Wuzzy wrote: > Assoicated with what?
x1 is the eigenvalue of linear transformation x2; x3 is the (generalized) eigenspace associated with eigenvalue x1 of linear transformation x3. x3 is the (generalized) eigenspace that belongs to x2 and x1 (in different slightly different, but related (obviously), senses for each). x1 is 'owns' all (eigen)vectors (and the zero vector) in (generalized) eigenspace x3 (of linear transformation x2).
So, x1 and x2 both are associated [in the backward direction from/wrt the previous usage of "associated with"] with (generalized) eigenspace x3; namely, x2 is the linear transformation that governs any possible (generalized) eigenspace x3, whereas x1 is the eigenvalue that which (one) of these possible (generalized) eigenspaces is being selected/referenced; elseways, x1 is an eigenvalue and can be associated with some (uncountably) infinite set of (generalized) eigenspaces (but definitely not ALL possible (generalized) eigenspaces), while the linear transformation x2 selects one such (generalized) eigenspace out of the crowd  and if we fix x1 and choose a bad x2, this (generalized) eigenspace will be trivial (id est: have only the zero vector (and possibly the infinite vector in each coordinate direction), which is/are in a sense (an) eigenvector(s) of all linear transformations).
Note, therefore, that the typical ordering of important information will make x3 depend on x1, x2, and x4. But, really, these are all mutually dependent objects: but exact specification may not be yielded in certain cases. For example, setting x4 = 1 (so that the eigenspace is not generalized (it is basic)), x1 = 0, and x3 = Transpose[(r,0)],+,* for real r [the vectorspace of (column) vectors of form (r,0) for real r], will tell us that any transformation that can work in x2 will be of form: a b 0 b = c d 0 d. We get constraints on a and c, but b and d are free (they could be any numbers at all, including ones that are not even real, if we so desire).
However, certain combinations of parameters (and/or constraints from context and previous definitions) will yield fully constrained parameters for the last x_i.
:) Make sense? (Also, do you understand what is meant by x4 and "generalized eigenspace"? How about the fact that the zero vector is an element of x3 (taken as a set), and therefore can be considered to be an eigenvector (for every generalization of eigenspace), in Lojbanic mathematics (culture), in some sense? "Degenerate" versus "nondegenerate"? Everything else?)


Comment #5:
Re: “associated with/”

Curtis W Franks (Mon Jan 13 05:11:58 2014)

Formatting messed up (I forgot about how the markup is handled). Hopefully this works.
> So, x1 and x2 both are associated [in the backward direction from/wrt the previous usage of "associated with"] ...
I was just bracketing out commentary information, indicating that the present usage of the word "associated" was essentially a semantic inverse of the previous usage of the same word. I was trying not to disrupt the flow of the explanation too much by this interjection of information, as well as distinguishing one type of parenthetical with this (more meta) usage of an aside. So, the hyperlinked text should be considered merely an aside.
> x3 = Transpose[(r,0)],+,* for real r
Should be: x3 is the vectorspace of column vectors of two entries, the first entry being any real number and the second entry being 0 identically. (Basically any vector pointing along a given line.)
> [the vectorspace of (column) vectors of form (r,0) for real r]
Same thing. I was just trying to explain what I meant the first time, but the explanation was hyperlinked.
> will tell us that any transformation that can work in x2 will be of form: > a b 0 b > = > c d 0 d.
I was making a square matrix of four entries (id est a 2by2 matrix). The entry in the first column and first row is a = 0, the entry in the second column and first row is b, the entry in the second row and first column is c = 0, and the entry in the second row and second column is d.



