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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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Notice that $HK=0$ gives that $PQ=QP=0$. Also, $P,Q$ are projections. So, we get that $P$ and $Q$ are orthogonal to each other. Further, rank of the matrix $P$ = rank $H$ = m. Similarly, rank $Q$= rank $K$= n-m. Now you should be able to conclude that $P+Q$ has rank n (proof ), and hence is invertible. Note that $(P+Q)^2=P+Q$. Together with the invertibility, it implies that $P+Q=I.$

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  • $\begingroup$ Thanks, but this is my problem. It is quite difficult to prove $P+Q$ is full rank $\endgroup$
    – Jlamprong
    Commented Jan 9, 2014 at 16:33
  • $\begingroup$ I have added a link for the proof of the statement. $\endgroup$
    – voldemort
    Commented Jan 9, 2014 at 17:14
  • $\begingroup$ Thanks @voldemort. But, the given link is different with my case. In there, we must have $PQ^*=Q^*P=0$ $\endgroup$
    – Jlamprong
    Commented Jan 13, 2014 at 12:23
  • $\begingroup$ $Q^{*}=Q^{t}=Q$ in your case-right? $\endgroup$
    – voldemort
    Commented Jan 13, 2014 at 16:07

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