Generalized Eigenvectors?

  • #1
I see that a generalized eigenvector can be represented as such:

(A - λI)xk+1 = xk, where A is a square matrix, x is an eigenvector, λ is the eigenvalue I is the identity matrix.


This might be used, for example, if we have duplicate eigenvalues, and can only derive one eigenvector from the characteristic equation, then we can use this eigenvector to find other eigenvectors.

Can someone explain this formula? I don't really get it. My textbook has a similar formula:

xk+1 = Axk

which I understand perfectly. But how does the former formula make sense?
 

Answers and Replies

  • #2
AlephZero
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Are you talking about an iterative method to find the eigenvalues and vectors? This looks like the inverse power method.

The eigenvectors of [itex]A[/itex] and [itex](A - \lambda I)[/itex] are identical, but the eigenvalues differ by [itex]\lambda[/itex].

The inverse power method converges to the eigenvalue with the smallest modulus.

If [itex]\lambda[/itex] is close to an eigenvalue of [itex]A[/itex], the corresponding eigenvalue of [itex](A - \lambda I)[/itex] will be close to 0 and the iteration will converge rapidly to that eigenpair.
 
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  • #3
HallsofIvy
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Suppose the characteristic equation for a matrix A is [itex](x- \lambda)^2= 0[/itex].

Then [itex]\lambda[/itex] is an eigenvalue having "algebraic multiplicity" 2. Now the "geometric multiplicity" of A, the number of independent eigenvectors corresponding to that eigenvalue, is either 1 or 2.

The crucial point is that every matrix satisfies its own characteristic equation. That is, [itex](A- \lambda I)^2= 0[/itex] so that [itex](A- \lambda I)^2v= 0[/itex] for every vector v. Now, it might happen that [itex](A- \lambda I)v= 0[/itex] for every vector, from which [itex](A- \lambda I)v= (A-\lambda I)((A- \lambda I)v)= (A- \lambda I)0= 0[/itex] follows immediately. That would be the case if there are two independent eigenvectors- if the geometric multiplicity were 0.

If not, if A has only one eigenvector, it must still be true that [itex](A- \lambda I)^2v= 0[/itex], even if v is NOT an eigenvector of A. But in that case, [itex]u= (A- \lambda I)v[/itex] is NOT 0 but we still must have [itex](A- \lambda I)^2 v= (A- \lambda I)((A- \lambda I)v)= (A- \lambda I)u= 0[/itex] which says that u is an eigenvector of A.

That is, in order to have [itex](A- \lambda I)^2 v= 0[/itex], either v is an eigenvector of A or [itex](A- \lambda v)[/itex] is an eigenvector.
 
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