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Math Help - Tricky matrix question

  1. #1
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    Tricky matrix question

    For a n*n skew-symmetric matrix K (K transpose = -K), show that K+I (identity) is invertible



    And show that (K-I)(K+I)^-1 is orthogonal (is it's own inverse) and has determinant 1
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  2. #2
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    Quote Originally Posted by steve_young View Post
    For a n*n skew-symmetric matrix K (K transpose = -K), show that K+I (identity) is invertible.
    first of all you should've mentioned that the entries of K come from real numbers. to prove that K+I is invertible, we need to prove that if K \bold{x} = -\bold{x}, for some vector \bold{x} \in \mathbb{R}^n, then \bold{x}=\bold{0}.

    this is easy to prove: -||\bold{x}||^2=K \bold{x} \cdot \bold{x}=\bold{x} \cdot K^T \bold{x} = - \bold{x} \cdot K \bold{x}=\bold{x} \cdot \bold{x}=|| \bold{x}||^2. thus || \bold{x} ||^2 = 0 and hence \bold{x} = \bold{0}. similarly we see that K-I is invertible too.


    And show that (K-I)(K+I)^-1 is orthogonal (is it's own inverse)
    let A=(K-I)(K+I)^{-1}. we must show that A^T=A^{-1}. well, we have K-I=A(K+I) and thus -K-I=(K-I)^T=(K+I)^TA^T=(-K+I)A^T. hence A^T=(K-I)^{-1}(K+I).

    therefore A^{-1}=(K+I)(K-I)^{-1}=(K-I)^{-1}(K+I)=A^T.


    and has determinant 1

    this is false! for example if K=0 is a 3 \times 3 matrix, then A=-I and thus \det A = -1. in general we can only say that \det A = \pm 1, which is obvious because A is orthogonal.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    and has determinant 1
    this is false! for example if K=0 is a 3 \times 3 matrix, then A=-I and thus \det A = -1. in general we can only say that \det A = \pm 1, which is obvious because A is orthogonal.
    In fact, we can say a bit more than that. The determinant of a matrix is equal to the determinant of the transpose. So \det(K+I) = \det(K^{\textsc t}+I) = \det(-K+I) = (-1)^n\det(K-I), from which \det\bigl((K-I)(K+I)^{-1}\bigr) = \det(K-I)\bigl(\det(K+I)\bigr)^{-1} = (-1)^n.

    Alternatively, you could reword the question to say that \det\bigl((I-K)(I+K)^{-1}\bigr) = 1.
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