Results 1 to 13 of 13
Like Tree2Thanks
  • 1 Post By FernandoRevilla
  • 1 Post By ymar

Thread: eigenvector

  1. #1
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    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 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member girdav's Avatar
    Joined
    Jul 2009
    From
    Rouen, France
    Posts
    678
    Thanks
    32

    Re: eigenvector

    The fact that the space is complex is important, since you now that the characteristic polynomial has a root.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: eigenvector

    Yes. I know that the roots of the characteristic polynomial are eigenvalue . but i show x has an eigenvector v\in V
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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 $\displaystyle B=\{v_1,\ldots,v_n\}$ of $\displaystyle V$ , if $\displaystyle \lambda$ is a root of the characteristic polynomial and $\displaystyle A=[T]_B$ then, the system $\displaystyle (A-\lambda I)x=0$ has at least a non zero solution $\displaystyle (\alpha_1,\ldots,\alpha_n)\in\mathbb{C}^n$ . Choose $\displaystyle v=\alpha_1v_1+\ldots+\alpha_nv_n$ .
    Thanks from vernal
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Feb 2011
    Posts
    147
    Thanks
    3

    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 $\displaystyle x$ does have an eigenvalue. By definition, an eigenvalue of a linear map $\displaystyle x$ is a scalar $\displaystyle \lambda$ such that there exists a vector $\displaystyle v$ in the space $\displaystyle V$ such that $\displaystyle xv=\lambda v.$ This is also the definition of the eigenvector $\displaystyle v.$

    There is a nice proof of this without the use of determinants/charcteristic polynomials. It's here, page 3.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    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 $\displaystyle x$ does have an eigenvalue. By definition, an eigenvalue of a linear map $\displaystyle x$ is a scalar $\displaystyle \lambda$ such that there exists a vector $\displaystyle v$ in the space $\displaystyle V$ such that $\displaystyle xv=\lambda v.$ This is also the definition of the eigenvector $\displaystyle v.$

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: eigenvector

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

  8. #8
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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?
    $\displaystyle \textrm{rank}\;(A-\lambda I)<n$ . Now, apply Rouche Fröbenius Theorem.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: eigenvector

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

    what is T?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Feb 2011
    Posts
    147
    Thanks
    3

    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 $\displaystyle v$ 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 $\displaystyle x$ has an eigenvalue? This:

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

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


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

    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 $\displaystyle x$ has an eigenvalue means this:

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

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

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

    the last formula being equvalent to

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

    and further equivalent to

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

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

    $\displaystyle xv=\lambda v,$

    just as we wanted.
    Thanks from vernal
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,163
    Thanks
    46

    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 $\displaystyle x$ instead of $\displaystyle T$ . Of course, it does not matter .
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: eigenvector

    Quote Originally Posted by FernandoRevilla View Post
    Sorry, I didn't notice that you call the linear map $\displaystyle x$ instead of $\displaystyle T$ . Of course, it does not matter .
    thanks:-D
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member vernal's Avatar
    Joined
    Dec 2011
    Posts
    75

    Re: eigenvector

    Quote Originally Posted by ymar View Post
    I made a small but significant mistake in my earlier post. The vector $\displaystyle v$ 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 $\displaystyle x$ has an eigenvalue? This:

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

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


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

    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 $\displaystyle x$ has an eigenvalue means this:

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

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

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

    the last formula being equvalent to

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

    and further equivalent to

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

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

    $\displaystyle xv=\lambda v,$

    just as we wanted.
    thanks for your help ymar
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Apr 11th 2011, 12:18 PM
  2. eigenvector
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Jan 22nd 2011, 09:39 AM
  3. Eigenvector
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 9th 2010, 03:43 AM
  4. eigenvector
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 21st 2010, 02:24 PM
  5. eigenvector of A+I
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Dec 9th 2009, 03:17 PM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum