Results 1 to 4 of 4

Thread: Linear Algebra, eigenvectors problem

  1. #1
    Newbie
    Joined
    Apr 2010
    Posts
    1

    Linear Algebra, eigenvectors problem

    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,779
    Thanks
    3028
    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.
    Very good!


    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
    I presume you meant (1) here.

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
    Well, what is (A^-1)Av? And, of course, because A and $\displaystyle A^{-1}$ are linear operators, $\displaystyle A^{-1}(\lambda v)= \lambda A^{-1}v$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,779
    Thanks
    3028
    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v
    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.
    Very good!


    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v
    I presume you meant (1) here.

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it
    Well, what is (A^-1)Av? And, of course, because A and $\displaystyle A^{-1}$ are linear operators, $\displaystyle A^{-1}(\lambda v)= \lambda A^{-1}v$.

    As long as $\displaystyle \lambda\ne 0$, you can divide both sides by $\displaystyle \lambda$. Of course, if $\displaystyle \lambda= 0$, then A does not have an inverse.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by ksheehan62 View Post
    If v is an eigenvector of an invertible matrix A, which of the following is/are necessarily true?

    (1) v is also an eigenvector of 2A
    (2) v is also an eigenvector of A^2
    (3) v is also an eigenvector of A^-1

    A) 1 only
    B) 2 only
    C) 3 only
    D) 1 and 3 only
    E) 1,2 and 3

    I am pretty sure 2 is true because if we look at Av = (lamba)v.
    A^2v = A(lambda)v = (Av)(lambda) = (lambda)v(lambda) = (lambda)^2 v

    So A^2v = (lambda)^2v. So that should be that v is an eigenvector for A^2 as well.

    2 is also true because if Av = (lamda)v then (2A)v = (2lambda)v

    Can someone help show 3? If Av = lambda v
    then (A^-1)Av =A^-1(lambda)v
    I am not sure how to show it


    $\displaystyle Av=\lambda v\Longrightarrow v=Iv=A^{-1}Av=A^{-1}(Av)=A^{-1}(\lambda v)=\lambda A^{-1}v\Longrightarrow $ multiplying now by $\displaystyle \lambda^{-1}$ in both extremes of the above equality (why

    can we do such a thing?) you get that $\displaystyle v$ is also an eigenvector of $\displaystyle A^{-1}$ .

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear algebra: Jordan forms & eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: May 3rd 2011, 05:14 AM
  2. Eigenvalues and eigenvectors [Linear Algebra]
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: Dec 28th 2010, 11:38 AM
  3. [SOLVED] Linear Algebra, basis of eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 21st 2010, 01:01 PM
  4. linear algebra with transformations and eigenvectors
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Nov 13th 2009, 11:40 AM
  5. Replies: 3
    Last Post: Nov 10th 2007, 05:15 AM

Search Tags


/mathhelpforum @mathhelpforum