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Math Help - Schur Complement problem

  1. #1
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    Schur Complement problem

    Let M be a real symmetric and positive definite matrix with blocks A, Bt, B and C.


     M= [[A,B^{t}] ; [B,C]]

    where  A is a  p\times p matrix;  B is  q\times p; and  C is  q\times q.

    Let  S=C-BA^{-1}B^{t} be the Schur complement. We prove that S is symmetric positive definite.


    I can prove that S is symmetric but I am having trouble proving that it is positive definite.
    I know that for S a symmetric matrix, S positive definite is equivalent to say that all eigen values of S are positive.

    I guess my question is how do we prove that all eigen values of S are positive?
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  2. #2
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    Quote Originally Posted by math8 View Post
    Let M be a real symmetric and positive definite matrix with blocks A, Bt, B and C.


     M= [[A,B^{t}] ; [B,C]]

    where  A is a  p\times p matrix;  B is  q\times p; and  C is  q\times q.

    Let  S=C-BA^{-1}B^{t} be the Schur complement. We prove that S is symmetric positive definite.


    I can prove that S is symmetric but I am having trouble proving that it is positive definite.

    I know that for S a symmetric matrix, S positive definite is equivalent to say that all eigen values of S are positive.

    I guess my question is how do we prove that all eigen values of S are positive?
    I think it's easier here to describe positive definiteness in terms of inner products rather than eigenvalues. If \langle x,y\rangle denotes the inner product in \mathbb{R}^p, and  A is a  p\times p matrix, then A is positive definite if (and only if) \langle Ax,x\rangle\geqslant0 for all x in \mathbb{R}^p.

    The condition for M to be positive definite is thus \textstyle \left\langle \bigl[ \begin{smallmatrix} A&B^{\textsc t}\\B&C \end{smallmatrix}\bigr]\bigl[{x\atop y}\bigr],\bigl[{x\atop y}\bigr]\right\rangle = \langle Ax,x\rangle + \langle B^{\textsc t}y,x\rangle + \langle Bx,y\rangle + \langle Cy,y\rangle \geqslant0 (for all x in \mathbb{R}^p and all y in \mathbb{R}^q). Now take x = -A^{-1}B^{\textsc t}y in that inequality and simplify the result. (You will need the property of an inner product, that \langle Bx,y\rangle = \langle x,B^{\textsc t}y\rangle.) You should end up with the inequality \langle Sy,y\rangle \geqslant0, which shows that S is positive definite.
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  3. #3
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    Schur Complement problem

    Thanks a lot, this was very helpful.

    When I used that specific value of x, it was very straightforward to get to

    I actually didn't need to use the property that
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