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Question on test, Eigenvalues

  1. Jul 2, 2008 #1
    I wasn't sure about this question on my exam. Let eigenvalues be 0, 1, -1

    Is it true that

    a. Not Invertible
    b. Diagonalizable
    c. Orthogonal

    a. True, since the determinant is 0
    b. I'm not sure, but I chose False b/c I had an eigenvalue of 0
    c. True, ah not sure either

    LOL!
     
  2. jcsd
  3. Jul 2, 2008 #2

    Dick

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    Right, it's not invertible. For c) orthogonal matrices are invertible, right? Look at the definition. b) would be true if the eigenvectors were complete and orthogonal. From the given information, I would have to answer "Can't say".
     
    Last edited: Jul 2, 2008
  4. Jul 2, 2008 #3
    Great, 13/15. Already a B.
     
  5. Jul 2, 2008 #4
    I had another weird question.

    It asked me for the dimension of the space that it occupied. I'm not sure how to answer that, usually it's dimension of the Column, Row, ... etc.

    I had a 5 by 6 matrix and found its rank to be 3.
     
  6. Jul 2, 2008 #5

    Dick

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    I don't know. What's the technical definition of 'space that it occupied'?? It represents a map from R^6 to R^5 and if it's rank is three then the range is a 3 dimensional subspace of R^5 and the kernel is a 3 dimensional subspace of R^6. Which one of those dimensions does it 'occupy'?
     
  7. Jul 2, 2008 #6
    I'm screwed. I think I will end up with a B :(
     
  8. Jul 2, 2008 #7

    matt grime

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    What has eigeinvalues 0,1,-1? A 3x3 matrix? A 4x4 matrix?


    N.B. Dick's comment about orthogonality of e-vectors is not relevant.
     
  9. Jul 2, 2008 #8

    Dick

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    Yeah, you're right. They just need to be complete. If it's 3x3 then that's true. I wrote exactly that first, then had a panic attack. Orthogonality is unnecessary. Thanks!
     
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