# Complicated Defective Eigenvalues

The usual, simple defective eigenvalue problem has algebraic multiplicity $$2$$ or $$3$$, and geometric multiplicity $$1$$. Then a complete set of generalized eigenvectors is obtained by using the lone rank $$1$$ generalized eigenvector (a.k.a., the eigenvector) to obtain a rank $$2$$ generalized eigenvector, the rank $$2$$ generalized eigenvector to obtain a rank $$3$$ generalized eigenvector, etc., peeling back the onion one layer at a time in a linear fashion.

What happens in more complicated cases where the geometric multiplicity is more than $$1$$? Suppose the geometric multiplicity is $$5$$ and the algebraic multiplicity is $$6$$. Will plugging any of the five rank $$1$$ generalized eigenvectors into $$Av_{r2} = v_{r1}$$ yield the same, correct $$v_{r2}$$, identical up to a scalar multiple?

Suppose the geometric multiplicity is $$5$$ and the algebraic multiplicity is $$10$$. Is picking a single $$v_{r1}$$ and proceeding linearly through the sequence $$v_{r1} \rightarrow v_{r2} \rightarrow v_{r3} \rightarrow v_{r4} \rightarrow v_{r5} \rightarrow v_{r6}$$ the way to go? Can we alternatively obtain a single $$v_{r2}$$ from each $$v_{r1}$$, or find any other total of five generalized eigenvectors of rank $$2$$ or greater? My intuition is that finding one $$v_{r6}$$ guarantees linear independence, while finding five $$v_{r2}$$'s only yields linearly independence for extremely rare matrices. If this is correct, it would imply that since any $$v_{r1}$$ can be used to derive a complete linearly independent set of generalized eigenvectors, a specific vector could occupy different ranks depending on the initial choice, i.e., it could be rank $$2$$ in one "stack" and rank $$5$$ in another.

Is this correct? If not, how does it in fact work?

• You should have learned (or will soon) that the number of Jordan blocks for an eigenvalue is equal to its geometric multiplicity. Each element of a basis for the eigenspace has its own chain; those chains can be of different lengths.
– amd
Apr 27 '20 at 5:41

I won’t go into too much detail here because I’m pretty sure that all of this will be covered in detail in whatever materials you’re studying, but I’ll offer some observations that seem relevant. Let $$m$$ be the algebraic multiplicity of the eigenvalue $$\lambda$$. By the Cayley-Hamilton theorem, $$(A-\lambda I)^m=0$$, so given any vector $$v$$, if we repeatedly apply $$A-\lambda I$$ to it the process reaches the zero vector after at most $$m$$ iterations. In fact, it might be fewer than that. The actual bound is given by the exponent $$r$$ of the corresponding factor of the minimal polynomial.
In other words, for any vector $$v$$, repeated applications of $$A-\lambda I$$ generate a unique chain of vectors $$v\to v_1\to v_2\to\dots\to 0$$ that consists of at most $$r$$ nonzero vectors. The last vector in this chain is, of course, an eigenvector of $$A$$. These chains can merge, but they can’t cross, as the latter would mean that $$(A-\lambda I)v$$ has two different values for some $$v$$. So, in particular, each generalized eigenvector corresponds to a unique eigenvector of $$A$$—it belongs to exactly one “stack,” to use your terminology, which also means that its position relative to the eigenvector end of the chain is fixed.