$\DeclareMathOperator{\rank}{rank}$
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!
Best Answer
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.