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Math Help - "Maximally diagonal" representation for a symmetric matrix

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    "Maximally diagonal" representation for a symmetric matrix

    In a given representation for a real symmetric positive-definite NxN matrix C, express it in the form

    C = S + K,

    where S is a diagonal positive-definite matrix of rank N, and K is a semipositive-definite matrix of the _smallest_ possible rank M < N. Do the solutions for S and K exist? If yes, are they unique? If yes, outline an algorithm to find S and K.
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    Quote Originally Posted by pmn1 View Post
    In a given representation for a real symmetric positive-definite NxN matrix C, express it in the form

    C = S + K,

    where S is a diagonal positive-definite matrix of rank N, and K is a semipositive-definite matrix of the _smallest_ possible rank M < N. Do the solutions for S and K exist? If yes, are they unique? If yes, outline an algorithm to find S and K.
    The answer certainly won't be unique. Take a very simple numerical example. If C = \begin{bmatrix}5&2\\2&3\end{bmatrix} then \begin{bmatrix}5&2\\2&3\end{bmatrix} = \begin{bmatrix}1&0\\0&2\end{bmatrix} + \begin{bmatrix}4&2\\2&1\end{bmatrix} = \begin{bmatrix}3&0\\0&1\end{bmatrix} + \begin{bmatrix}2&2\\2&2\end{bmatrix}.

    In fact, for any x in the interval 4/3 < x < 5 we can write \begin{bmatrix}5&2\\2&3\end{bmatrix} = \begin{bmatrix}5-x&0\\0&3-\tfrac4x\end{bmatrix} + \begin{bmatrix}x&2\\2&\tfrac4x\end{bmatrix}, with the last matrix having rank 1.

    In fact, I suspect that the only case where the answer is unique will be when C is a multiple of the identity. In that case, the unique solution is C = C + 0. For any other positive definite matrix C, we can write C = \lambda I + (C-\lambda I), where \lambda is the smallest eigenvalue of C. Then \lambda I is positive definite, and C-\lambda I is positive semi-definite. Of course, it may not have minimal rank. But even if it does, then (as in the numerical example above) I wouldn't expect that to be the unique solution.
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