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Given $m<n$. Suppose that $H$ and $K$ be $m \times n$ and $n\times (n-m)$ matrices such that rank$(H)=m$, rank$(K)=n-m$, and $HK=0$. For fixed non singular symmetric matrix $A$ define \begin{equation} P=A^{-1}H^T(HA^{-1}H^T)^{-1}H\ \text{and} \ Q=K(K^TAK)^{-1}K^TA. \end{equation}

I'd like to have $P+Q$ is the $n\times n$ identity matrix.

It can be verified that $[H^T,A^{-1}K](P+Q)=[H^T,A^{-1}K].$ So, if $[H^T,A^{-1}K]$ is invertible then $P+Q=I$. However, it is quite difficult the invertibility of $[H^T,A^{-1}K]$. Therefore, I wonder $P+Q=I$ not need to be true. Can anyone help me? Any help will be appreciated. Thanks

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  • $\begingroup$ By the way, is the matrix $A$ by chance also positive definite? If so, is it necessary to work with the orthogonal complement of the nullspace of $H$ with respect to the Euclidean inner product (spanned by the columns of $K$) and consider instead the $A$-orthogonal complement? In that case, you don't need to bother with the (non-)singularity of $[H^T,A^{-1}K]$. $\endgroup$ – Algebraic Pavel Jan 9 '14 at 15:56
  • $\begingroup$ Thanks @AlgebraicPavel. Yes, $A$ is positive semidefinite. I am sorry, I don't understand with your second question? Could you make me clearly please? $\endgroup$ – Jlamprong Jan 9 '14 at 16:09
  • $\begingroup$ First, if it is nonsingular and semidefinite, then it is for sure definite. Second, you seem to work with orthogonal projectors in generalised inner products (induced by $A$ and $A^{-1}$), while for the space spanned by $K$ you consider the Euclidean inner product. My question was, if it is really necessary. But without more context (what is given, what is to be obtained) it is hard to guess. $\endgroup$ – Algebraic Pavel Jan 9 '14 at 16:17
  • $\begingroup$ Yes, it is necessary. Actually, what given is only $H$ and $A$. I defined $K$ by a matrix such that $HK=0$. Therefore, I'd like to express $Q$ in term of $H$. Do I understand correctly your question? $\endgroup$ – Jlamprong Jan 9 '14 at 16:24
  • $\begingroup$ Could you then instead of $K$ such that $HK=0$ consider $K$ such that $HAK=0$? Then, given $Q$, $P$ such that $P+Q=I$ would be simply $P=H^T(HAH^T)^{-1}HA$ (if I've put the formula correctly). $\endgroup$ – Algebraic Pavel Jan 9 '14 at 17:16

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