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Math Help - singular values

  1. #1
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    singular values

    Suppose T ∈ L(V). Prove that T is invertible if and only if 0 is
    not a singular value of T.

    I kinda have a love-hate feeling about proofs like these. I mean I have
    so many ideas, but I don't know which one to use.
    I have three routes for the reverse direction using a converse.
    All three routes will use an orthonormal basis under sqrt(T*T)
    since sqrt(T*T) is positive.

    1. Show that there is at least one eigenvector in null sqrt(T*T)'s basis
    and considering that dim rangeT=dim range sqrt(T*T), and both
    dim range T and dim range sqrt(T*T) is <dim range V, range T is not
    invertible.
    2. Show that if 0 is a value on sqrt(T*T)'s diagonal, 0 must also be a
    value on T's diagonal.

    3. State that since 0 is on sqrt(T*T)'s diagonal, and
    any value on the isometry S's diagonal (when multiplied by 0) will
    give 0, S(sqrt(T*T)) will map ej to a 0 vector and since Ssqrt(T*T)=T,
    ej must be a vector in nullT (or null T must have at least one eigenvector in
    its basis.
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  2. #2
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    Opalg's Avatar
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    I'm not sure how this would fit in with your approach to singular values. My definition is that they are the diagonal entries of the matrix D in the decomposition T=U^*DV, where U and V are unitary matrices (and the transformation T is identified with its matrix with respect to some basis). In that case, all you have to do to prove the result is to take determinants, because U and V have nonzero determinants, so \det(T)=0\ \Leftrightarrow\ \det(D)=0, and this holds if and only if one of the diagonal elements of D is zero.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    I'm not sure how this would fit in with your approach to singular values. My definition is that they are the diagonal entries of the matrix D in the decomposition T=U^*DV, where U and V are unitary matrices (and the transformation T is identified with its matrix with respect to some basis). In that case, all you have to do to prove the result is to take determinants, because U and V have nonzero determinants, so \det(T)=0\ \Leftrightarrow\ \det(D)=0, and this holds if and only if one of the diagonal elements of D is zero.
    Oh hate to argue, but my book doesn't introduce determinants until later
    on. It says that all singular values are eigenvalues of sqrt(T*T).


    wait I figured something out...
    Assume 0 is an eigenvalue of sqrt(T*T). Now we know that
    llTvll=llS(sqrt(T*T))vll=llsqrt(T*T)vll. Now llTvll=llsqrt(T*T)vll.
    If v is an eigenvector of 0, then llsqrt(T*T)vll=ll0*vll=0.
    This implies that llTvll=0. But v is an eigenvector=/=0.
    Therefore llTvll=0 if T maps the nonzero v to 0. Therefore T is not invertible.

    Not saying your method is wrong, just saying that I'm not familiar with
    determinants and that my book has taught me about singular values
    from a different perspective.
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