> 1. A linear mapping is not a "kind of equivalence" by any reasonable definition. For instance, the function that maps every vector to 0 is a linear mapping, and it has plenty of eigenvectors. (All with eigenvalue 0.)
Linear mappings are the homomorphisms between vector spaces. So they are a "kind of equivalence".
I know what an equivalence class is, thank you. How does that make the map from (say) R^3 to itself that sends everything to 0 a "kind of equivalence"?
It gives rise to an equivalence relation on R^3 in a fairly natural way, as indeed any function gives rise to an equivalence relation on its domain: x~y iff f(x)=f(y). But that doesn't mean that it is a "kind of equivalence".
Now, obviously, "kind of" is vague enough that saying "an endomorphism of a vector space simply Is Not a 'kind of equivalence'" would be too strong. But I would like to know what, exactly, you mean by calling something a "kind of equivalence", because I'm unable to think of any meaning for that phrase that (1) seems sensible to me and (2) implies that endomorphisms of vector spaces are "kinds of equivalence".
(Looking back at what bonsaitree wrote, I see s/he said "of any kind of equivalence", which I unconsciously typo-corrected to "or any kind of equivalence". But perhaps I misunderstood and bonsaitree meant something else, though I can't think what it might be. bt, if you're reading this: my apologies if I misunderstood, and would you care to clarify if so?)
Linear mappings are the homomorphisms between vector spaces. So they are a "kind of equivalence".
See for example http://en.wikibooks.org/wiki/Linear_Algebra/Definition_of_Ho...