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Eigenvectors and eigenvalues  how to find the column vector 
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#1
Nov1912, 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 :) 


#2
Nov1912, 02:09 PM

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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. 


#3
Nov1912, 02:30 PM

P: 43

When I solve it I get
v = (1 0) (0 0) (0 0) ? 


#4
Nov1912, 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.



#5
Nov1912, 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 nontrivial (i.e. nonzero) vector v such that [itex]Av= \lambda v[/itex]. Since two is not an eigenvalue, there cannot exist a nontrivial vector but, as Alephzero said, "There IS a vector that satisfies A*v = 2 v, but you might think it's not a very interesting vector.". 


#6
Nov1912, 03:04 PM

P: 43

So the vector is (0,0,0) ?
Or have I completely misunderstood this ? 


#7
Nov1912, 03:30 PM

P: 345




#8
Nov1912, 04:05 PM

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


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