Let A be a linear transformation from V to V. Suppose it has an eigenvalue lambda knot. Prove that the set S = {x in V such that x is an eigenvector with the eigenvalue lambda knot} is a subspace of V. I'm assuming S is the eigenspace corresponding to lambda knot? I'm having difficulty understanding how the set S can be a subspace (must contain the zero vector) yet the zero vector cannot be an eigenvector?

Question

zzr0ck3r · Answer

you are correct

zzr0ck3r · Answer

Does it mean the set spanned by elements in S?

tiffany_rhodes · Answer

No, it just says S = {x in V such that "x is an eigenvector with the eigenvalue lambda knot" }.

tiffany_rhodes · Answer

sorry, hope that's clear.

kainui · Answer

Haha cute mistake, it's not "knot" it's "naught" in $\lambda_0$.

tiffany_rhodes · Answer

lol, oops. Never seen it actually written out in words

zzr0ck3r · Answer

also, why a "naught" at all.... only one lambda

tiffany_rhodes · Answer

It's just how my professor denoted the eigenvalue

zzr0ck3r · Answer

yeah yeah all good.

zzr0ck3r · Answer

well you are right and there must be some mistake.

kainui · Answer

Well presumably the linear map A has multiple eigenvaues and eigenvectors.

zzr0ck3r · Answer

read this http://math.stackexchange.com/questions/990016/can-the-zero-vector-be-an-eigenvector-for-a-matrix It may clear it up

tiffany_rhodes · Answer

from what I'm reading online, it says that an eigenspace is all eigenvectors associated with an eigenvalue plus the zero vector ?

zzr0ck3r · Answer

just show closure of addition and multiplication

tiffany_rhodes · Answer

okay!

kainui · Answer

I think the problem with "zero vector being an eigenvector" is that if you consider it as an eigenvector then it'd have infinitely many possible eigenvalues. In this case, we're really just focused on the space spanned by this single vector so it's not a big deal I think. Interesting kinda exception for why 0 shouldn't be included though, good eye @tiffany_rhodes

inkyvoyd · Answer

Yeah, nice catch - for our crappy applied linear algebra class our teacher was very clear about the eigenvectors being non-zero, and even added that this theorem in particular you're looking to prove has to specifically add in the zero vector for the set to actually be a subspace. I would just assume that semantic mistake, point it out before doing the proof, assume what they meant, and continue carrying on I guess.

tiffany_rhodes · Answer

thanks guys! I appreciate the help. @inkyvoyd @Kainui @zzr0ck3r

zzr0ck3r · Answer

or don't point it out because they probably said it :)

zzr0ck3r · Answer

are you cool on the closure parts?