For an complex matrix , does not necessarily have a basis consisting of eigenvectors of . But it will always have a basis consisting of generalized eigenvectors of .
However this is not the end of the story. Letting , we see that is the zero matrix. Moreover, , where . We then see that is not an eigenvector of , but is. There is an inclusion In this example, the vector is referred to as a generalized eigenvector of the matrix ; it satisfies the property that the vector itself is not necessarily an eigenvector of , but is for some . One other observation worth noting: in this example, the smallest exponent of satisfying the property is , the algebraic multiplicity of the eigenvalue . This is not an accident.
Letting , we have a sequence of inclusions
The generalized eigenvalue problem is to find a basis for each generalized eigenspace compatible with this filtration. This means that for each , the vectors of lying in is a basis for that subspace.
This turns out to be more involved than the earlier problem of finding a basis for , and an algorithm for finding such a basis will be deferred until Module IV.
One thing that can often be done, however, is to find a Jordan chain. We will first need to define some terminology.
By the above Theorem, such an always exists. A generalized eigenvector of , then, is an eigenvector of iff its rank equals . For an eigenvalue of , we will abbreviate as .
By definition of rank, it is easy to see that every vector in a Jordan chain must be non-zero. In fact, more is true
In this way, a rank generalized eigenvector of (corresponding to the eigenvalue ) will generate an -dimensional subspace of the generalized eigenspace with basis given by the Jordan chain associated with .
Thus is a generalized eigenvector of of rank , and the Jordan chain is a basis for (in fact, it is the standard basis).
Thus there is a gap of two between the dimension of the generalized eigenspace , and that of the regular eigenspace . Note also that is an eigenvector for .
Can we find a Jordan chain which provides a basis for the generalized eigenspace , which we see is all of ? If a single Jordan chain is going to do the job, it must have length , and therefore be the Jordan chain associated to a generalized eigenvector of rank .
Now , , with . We are then looking for a vector with (which is automatically the case), but with . We see that this last condition is satisfied iff the third coordinate of is non-zero.
There is clearly a choice involved. The simplest choice here is to take . Then , and . So in this case we see
The previous examples were designed to be able to easily find a Jordan chain. The following is a bit more involved.
For , we cannot have two linearly independent Jordan chains of length , because that would give . So we must have a single Jordan chain of length 2. Now . Inspection shows the vector is in but not in . Letting yields a Jordan chain of length 2: which is then a basis for .