Results 1 to 5 of 5

Thread: eigen vector problem....

  1. #1
    Junior Member
    Joined
    Jan 2009
    From
    india
    Posts
    51

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    $\displaystyle Au=u$
    $\displaystyle Av=3v$
    $\displaystyle A(u+v)=u+3v$
    Think
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    From
    india
    Posts
    51
    Quote Originally Posted by Gamma View Post
    $\displaystyle Au=u$
    $\displaystyle Av=3v$
    $\displaystyle A(u+v)=u+3v$
    Think
    really very silly problem...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    5
    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 $\displaystyle Au=u$ and $\displaystyle Av=3v$, where $\displaystyle u,v\neq0$

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

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

    It is now evident that $\displaystyle u+v$ has no eigenvalue, and thus it is not an eigenvector of A.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 $\displaystyle Au=ru, \ Av=sv,$ where $\displaystyle u,v$ are non-zero vectors and $\displaystyle r \neq s$ are scalars. we'll show that there's no scalar $\displaystyle t$ satisfying $\displaystyle A(u+v)=t(u+v).$

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

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


    Note: the above is a trivial thing if you know that distinct eigenvalues correspond to linearly independent eigenvectors.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Eigen Vector
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Aug 18th 2011, 01:41 AM
  2. Simple eigen value / vector problem needed
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 20th 2010, 04:48 AM
  3. Replies: 5
    Last Post: Apr 5th 2010, 02:41 AM
  4. Eigen value , eigen vector
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: Jan 5th 2010, 04:00 AM
  5. Find the eigen values and eigen vector
    Posted in the Advanced Algebra Forum
    Replies: 13
    Last Post: Nov 24th 2009, 07:01 PM

Search Tags


/mathhelpforum @mathhelpforum