## Eigenvectors and eigenvalues - how to find the column vector

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 :)
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 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.
 When I solve it I get v = (1 0) (0 0) (0 0) ? Attached Thumbnails

## 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.".
 So the vector is (0,0,0) ? Or have I completely misunderstood this ?

 Quote by Tala.S So the vector is (0,0,0) ?
Right. But it is not counted as an eigenvector.

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