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Linear algebra: determinants and eigenvalues

  1. May 26, 2005 #1
    i'm reading and doing some work in introduction to linear algebra fifth edition, and i came across some problems that i had no clue.

    1. An (n x n) matrix A is a skew symmetric (A(transposed) = -A). Argue that an (n x n) skew-symmetrix matrix is singular when n is an odd integer.

    2. Prove that if A is nonsingular, then 1/(eigenvalue symbol) is an eigenvalue of A^-1

    can someone explain some of these for me. thank you
     
  2. jcsd
  3. May 26, 2005 #2

    matt grime

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    The second follows from considering the characteristic polynomial (as does the first now I imagine).
     
  4. May 26, 2005 #3
    can you please explain it in more details?
     
  5. May 26, 2005 #4

    Galileo

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    For the first, you can show that det(A) is zero. Just use what you know about determinants and A.
     
  6. May 26, 2005 #5

    matt grime

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    Yep, that seems better.

    But the second is the char poly. e-values are the roots of... relate the char poly of A to that of A^{-1}
     
  7. May 26, 2005 #6
    i dont understand why det(A) has to be zero for the skew symmetric matrix. i know that if n is odd then det(-A)=-det(A) but i dont get why it has to equal zero. and why cant it be zero if n is even??
     
    Last edited: May 26, 2005
  8. May 26, 2005 #7

    Galileo

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    You're almost there. You haven't used the fact that A is skew symmetric yet.
     
  9. May 26, 2005 #8
    Eh? Nothing written in this thread implies that a skew-symmetric matrix of even order is always invertible (consider the zero matrix...).
     
  10. May 26, 2005 #9
    arg.. this is confusing. so since its skew-symmetric matrix det(-A)=-det(A)=det(A(transposed)). but why must it be zero :(
     
  11. May 26, 2005 #10
    Forget about skew-symmetric matrices for a while. In general, how are det(A') and det(A) related (A is any square matrix, A' the transpose of A)?
     
  12. May 26, 2005 #11
    det(A') and det(A) are equal. so that means the two determinants are zero?? man i suck
     
  13. May 26, 2005 #12
    Right:

    det(A') = det(A).

    But as already noted, for a skew-symmetric matrix of odd order, we also have that

    det(A') = -det(A).

    Hence...?
     
  14. May 26, 2005 #13

    matt grime

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    Hopefully that's that cleared up for that one.

    How about the other: if Av=tv (where t is the eigenvalue and v is the eigenvector) then v=A^{-1}tv=tA^{-1}v

    can you complete that from there?
     
  15. May 26, 2005 #14
    i know that det(A') = det(A) = -det(A). i think i also mentioned this in the earlier post. But i still dont get why all the determinants are zero.

    oh yeah. i figured out number 2. thanks
     
  16. May 26, 2005 #15

    matt grime

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    Oh let D bet Det(A), you know D, a number, is equal to minus D. How many real (or complex) numbers satisfy D=-D? Or 2D=0?
     
  17. May 26, 2005 #16
    ah ha! that makes more sense. and thats so simple too. thanks alot!!
     
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