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Simple quadratic optimization problem

  1. Feb 27, 2009 #1
    Let [tex]P^ + ,P^ - ,I,Q \in R^{n\times n}, K\in R^{n\times 1}, M\in R^{1 \times n}[/tex], and assume that [tex]Q[/tex] is positive definite, [tex]P^ -[/tex] is positive semidefinite whence [tex](MP^ - M^T + Q)^{ - 1}[/tex] exists (where [tex]T[/tex] denotes transpose).

    In what sense does [tex]K = P^ - M^T(MP^ - M^T + Q)^{ - 1}[/tex] minimize the quadratic expression [tex]P^ + : = (I - KM)P^ - (I - KM)^T + KQK^T[/tex], over [tex]K[/tex]?
    Is this minimization of [tex]P^ +[/tex] over all vectors [tex]K[/tex] with respect to the usual ordering for positive semidefinite matrices [tex]A\le B[/tex] iff [tex]B - A [/tex]is positive semidefinite?

    Next consider the extension [tex]P^ + : = (I - KAM)P^ - (I - KAM)^T + KAQA^TK^T[/tex], where [tex]A\in R^{n\times 1}, K\in R^{n\times n}[/tex] and [tex]K[/tex] diagonal, where all other matrices are as above.
    What is the minimum over [tex]K[/tex] (with respect to the previous ordering or something )??

    Any help will be deeply appreciated.
  2. jcsd
  3. Feb 28, 2009 #2
    I think maybe there is a typo, because if K is a column vector (in Rnx1) and Q a square matrix then KQK* is meaningless, you can't do that matrix multiplication. However, if you meant that K is a row vector, then that implies KQK* is a scalar, so it cannot possibly add to anything to equal the nxn matrix P+.
  4. Feb 28, 2009 #3
    That is correct [tex]Q > 0[/tex], a scalar.

    Please help with this.
  5. Feb 28, 2009 #4
    What are your thoughts on the problem so far?
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