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Math Help - Prove the following for real n x n matirices:

  1. #1
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    Post Prove the following for real n x n matirices:

    i) If A is an orthogonal matrix whose eigenvalues are all different from 1 , then I +A is non singular and S = (I A)(I + A)-1 is skew symmetric.
    ii)If B is a skew-symmetric matrix ,then C = (I B)( I B)-1 is an orthogonal matrix with no eigenvalue equal to 1.


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  2. #2
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    Quote Originally Posted by puneet View Post
    i) If A is an orthogonal matrix whose eigenvalues are all different from –1 , then I +A is non singular and S = (I – A)(I + A)-1 is skew symmetric.
    if (I+A)v=0, for some non-zero vector v, then Av=-v, which means -1 is an eigenvalue of A. contradiction! so I+A is non-singular. to show that S is skew symmetric:

    S(I+A)=I-A \Longrightarrow (I+A^{T})S^{T}=I-A^{T} \Longrightarrow (A+AA^{T})S^{T}=A-AA^{T} \Longrightarrow

    (A+I)S^{T}=A-I \Longrightarrow S^{T}=-(I+A)^{-1}(I-A)=-(I-A)(I+A)^{-1}=-S. \ \ \ \square

    ii)If B is a skew-symmetric matrix ,then C = (I – B)( I + B)-1 is an orthogonal matrix with no eigenvalue equal to –1.
    C(I+B)=I-B \Longrightarrow (I+B^{T})C^{T}=I-B^{T} \Longrightarrow (I-B)C^{T}=I+B. thus: (I+B)^{-1}(I-B)C^{T}=I. but: (I+B)^{-1}(I-B)=(I-B)(I+B)^{-1}=C. thus: CC^{T}=I, i.e. C is

    orthogonal. to show that C has no eigenvalue equal to -1 let D=\frac{I+B}{2}. then: (I+C)D=\frac{I+B+I-B}{2}=I. hence I+C is non-singular and so C has no eigenvalue equal to -1.
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