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We consider a symmetric, nonsingular matrix: \begin{equation*} H=\left[\begin{array}{cc} 0 & B\\ B^T & A \end{array}\right] \end{equation*} Where $B$ has dimensions $m \times n$ with $m \geq n$. Give a backward stable finite (actually, it means we shouldn't compute the eigen value or singular value of $H$) algorithm computinf the factorization: $H=QMQ^T$. With $Q$ orthognal and $M$ has form: \begin{equation*} \left[\begin{array}{ccc} 0 & 0 & Y^T \\ 0 & X & Z^T \\ Y & Z & W \\ \end{array}\right] \end{equation*} Where $X$ is symmetric positive definite, $W$ is symmetric, and $Y$ is square and lower antitriangular.

I have been thinking about this problem for a long time but with no idea. I know that probably we need to use backward stable methods (like Householder etc.) to make QR decomposition to $B$. But I have no idea that how can we generate a positive definite part $X$ from the symmetric matrix $A$. Can anyone help me ?

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    How can we have $m\ge n$ and still $H$ is non-singular?? – 2017-02-19
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    I mean, if $H$ is non-singular then $m$ must be _equal_ to $n$ – 2017-02-19
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    Does $Y$ lower antitriangular (for a $p \times p$ matrix) means $i+j – 2017-02-19
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    Yes, that's correct. – 2017-02-19

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Consider an example: $$ H = \left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & -1 & 0 \\ 1 & 0 & 1 \end{array} \right) $$ Since you cannot change the determinant by orthogonal transformation, it is not possible to get positive H(2,2) preserving the considered block structure. So, I guess, you have missed the p.s.d condition on $H$ probably.