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Thread: 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.


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

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

    Let $\displaystyle S=C-BA^{-1}B^{t}$ 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.


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

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

    Let $\displaystyle S=C-BA^{-1}B^{t}$ 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 $\displaystyle \langle x,y\rangle$ denotes the inner product in $\displaystyle \mathbb{R}^p$, and $\displaystyle A$ is a $\displaystyle p\times p$ matrix, then A is positive definite if (and only if) $\displaystyle \langle Ax,x\rangle\geqslant0$ for all x in $\displaystyle \mathbb{R}^p$.

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