generalised vectors, and why exactly does it guarantee linear independence? If we have, say, n different eigenvalues with certain algebraic multiplicities, what guarantees that their respective generalised vectors do not overlap? Is it something to do with the way the nullspace maps these things?

Not Legato
  • 180
  • 2
    These generalised eigenvectors, together with the ‘true’ eigenvectors, make up a Jordan basis for the Jordan normal form of the matrix. – Bernard May 06 '19 at 19:04
  • Ah. That's a problem. I'm wholly unfamiliar with Jordan forms. I suppose I'll have to delve into them eventually, but my understanding of linear algebra at this point might not warrant going there yet. – Not Legato May 06 '19 at 19:17

1 Answers1

2

The chain of generalized eigenvectors built up in this way spans an invariant subspace relative to $A$. This lets you decompose the ambient space into the direct sum of subspaces similarly to the way that you can decompose the ambient space into a direct sum of eigenspaces when $A$ is diagonalizable. The action of $A$ on each of these subspaces has a relatively simple description, and there’s no “crosstalk” between them. This is all covered, more or less, in that Wikipedia article. There are standard proofs of this in any competent linear algebra text. You can find a proof of an equivalent proposition in the Wikipedia article on Jordan normal form.

amd
  • 53,693
  • Can you elaborate on the decomposition of these spaces a little more symbolically? I'm unfamiliar with the term 'ambient space'. – Not Legato May 06 '19 at 19:20
  • @NotLegato The ambient space is simply the domain $V$ of the linear map $T:V\to V$ represented by $A$—the space in which all of these vectors we’re talking about live. – amd May 06 '19 at 20:44