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Eigenvalue, Eigenvector and Eigenspace 
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#1
May1014, 01:17 AM

P: 819

Let's say my eigenvalue λ=1 and we assume eigenvector of zero are noneigenvector.
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 


#2
May1014, 03:00 AM

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You mean you are trying to construct an eigenspace from a single vector?



#3
May1014, 04:29 AM

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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) 


#4
May1014, 06:56 AM

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Eigenvalue, Eigenvector and Eigenspace



#5
May1014, 09:34 AM

P: 819

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] 


#6
May1014, 10:09 AM

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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. 


#7
May1114, 02:06 PM

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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" 


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