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Eigenvector question

  1. Aug 3, 2005 #1
    Hi all,

    I have the following question.

    A = nxn non singular matrix
    I = nxn identity matrix
    li = eigevalues of A i=1,2...n
    ui = eigenvectors corresponding to the previous eigenvalues.

    It true that

    ( A - l1 * I ) * x =0

    is satisfied by any vector of the form x = a1 * u1 where a1= arbitrary real number

    Lanczos in his book Applied Analysis p. 61 claims that the following quadratic equation in A

    ( A - l1 * I ) * ( A - l2 * I ) * x = 0

    is satisfied by an arbitrary linear combination of the first two eigenvectors

    x = a1 * u1 + a2 * u2

    It is not very obvious to me why this happens.
    (Extending this to include n eigenvectors and eigenvalues will eventually lead to the so called Cayley-Hamilton theorem.)

    I was wondering if you could give me a hint starting from first principals.

    Thanks
     
  2. jcsd
  3. Aug 3, 2005 #2

    lurflurf

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    ( A - l1 * I ) * ( A - l2 * I ) = ( A - l2 * I ) *( A - l1 * I )
    so if
    T=( A - l1 * I ) * ( A - l2 * I ) = ( A - l2 * I ) *( A - l1 * I )
    T(a1 * u1)=a1*T(u1)=0
    T(a2 * u2)=a1*T(u2)=0
    hence by linearity
    T(a1 * u1 + a2 * u2)=0
     
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