So far I've been seeing that the vector that makes up a basis for the null space of the matrix $A-Ix$ (where x is an eigenvalue) is the eigenvector corresponding to the eigenvalue $x$. But I've never run into a situation where the basis had more than one vector in it.
Let's say the basis of the null space of $A-2*I$ was made up of two vectors. Would that mean those two (and only those two) vectors are the eigenvectors corresponding to the eigenvalue $2$ for the matrix $A$?
One thing of note is that, from what I understand, the identity transformation (in any space) has only one eigenvalue, $1$, but it has infinitely many eigenvectors (not sure if there is a vector space where this wouldn't be true, but it's true for $R^n$ at least). Not sure how this fact fits in here.
Just wanting to be sure I'm not misunderstanding something. I'm grateful for any help.