If I have an 3 eigen values E1 = 1 , E (2,3) = 2 <--- Doubly Degenerate Value, w

Question

If I have an 3 eigen values E1 = 1 , E (2,3) = 2 <--- Doubly Degenerate Value, where the corresponding eigenvectors are

V1 = [-2 1 1]^t and V2 = [-1 0 1]^t, how do I calcualte the third Eigenvalue if my final reduction for the calculate eigenvalue E(2,3) is

1  0  1  : 0

0  0  0  : 0

0  0  0  : 0

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How do I gain two eigenvectors from this?

Can I simply say this

-->  Let x1 = r

therefore x3 = -r

let x2 = s

Therefore choosing parameters to create two eigenvectors:

let r = 1, s = 0

therefore, first eigenvector corresponding to eigenvalue is

V = (1  0  1) ^t

Where, V = ( r  s -r) ^t

and the second is simply created by choosing the following parameters:

let r = 0, and s =1

V = (0  1  0)^t

Therefore, we can conclude that our third eigenvector is V3 = ( 0  1  0)^t

Is this correct?

Heres the original problemstyle="text-decoration: underline;">

http://www.mediafire.com/view/bode58n8372k7ji/Untitled-2.jpg

Im mostly interested in how to properly calculate the third eigenvector! Ive already done EVERYTHING up to the point of calculating it myself, but Im not sure if my answer above is correct!!!

If I have an 3 eigen values E1 = 1 , E (2,3) = 2 leftarrow Doubly Degenerate Value, where the corresponding eigenvectors are V1 = [-2 1 1]^t and V2 = [-1 0 1]^t, how do I calculate the third Eigenvalue if my final reduction for the calculate eigenvalue E(2,3) is How do I gain two eigenvectors from this?

Explanation / Answer

absoultely..that's the approach. but i am just saying that the method is correct but i don't know wheather u have a calculation mistake or not. for example if r=1 , then it sholud be (1 0 -1)^t , but u have written (1 0 1)^t.. so just check ur calculation , the method is absolutely corrcet.

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