Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Eigenvalues and eigenvectors

  1. May 5, 2013 #1
    can someone PLEASE explain eigenvalues and eigenvectors and how to calculate them or a link to a site that teaches it simply?
     
  2. jcsd
  3. May 5, 2013 #2
    Ive already read the wiki and asked my math teacher, he doesnt even know what they are.
     
    Last edited by a moderator: May 5, 2013
  4. May 5, 2013 #3

    UVW

    User Avatar

  5. May 5, 2013 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you know what "vectors" and "linear transformations" are? Do you know what a "linear vector space" is?
     
  6. May 5, 2013 #5
    That is sad to hear, eigenvectors and eigenvalues are very basic maths. Teachers are very underqualified these days.

    A linear operator is a function that maps one vector space into another, there are certain vectors which when transformed by the linear operator, comes out as a scalar multiple of itself, the vector is the eigenvector and the multiple is the eigenvalue.
     
  7. May 5, 2013 #6
    dont judge me but how does one map one vector space onto another?
     
  8. May 5, 2013 #7
    he doesnt know what a vector space even is! And the wiki doesnt do much to even seperate it from vectors.
     
  9. May 5, 2013 #8

    WannabeNewton

    User Avatar
    Science Advisor

    Let ##V## be a vector space over ##F## and let ##T:V\rightarrow V## be a linear operator. We say ##v\in V\setminus \left \{ 0 \right \}## is an eigenvector of ##T## if there exists a ##\lambda\in F## such that ##T(v) = \lambda v##. We call ##\lambda## an eigenvalue of ##T##.

    As an example, let ##V = M_{n\times n}(\mathbb{R})## and let ##T:V\rightarrow V,A \mapsto A^{T}##. We want to find the eigenvalues of ##T##. Let ##A\in V## such that ##T(A) = A^{T} = \lambda A##. Note that ##T(T(A)) = \lambda ^{2}A = (A^T)^T = A## hence ##A(\lambda^{2} - 1) = 0## and since eigenvectors have to be non-zero, this implies ##\lambda = \pm 1##.
     
  10. May 6, 2013 #9
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Eigenvalues and eigenvectors
Loading...