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Math Help - Finding Eigenvectors and Eigenvalues of a reflection about a Plane

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    Finding Eigenvectors and Eigenvalues of a reflection about a Plane

    Find all eigenvalues and eigenvectors of the linear transformation. Find a basis consisting of eigenvectors if possible.

    Reflection about a plane V in R^3.


    I'm not sure how to set up this problem, any help would be appreciated.
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by vcf323 View Post
    Find all eigenvalues and eigenvectors of the linear transformation. Find a basis consisting of eigenvectors if possible.

    Reflection about a plane V in R^3.

    We can choose a basis B = { e_1 , e_2 , e_3 } of IR^3 such that e_1 , e_2 belongs to V and e_3 is orthogonal to V. So, if R is the reflexion over V,

    R ( e_1 ) = e_1

    R ( e_2 ) = e_2

    R ( e_3 ) = - e_3

    That is, B is a basis of IR^3 whose elements are eigenvectors of R .
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    For any reflection about a (hyper)plane (subspace of dimension one less than the dimension of the vector space), you can always set up an orthonormal basis where one of the basis vectors is perpendicular to the given plane and the others are in that plane. If "u" is that basis vector perpendicular to the plane, then R(u)= -u and if v is any basis vector in the plane, then R(v)= v. That is, -1 is one eigenvalue with the vector perpendicular to the plane being a corresponding eigenvector and 1 is the only other eigenvalue with all the basis vectors in the plane being independent eigenvectors corresponding to eigenvalue 1.

    For example, if the plane is x+2y+ 3z= 0, then a vector perpendicular to that is <1, 2, 3> which has length sqrt(1+ 4+ 9)= sqrt(14) so <1/sqrt(14), 2/sqrt(14), 3/sqrt(14)> is a unit vector in that direction. Any muliple of that is an eigenvector corresponding to eigenvalue -1.

    Any vector, < x, y, z> in the plane must satisfy x+ 2y+ 3z= 0 or x= -2y -3z so that <x, y, z>= <-2y- 3z, y, z>= <-2y, y, 0>+ <-3z, 0, z>= y<-2, 1, 0>+ z<-3, 0, 1> so that <-2, 1, 0> and <-3, 0, 1> span the space. Those vectors are mapped into themselves by the reflection and so are eigenvectors corresponding to eigenvalue 1. If you are not worried about "orthonormal", {<1,2,3>, <-2,1,0>, <-3,0,1>} are already as basis for the vector space consisting of eigenvectors. If you want "orthonormal" use <1/sqrt(14), 2/sqrt(14), 3/\sqrt(14)> and use "Gram-Schmidt" to "orthonrmalize" the other two vectors. (Obviously <1,2,3> and any multiple of it are already orthogonal to <-2,1,0> and <-3,0,1> and so to any linear combination of them.)
    Last edited by HallsofIvy; April 20th 2011 at 03:19 PM.
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    MHF Contributor FernandoRevilla's Avatar
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    Also we can directly use a well known theorem:

    If H : a_1 x_1 + ... + a_n x_n = 0 is a hyperplane of of IR^n then, the projection matrix over H is


    P = I - ( N^t N ) ( N N^t ) ( being N = ( a_1 , ... , a_n )^t )

    As a consequence the matrix of the reflection over H is

    R = 2 P - I = I - 2 ( N^t N ) ( N N^t )


    Edited: Of course my comments are only related to find the matrix of the reflection.
    Last edited by FernandoRevilla; April 21st 2011 at 02:42 AM.
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