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Eigenvectors and eigenvalues - how to find the column vector

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Tala.S
#1
Nov19-12, 01:07 PM
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 :)
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AlephZero
#2
Nov19-12, 02:09 PM
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Quote Quote by Tala.S View Post
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.
Tala.S
#3
Nov19-12, 02:30 PM
P: 43
When I solve it I get

v =

(1 0)
(0 0)
(0 0)

?
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Erland
#4
Nov19-12, 02:58 PM
P: 345
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.
HallsofIvy
#5
Nov19-12, 02:59 PM
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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 [itex]\lambda[/itex] is an eigenvalue for matrix A if and only if there exist a non-trivial (i.e. non-zero) vector v such that [itex]Av= \lambda v[/itex].

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.".
Tala.S
#6
Nov19-12, 03:04 PM
P: 43
So the vector is (0,0,0) ?

Or have I completely misunderstood this ?
Erland
#7
Nov19-12, 03:30 PM
P: 345
Quote Quote by Tala.S View Post
So the vector is (0,0,0) ?
Right. But it is not counted as an eigenvector.
AlephZero
#8
Nov19-12, 04:05 PM
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Quote Quote by Tala.S View Post
So the vector is (0,0,0) ?

Or have I completely misunderstood this ?
Correct. The question asked you to find a column vector, not a non-zero column vector!

You only get non-zero solutions of Av = λv when λ is an eigenvalue, but v = 0 is a solution for any value of λ.


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