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    Gramian matrix

    Could anyone give me some help on this one problem?

    An n by n matrix A is called Gramian matrix if there exists a real (square) matrix B such that A=tranpose of B times B. Prove that A is Gramian matrix if and only if A is symmetric and all of its eigenvalues are non-negative.
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by namelessguy View Post
    Could anyone give me some help on this one problem?

    An n by n matrix A is called Gramian matrix if there exists a real (square) matrix B such that A=tranpose of B times B. Prove that A is Gramian matrix if and only if A is symmetric and all of its eigenvalues are non-negative.
    ( \implies )
    suppose A_{nxn} is a Gramian Matrix.
    Then there exists a matrix B_{nxn} such that A = B^T B

    Let B be the matrix such that A = B^T B
    A^T = (B^T B)^T = B^T B = A *
    hence, A is symmetric. (to show that all of its eigenvalues are non-negative, take an arbitrary symmetric matrix A = B^T B and show that its eigenvalues are non-negative.)

    * have you proven that (AB)^T = B^T A^T?
    Last edited by kalagota; November 15th 2007 at 07:18 AM.
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    Quote Originally Posted by kalagota View Post
    \( \implies \)
    suppose A_{nxn} is a Gramian Matrix.
    Then there exists a matrix B_{nxn} such that A = B^T B

    Let B be the matrix such that A = B^T B
    A^T = (B^T B)^T = B^T B = A *
    hence, A is symmetric. (to show that all of its eigenvalues are non-negative, take an arbitrary symmetric matrix A = B^T B and show that its eigenvalues are non-negative.)

    * have you proven that (AB)^T = B^T A^T?
    Thanks for your help. Yes, we have prove that (AB)^t = B^t A^t . I'm sorry that I'm kinda slow why if A^t = A * then A is symmetric? I thought I had to prove A^t = A
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by namelessguy View Post
    Thanks for your help. Yes, we have prove that (AB)^t = B^t A^t . I'm sorry that I'm kinda slow why if A^t = A * then A is symmetric? I thought I had to prove A^t = A
    oww, what was your definition of a symmetric matrix?

    here it is.. A is symmetric if and only if  a_{ij} = a_{ji} for all i,j..
    and you will see that using that definition, and form the transpose of A, that transpose is same as A..

    besides, the transpose of A, for A=[a_{ij}] is  A^T = [b_{ij}] where  b_{ij} = a_{ji}
    Last edited by kalagota; November 15th 2007 at 07:31 AM.
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