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Finding the Eigenstuff of a Orthogonal Projection onto a plane

  1. Aug 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Let S be the subspace of R3 defined by x1 - x2 + x3 = 0. If L: R3 -> R3 is an orthogonal projection onto S, what are the eigenvalues and eigenspaces of L?


    2. Relevant equations



    3. The attempt at a solution
    First off, I hadn't seen the term eigenspace before. From looking it up, it seems like it is the set of all eigenvectors with the same eigenvalue together with the zero vector (according to wikipedia). Well certainly any vector in the plane defined by x1 - x2 + x3 = 0 will be projected to itself and thus has an eigenvalue of 1. So I would want to say that the eigenspace is simply that plane. It seems like no other eigenvectors should exist and no other eigenvalues. Is it adequate to say that the only vectors that won't change direction under a projection are those already in the space being projected on and those will have an eigenvalue of 1 since they are unchanged?

    Additionally, I had not seen an orthogonal project previously. Looking it up I came under the impression that this means the corresponding matrix is symmetric or hermitian. But I thought this would mean the L must be square and by the spectral theorem L can be written as QDQT where D is the diagonal matrix with the respective eigenvectors. But from above it appears it has infinitely many eigenvectors. So now I am just rather confused.

    This is from a past graduate entrance placement exam
     
  2. jcsd
  3. Aug 13, 2012 #2

    vela

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    Consider vectors perpendicular to the plane. Are they eigenvectors?
     
  4. Aug 13, 2012 #3

    Bacle2

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    Have you tried writing L with respect to some basis, i.e., as a matrix?

    If L is diagonalizable, then one can find a basis for ℝ3 consisting of

    eigenvectors of L . So you need at least 3 L.I eigenvectors associated to L.
     
  5. Aug 13, 2012 #4
    Vela: I though any vector perpendicular which I thought is just the normal vector (1, -1, 1) would be mapped to the 0 vector and then not a eigenvector. Is that wrong?

    Bacle2: My initial thought was to find L directly by seeing how the standard basis would be projected. I don't know the name of the corresponding theorem but that is the only approach I know for finding the transformation matrix. But then the only way I know how to do that is using vector calculus and doing each one out long handed. (Is there a quicker approach?). Moreover, since the question only asked for the eigenspace and eigenvalue, then it seemed to me like the answer is simply the plane. Is that incorrect?

    I also thought that I could just choose two orthogonal vectors in the plane and this would be a basis for the eigenspace ie (1,1,0) and (1, -1, -2) but again then it falls short of the 3 needed to make it square.

    Thanks so much!
     
  6. Aug 13, 2012 #5

    vela

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    It's an eigenvector with eigenvalue 0.
     
  7. Aug 13, 2012 #6
    That just blew my mind. So does that mean that any vector that is in the nullspace is an eigenvector with an eigenvalue of 0?
     
  8. Aug 13, 2012 #7

    vela

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    Yup, except for x=0 of course.
     
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