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Math Help - eigen vector problem....

  1. #1
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    eigen vector problem....

    Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
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  2. #2
    Super Member Gamma's Avatar
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    Au=u
    Av=3v
    A(u+v)=u+3v
    Think
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  3. #3
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    Quote Originally Posted by Gamma View Post
    Au=u
    Av=3v
    A(u+v)=u+3v
    Think
    really very silly problem...
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathventure View Post
    Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
    Since (1,u) and (3,v) are eigenpairs, then Au=u and Av=3v, where u,v\neq0

    Suppose that u+v is an eigenvector. Thus A(u+v)=\lambda(u+v) where u+v\neq0

    But A(u+v)=Au+Av=u+3v and this is equal to \lambda u+\lambda v. Comparing both sides, we get 1=\lambda=3\implies 3=1, which is impossible!

    It is now evident that u+v has no eigenvalue, and thus it is not an eigenvector of A.
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  5. #5
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    Quote Originally Posted by Mathventure View Post

    Q:Let u and v be the eigen vectors of A corresponding to eigen values 1 and 3 respectively.Prove that u+v is not an eigen vector of A?
    let's do the general case: suppose Au=ru, \ Av=sv, where u,v are non-zero vectors and r \neq s are scalars. we'll show that there's no scalar t satisfying A(u+v)=t(u+v).

    so suppose there's such t. then ru+sv=t(u+v) and thus (r-t)u=(t-s)v. call this 1). we also have r(r-t)u=A(r-t)u=A(t-s)v=s(t-s)v, which with 1) gives us:

    r(r-t)u=s(r-t)u, and hence r=t because u \neq 0 and r \neq s. this with 1) gives us r=s, which is a contradiction!


    Note: the above is a trivial thing if you know that distinct eigenvalues correspond to linearly independent eigenvectors.
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