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Finding Eigenvectors by inspection

  1. Nov 16, 2006 #1
    Would someone please explain to me how I can find eigenvalues and eigenvectors by inpection of simple symmetric matrices? I just can't figure it out.

    He is an example:
    By looking at [tex]A=\left(\begin{matrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{matrix}\right)
    [/tex] I should be able to guess that [1,1,1] is an eigenvector.
     
    Last edited: Nov 17, 2006
  2. jcsd
  3. Nov 17, 2006 #2

    quasar987

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    Because the elements in the lines sum to the same thing. when you multiply the matrix by the vector (1,1,1), you get a new vector whose components are the sum of each lines in the matrix. as soon as the lines of a matrix sum to the same thing, then (1,1,1) is an eigenvector with associated eigenvalue the sum of the lines, in this case 2-1-1=0.
     
  4. Nov 17, 2006 #3
    How do I know how many eigenvalues there will be. In other words, if I could see that [1,1,1] is an eigenvector just by inspection, then how do I know if there are any other eigenvectors?
     
  5. Nov 17, 2006 #4
    The eigenvectors are spanned by [1,1,1]. An example would be [2,2,2] is an eigevector of A. Also, an nxn matrix has at most n eigenvalues (counted with their algebraic multiplicities). To find the eigenvalues you just have to solve the characteristic equation. Then the eigenvectors are the nonzerovectors in the eigenspace associated with that eigenvalue. Eigenspace with eigenvalue L is ker(A-LI) where I is the nxn identity
     
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