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Thread: Symmetric matrix proof

  1. #1
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    Symmetric matrix proof

    Donīt know if the descriptive title is misleading but here goes.

    Show that for every m/nmatrix A with a range r there exists a matrix AA^T which is symmetric and has a range r.

    This is what I have:

    The matrix AA^T is a square m/m matrix and also symmetric because (AA^T)^T=A^{TT}A^T=AA^T.

    The part with range is still confusing. Now this should be true

    dimR(A)=dimR(A^T)=r

    but how should I pursue dimR(AA^T)=r? Im thinking about Orthogonal matrices. Because if AA^T is symmetric which it is, then there exists an orthogonal matrix Q such that Q^TAA^TQ=D where D is a diagonal matrix with the eigenvalues of AA^T in the diagonal.

    any help appreciated
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  2. #2
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    Quote Originally Posted by wilbursmith View Post
    Donīt know if the descriptive title is misleading but here goes.

    Show that for every m/nmatrix A with a range r there exists a matrix AA^T which is symmetric and has a range r.

    This is what I have:

    The matrix AA^T is a square m/m matrix and also symmetric because (AA^T)^T=A^{TT}A^T=AA^T.

    The part with range is still confusing. Now this should be true

    dimR(A)=dimR(A^T)=r

    but how should I pursue dimR(AA^T)=r?
    Use the rank + nullity theorem. If A^{\textsc{t}}x = 0 then clearly AA^{\textsc{t}}x = 0. Conversely, if AA^{\textsc{t}}x = 0 then 0 = x^{\textsc{t}}AA^{\textsc{t}}x = (A^{\textsc{t}}x)^{\textsc{t}}(A^{\textsc{t}}x), which implies that A^{\textsc{t}}x=0. So AA^{\textsc{t}} and A^{\textsc{t}} have the same kernel, hence the same nullity, hence the same rank.
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