$\DeclareMathOperator{\rank}{rank}$ First off I'm sorry I'm still not able to make of use the built in formula expressions, I don't have time to learn it now, I'll do it before my next question.

I have a couple of questions regarding eigenvectors and generalized eigenvectors. To some of these questions I know the answer partially or there are some uncertainties so I will just ask in the most general form, but I can really appreciate precise answers.

How do I know how many eigenvectors to expect for each eigenvalue?

How do I know how many generalized eigenvectors to expect for each of those eigenvectors? Consider a matrix $A$ whose eigenvalues and vectors I'd like to compute. Do basic row and column operations on either $A$ or $(A - \lambda I)$ (lambda be an eigenvalue) change any of the eigenvalues, -vectors or determinants of the two corresponding matrices?

Any of the following statements my be wrong and I'd appreciate it if you could point out where the errors are.

Consider this special case for the matrix A:

Its rank is $4$. The characteristic polynomial tells me there is an eigenvalue lambda with algebraic multiplicity $4$. In order to determine the geometric multiplicities to the corresponding eigenvalues (which there is just one of) I can determine the rank of $(A - \lambda I) = (A - 2I)$. Said matrix looks like this

Operating with basic row and column operations on this matrix $(A - 2I)$ I can reduce down to a matrix with just one $1$ and all the other elements will be zero. Thus the rank of this matrix is $1$. In order to get the geometric multiplicity corresponding to this eigenvalue I compute $$ \rank(A) - \rank(A-2I) = 4 - 1 = 3 $$ So the geometric multiplicity of this eigenvalue is $3$, which means I can expect $3$ eigenvectors.

If so far no errors have been made and no corrections have been given, consider the following: Are the eigenvectors to this specific problem unique? Clearly I can reduce the matrix $(A - 2I)$ down to a matrix with one 1 at any element I like.

Let's say we picked the 1 as the first element of said matrix; are my eigenvectors just $(0,1,0,0)^T;(0,0,1,0)^T;(0,0,0,1)^T$? (the T stands for transposed)

How do I compute the generalized eigenvectors, which eigenvector do I pick, how do I determine which one to choose and what is it?

Thanks for your time!

  • $\begingroup$ I thank you very kindly for your edits! One more question: What is the relationship between algberaic/geometric multiplicities and eigenvalues/eigenvectors/generalized eigenvectors? $\endgroup$ – axin Mar 2 '14 at 22:35
  • $\begingroup$ Row operations will certainly change eigenvalues and eigenvectors in general. For example, the $n\times n$ identity matrix has all nonzero vectors as eigenvectors, with unique eigenvalue 1. But you can perform row operations on it to obtain any invertible $n\times n$ matrix you like, which will have other eigenvalues and fewer eigenvectors than the identity matrix. The one thing that doesn't change after row operations is the null space, i.e. the eigenspace with eigenvector $0$. $\endgroup$ – bradhd Mar 2 '14 at 22:35
  • $\begingroup$ So the operations I performed on the matrix skewed the results? That is to say it is wrong. I'm really keen on finding the correct way to determine the numbers of such vectors (how many there are of for each eigenvalue) and how to compute them precisely. $\endgroup$ – axin Mar 2 '14 at 22:45
  • $\begingroup$ The eigenvectors of $A$ with eigenvalue $\lambda$ are precisely the nonzero elements of the null space of $A-\lambda I$. So if you can calculate null spaces of matrices (which you can do using row operations) then you can calculate eigenspaces too. $\endgroup$ – bradhd Mar 3 '14 at 0:03
  • $\begingroup$ Note that there will always be either zero or infinitely many eigenvectors of $A$ with a given eigenvalue, so it doesn't make sense to ask "how many eigenvectors" there are. I suspect that you intend to ask instead what the dimension of a given eigenspace is. $\endgroup$ – bradhd Mar 3 '14 at 0:04

To find the geometric multiplicity of an eigenvalue $\lambda$, you want to find $nullity(A-\lambda I)=n-rank(A-\lambda I)$, where n is the number of columns of A.

(Notice that you want to use n instead of rank(A).)

Any basis of $\ker(A-\lambda I)$ will give you a set of linearly independent eigenvectors for the eigenvalue $\lambda$.

In your example, you can find a generalized eigenvector w for $\lambda=2$ by either selecting an eigenvector v and then solving $(A-2I)w=v$ for w, or by choosing any vector w which is not in $\ker(A-2I)$ and then taking $v=(A-2I)w$ as one of your eigenvectors.

  • $\begingroup$ These have been very good answers so far. Are the three eigenvectors I computed correct? If so, the only generalized eigenvector to any of these three eigenvectors I can think of is the zero vector. Is this allowed? Also is there any rule on which of the eigenvectors to pick to find the one gen eigenvector making it 4 vectors in total? Or did I make any prohibited matrix operations and thus arrived at the wrong set of eigenvectors? $\endgroup$ – axin Mar 3 '14 at 8:59
  • $\begingroup$ $v=(x,y,z,u)$ is an eigenvector if $x=y+u$ (and v is not the zero vector), so only your second vector is an eigenvector. (A zero vector is not allowed.) You can see a set of 3 LI eigenvectors in the link you provided. If you solve $(A-2I)w=v$ to get w, you need to start with an arbitrary eigenvector $v=(y+u,y,z,u)$, so you might find that the second method is easier. $\endgroup$ – user84413 Mar 3 '14 at 16:51
  • $\begingroup$ My script does not mention this method. I have stumbled over this on the internet before, but it's never been explained very well. So how does it work exactly? Why and when am I allowed to do this? And why is it exactly x that has to fulfill this requirement (y + u, that is)? $\endgroup$ – axin Mar 3 '14 at 16:57
  • $\begingroup$ I got the equation $x=y+u$ by solving $(A-2I)v=0$ using row reduction. Let me see if I can find a good online reference for finding the generalized eigenvectors, and then I'll send it to you if I find one. $\endgroup$ – user84413 Mar 3 '14 at 19:17
  • $\begingroup$ I already figured that part out, so I know now how to compute all the eigenvectors. Only thing that still keeps me wondering is how to get the correct generalized eigenvector. $\endgroup$ – axin Mar 3 '14 at 19:23

Regarding counting eigenvectors:

Algebraic multiplicity of an eigenvalue = number of associated (linearly independent) generalized eigenvectors. That is, the characteristic polynomial of $A$ will be of the form $$ p(x) = (x - \lambda_1)^{j_1} \cdots (x - \lambda_n)^{j_n} $$ and $j_i$ is the number of generalized eigenvectors associated with $\lambda_i$.

Geometric multiplicity of an eigenvalue = number of associated (linearly independent) standard eigenvectors. For eigenvalue $\lambda_i$, this is the same as $\dim(\ker(A - \lambda I))$.


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