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Homework Help: Zero eigenvector?

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    I can calculate the proper eigenvalues, but when I plug them back into the matrix, I get x1=0 and x2=0. But this is not the answer Maple gives me! How do I solve for the eigenvector when it appears that a zero vector is the only solution?

    2. Relevant equations

    For example, for the matrix {1,1},{1,-1} (rows shown), Maple gives me: eigenvalue of sqrt(2) with eigenvector {1/((sqrt(2)-1),1} and eigenvalue of -sqrt(2) with eigenvector {1/(-(sqrt(2)-1),1}

    3. The attempt at a solution

    But I can't get these eigenvectors when I try to solve by hand! How do you solve in these situations?
    What are these situations called?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 13, 2010 #2
    For sqrt(2), You end up with the eigenvector equation:

    [tex]\left(\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right)\binom{x}{y}=\sqrt{2}\binom{x}{y}[/tex]

    or:

    [tex](1-\sqrt{2}) x+y=0[/tex]

    [tex]x-y(1+\sqrt{2})=0[/tex]

    which are redundant so any vector (x,y) that satisfies either equation is a suitable eigenvector like:

    [tex]\binom{1+\sqrt{2}}{1}[/tex]

    and even:

    [tex]\binom{\frac{1}{\sqrt{2}-1}}{1}[/tex]
     
  4. Sep 13, 2010 #3
    Thanks! Oh duh! I didn't realize that the 2 equations were the same. That's why I was getting x=0 and y=0 as solutions. I had to multiply through by the appropriate constant to make the 2 equations look the same.
     
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