Let's say my eigenvalue λ=-1 and we assume eigenvector of zero are non-eigenvector. An eigenspace is mathematically represented as E_{λ} = N(λ.I_{n}-A) which essentially states, in natural language, the eigenspace is the nullspace of a matrix. N(λ.I_{n}-A) is a matrix. Would it then be valid to say that the eigenspace, E_{λ}, whose eigenvalue, λ=-1, is the nullspace of the matrix, N(λ.I_{n}-A), is equivalent to the the vector , v, where Av = 0. If v is the nullspace of the matrix A then Av = 0, and similarly, if E_{λ} is the nullspace of a matrix, N(λ.I_{n}-A), then, it must equally be true that [ N(λ.I_{n}-A) ] [E_{λ=-1}] = 0
It is not clear exactly what you mean by [ N(λ.I_{n}-A) ] [E_{λ=-1}] = 0 Certainly some interpretations would make the statement true, but [ N(λ.I_{n}-A) ] =[E_{λ=-1}] is also true (some interpretations)
"We assume eigenvector of zero are non-eigenvector"? I don't know what this means. Do you mean that you are assuming that 0 is not an eigenvector or are you asserting that an "eigenvector" must not be the 0 vector? No. First, it is a set of vectors, not a single vector. Second, it is the set of vectors such that (λ.I_{n}-A)v= 0, or equivalently, λ.I_{n}v= Av, as you said, not "Av= 0". I cant make sense out of that. You have, as part of your hypothesis, "If v is (in) the nullspace of the matrix A" but you conclusion has no "v" in it.
You probably can't because it doesn't make sense. It's still a new topic to me so I'd figure the fastest way to refine my understanding is to state what I know and have it corrected where the understanding is flawed. I'm stating that a zero vector is not within the consideration of an eigenvector as I've learnt. Let's see: λ is an eigenvalue of the eigenvetor V IFF det(λI_{n}-A) = 0 Say we have a matrix [1 2;4 3] and det((λI_{n}-A) = 0 gives λ_{1}=5 and λ_{2} = -1 1) Does this then states that λ_{1}=5 and λ_{2} = -1 are eigenvalues corresponding to the eigenvector V? 2) E_{λ1,λ2}=Null(λI_{n}-A) is interpreted as the set of vector with eigenvalues of λ_{1} and λ_{2} that maps the matrix (λI_{n}-A) to the zero vector. True? 3) where λ_{1}=-1: I have E_{λ1}=N(λI_{n}-A) = N(-1[1 0; 0 1] - [1 2; 4 3]) = N([-2 -2; -4 -4]) 4) (I shall take only λ_{1} as an example. Since E_{λ1} maps the matrix [-2 -2; -4 -4] to the zero vector then could it then be written as AV=0 where A = [1 2; 4 3] and V is the set of vector V = {v_{1}, v_{2},....v_{n}} that maps the matrix A to the zero vector? If I could, then, matrix A after RREF(is it a necessarily condition to perform RREF on matrix A? what is the implication if I perform REF instead of RREF?) becomes [ 1 1; 0 0 ] and V = {v_{1}, v_{2}} and so, [ 1 1; 0 0] [v_{1};v_{2}] = [0;0]
By definition, an eigenvector is nonzero. These are eigenvalues of that matrix. Each eigenvalue has its own eigenvector. For the eigenvalue λ_{1} = 5, an eigenvector associated with this eigenvalue is any nonzero solution x of the matrix equation Ax = 5x. Equivalently, x is a nonzero solution of (A - 5I)x = 0. Not sure. This is a pretty cumbersome way to say it. For a given eigenvalue λ of a matrix A, x is an eigenvector if it is a nonzero solution of the equation Ax = λx. This is equivalent to (A - λI)x = 0, so x is any nonzero vector in the nullspace of A - λI. No. You're not looking for solutions of Ax = 0 - you're looking for solutions of (A - λI)x = 0.
I'm going to jump in here- Yes, many textbooks define an eigenvector to be non-zero. But many others define "Eigenvalue" by "[itex]\lambda[/itex] is an eigenvalue for linear operator A if there exist a non-zero vector, v, such that [itex]Av= \lambda v[/itex]" and then define an Eigenvector, corresponding to eigenvalue [itex]\lambda[/itex], to be any vector, v, such that [itex]Av= \lambda v[/itex], which includes the 0 vector. Personally, I prefer that because it allows us to say "the set of all eigenvectors, corresponding to eigenvalue [itex]\lambda[/itex], is a subspace" rather than having to say "the set of all eigenvectors, corresponding to eigenvalue [itex]\lambda[/itex], together with the 0 vector, is a subspace"