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


1 Answer 1


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 $P+Q=I.$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.