# Eigenvectors and eigenvalues - how to find the column vector

 P: 43 Hi We have a matrix A (picture), the eigenvalues are λ1 = 4 and λ2 = 1 and the eigenvectors are λ1 : t(1,0,1) λ2 : t1(1,0,2) + t2(0,1,0) I have to examine if there's a column vector v that satifies : A*v = 2 v I would say no there doesn't exist such a column vector v because 2 isn't an eigenvalue: Av = λv so Av = 2v but we know that λ is 4 or 1 and not 2 Am I wrong ? I would be nice if someone could give me their opinion :) Attached Thumbnails
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P: 7,285
 Quote by Tala.S I have to examine if there's a column vector v that satifies : A*v = 2 v I would say no there doesn't exist such a column vector v because 2 isn't an eigenvalue:
You seem to understand what eigenvalnes and vectors are, which is good.

But you aren't quite right. There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.

Try solving (A - 2I)v = 0, and see what you get.
 P: 43 When I solve it I get v = (1 0) (0 0) (0 0) ? Attached Thumbnails
 P: 353 Eigenvectors and eigenvalues - how to find the column vector You cannot use Maple in this way. You should use linsolve(A1,b) or LinearSolve(A1,b). Or better, do it by hand.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,682 I don't understand what you mean by that. Your vectors before were single columns of three numbers. How could v be two columns? You were close to right when you said before "doesn't exist such a column vector v because 2 isn't an eigenvalue". What is true is that $\lambda$ is an eigenvalue for matrix A if and only if there exist a non-trivial (i.e. non-zero) vector v such that $Av= \lambda v$. Since two is not an eigenvalue, there cannot exist a non-trivial vector but, as Aleph-zero said, "There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.".
 P: 43 So the vector is (0,0,0) ? Or have I completely misunderstood this ?
P: 353
 Quote by Tala.S So the vector is (0,0,0) ?
Right. But it is not counted as an eigenvector.
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