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Math Help - eigenvector

  1. #1
    Member vernal's Avatar
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    eigenvector

    Let V be an n-dimensional complex vector space where and let be a linear map.
    show that has an eigenvector .

    i can't solve this question . how can ?
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  2. #2
    Super Member girdav's Avatar
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    Re: eigenvector

    The fact that the space is complex is important, since you now that the characteristic polynomial has a root.
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  3. #3
    Member vernal's Avatar
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    Re: eigenvector

    Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigenvector

    Quote Originally Posted by vernal View Post
    Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
    Fix a basis B=\{v_1,\ldots,v_n\} of V , if \lambda is a root of the characteristic polynomial and A=[T]_B then, the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n . Choose v=\alpha_1v_1+\ldots+\alpha_nv_n .
    Thanks from vernal
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    Re: eigenvector

    Quote Originally Posted by vernal View Post
    Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
    If you already know that the roots of the characteristic polynomial are eigenvalues, then you are done. The characteristic polynomial does have a complex root and therefore x does have an eigenvalue. By definition, an eigenvalue of a linear map x is a scalar \lambda such that there exists a vector v in the space V such that xv=\lambda v. This is also the definition of the eigenvector v.

    There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.
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  6. #6
    Member vernal's Avatar
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    Re: eigenvector

    Quote Originally Posted by ymar View Post
    If you already know that the roots of the characteristic polynomial are eigenvalues, then you are done. The characteristic polynomial does have a complex root and therefore x does have an eigenvalue. By definition, an eigenvalue of a linear map x is a scalar \lambda such that there exists a vector v in the space V such that xv=\lambda v. This is also the definition of the eigenvector v.

    There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.
    thanks..but in page 3 :Theorem 2.1 Every linear operator on a finite-dimensional complex vector space
    has an eigenvalue., but my question eiguenvector.
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  7. #7
    Member vernal's Avatar
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    Re: eigenvector

    Quote Originally Posted by FernandoRevilla View Post
    Fix a basis B=\{v_1,\ldots,v_n\} of V , if \lambda is a root of the characteristic polynomial and A=[T]_B then, the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n . Choose v=\alpha_1v_1+\ldots+\alpha_nv_n .
    thanks fernandorevilla..
    you said the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n , why?
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  8. #8
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigenvector

    Quote Originally Posted by vernal View Post
    thanks fernandorevilla.. you said the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n , why?
    \textrm{rank}\;(A-\lambda I)<n . Now, apply Rouche Fröbenius Theorem.
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  9. #9
    Member vernal's Avatar
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    Re: eigenvector

    Quote Originally Posted by FernandoRevilla View Post
    Fix a basis B=\{v_1,\ldots,v_n\} of V , if \lambda is a root of the characteristic polynomial and A=[T]_B then, the system (A-\lambda I)x=0 has at least a non zero solution (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n . Choose v=\alpha_1v_1+\ldots+\alpha_nv_n .
    excuse me

    what is T?
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  10. #10
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    Re: eigenvector

    Quote Originally Posted by vernal View Post
    thanks..but in page 3 :Theorem 2.1 Every linear operator on a finite-dimensional complex vector space
    has an eigenvalue., but my question eiguenvector.
    I made a small but significant mistake in my earlier post. The vector v is supposed to be non-zero.

    The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that x has an eigenvalue? This:

    1) \left (\exists \lambda\in\mathbb{C}\right)\left( \exists v\in V\setminus\{0\}\right ) xv=\lambda v.

    What does it mean that x has an eigenvector? This:


    2) \left( \exists v\in V\setminus\{0\}\right )\left (\exists \lambda\in\mathbb{C}\right) xv=\lambda v.

    These formulas are equivalent. The order of the existential quantifiers doesn't matter.

    However, Sheldon Axler uses a different definition of an eigenvalue. For him, that x has an eigenvalue means this:

    3) There exists \lambda\in \mathbb{C} such that x-\lambda I is not injective.

    We want to see that 3) implies 2). What does it mean that x-\lambda I is not injective? It means that we have two vectors v_1,v_2\in V, such that v_1\neq v_2 and

    (x-\lambda I)v_1=(x-\lambda I)v_2,

    the last formula being equvalent to

    xv_1 -\lambda v_1 = xv_2 -\lambda v_2,

    and further equivalent to

    x(v_1-v_2)=\lambda (v_1-v_2).

    Since v_1\neq v_2, we have that the vector v_1-v_2 is non-zero. Therefore, we have found a non-zero vector v such that

    xv=\lambda v,

    just as we wanted.
    Thanks from vernal
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  11. #11
    MHF Contributor FernandoRevilla's Avatar
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    Re: eigenvector

    Quote Originally Posted by vernal View Post
    excuse me what is T?
    Sorry, I didn't notice that you call the linear map x instead of T . Of course, it does not matter .
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  12. #12
    Member vernal's Avatar
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    Re: eigenvector

    Quote Originally Posted by FernandoRevilla View Post
    Sorry, I didn't notice that you call the linear map x instead of T . Of course, it does not matter .
    thanks:-D
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  13. #13
    Member vernal's Avatar
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    Re: eigenvector

    Quote Originally Posted by ymar View Post
    I made a small but significant mistake in my earlier post. The vector v is supposed to be non-zero.

    The existence of an eigenvalue implies the existence of an eigenvector. With the definitions from my earlier post it's immediate. What does it mean that x has an eigenvalue? This:

    1) \left (\exists \lambda\in\mathbb{C}\right)\left( \exists v\in V\setminus\{0\}\right ) xv=\lambda v.

    What does it mean that x has an eigenvector? This:


    2) \left( \exists v\in V\setminus\{0\}\right )\left (\exists \lambda\in\mathbb{C}\right) xv=\lambda v.

    These formulas are equivalent. The order of the existential quantifiers doesn't matter.

    However, Sheldon Axler uses a different definition of an eigenvalue. For him, that x has an eigenvalue means this:

    3) There exists \lambda\in \mathbb{C} such that x-\lambda I is not injective.

    We want to see that 3) implies 2). What does it mean that x-\lambda I is not injective? It means that we have two vectors v_1,v_2\in V, such that v_1\neq v_2 and

    (x-\lambda I)v_1=(x-\lambda I)v_2,

    the last formula being equvalent to

    xv_1 -\lambda v_1 = xv_2 -\lambda v_2,

    and further equivalent to

    x(v_1-v_2)=\lambda (v_1-v_2).

    Since v_1\neq v_2, we have that the vector v_1-v_2 is non-zero. Therefore, we have found a non-zero vector v such that

    xv=\lambda v,

    just as we wanted.
    thanks for your help ymar
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