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November 17th, 2013, 08:10 AM  #1 
Newbie Joined: Sep 2012 Posts: 14 Thanks: 1  NonParallel Eigenvectors with the same Eigenvalue
Hello, I have the following matrix: \left[ \begin{array}{ccc} 3 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \\ \end{array} \right] Using the characteristic equation, I have found that the eigenvalues are 1 and 2. I am then asked to find two nonparallel eigenvectors with eigenvalue = 2. I have found that one one eigenvector is: \left[ \begin{array}{c} 1 \\ 0 \\ 1 \\ \end{array} \right] How would I go about finding the other nonparallel eigenvector with eigenvalue=2? Any help would be greatly appreciated. 
November 17th, 2013, 08:12 AM  #2  
Newbie Joined: Sep 2012 Posts: 14 Thanks: 1  Re: NonParallel Eigenvectors with the same Eigenvalue Quote:
 
November 17th, 2013, 05:20 PM  #3 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: NonParallel Eigenvectors with the same Eigenvalue
The definition of "eigenvalue" and "eigenvector" is that " is an eigenvalue of A if and only if there exist a nonzero vector v such that " and "v is an eigenvector corresponding to eigenvalue of linear transformation, A, if and only if ". If is an eigenvalue for , then there exist vector such that . That is equivalent to the three equations 3x+ y z= x, x+ y+ z= y, and x+ y+ z= z. Those then reduce to 2x+ y z= 0, x+ z= 0, and x+ y= 0. From x+ z= 0 we get z= x. From x+ y= 0 we get y= x. Putting those into 2x+ y z= 0 we have 2x x x= 0 which is true for all x. Because y= x and z= x, any eigenvector corresponding to eigenvalue 1 must of the form <x, x, x>= x<1, 1, 1>. The set of all eigenvectors corresponding to eigenvalue 1 make up the one dimensional vector space spanned by <1, 1, 1>. If is an eigenvalue for , then there exist vector such that That is equivalent to the three equations 3x+ y z= 2x, x+ y+ z= 2y, and x+ y+ z= 2z. Those then reduce to x+ y z= 0, x y+ z= 0, and x+ y z= 0. Those three equations are all equivalent! The first and third are identical and the second is just the first multiplied by 1. Any x, y, z that satisfy them must have z= x+ y so such an eigenvector is of the form < x, y, x+ y>= <x, 0, x>+ <0, y, y>= x<1, 0, 1>+ y<0, 1, 1>. Now do you see the two nonparallel eigenvectors corresponding to eigenvalue 2? (Of course, this can only happen because eigenvalue 2 has "algebraic multiplicity" 2 (2 is a double root of the characteristic polynomial). The fact that the eigenspace has dimension 2 means it has "geometric multiplicity" 2. The geometric multiplicity of an eigenvalue must be less than or equal to its algebraic multiplicity.) 

Tags 
eigenvalue, eigenvectors, nonparallel 
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