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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=AH^T(HAH^T)^{-1}H\ \text{and} \ Q=K(K^TAK)^{-1}K^TA. \end{equation}

Prove that $P+Q$ is the $n\times n$ identity matrix.

I have proved $(P+Q)K=K$ and $H(P+Q)=H$. However, I haven't found the proof completely. Can anyone help me? Any help will be appreciated. Thanks

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  • $\begingroup$ Yes, I did typo. Matrix $A$ should be a non singular. It has been corrected now. Thanks $\endgroup$ – Jlamprong Nov 29 '13 at 22:41
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I think this is a counterexample: $$ A=\begin{bmatrix}1&3\\3&1\end{bmatrix},\ \ H=\begin{bmatrix}1&0\end{bmatrix},\ \ K=\begin{bmatrix}0\\1\end{bmatrix}. $$ Then $$ P=AH^T(HAH^T)^{-1}H=\begin{bmatrix}1&3\\3&1\end{bmatrix}\,\begin{bmatrix}1\\0\end{bmatrix}\left(\begin{bmatrix}1&0\end{bmatrix}\begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix} \right)^{-1}\begin{bmatrix}1&0\end{bmatrix}\\ =\begin{bmatrix}1&3\\3&1\end{bmatrix}\,\begin{bmatrix}1\\0\end{bmatrix}\begin{bmatrix}1&0\end{bmatrix}=\begin{bmatrix}1&0\\3&0\end{bmatrix} $$ and $$ Q=K(K^TAK)^{-1}K^TA=\begin{bmatrix}0\\1\end{bmatrix}\left( \begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}0\\1\end{bmatrix}\right)^{-1}\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}1&3\\3&1\end{bmatrix}\\ =\begin{bmatrix}0\\1\end{bmatrix}\,\begin{bmatrix}0&1\end{bmatrix}\begin{bmatrix}1&3\\3&1\end{bmatrix}=\begin{bmatrix}0&0\\3&1\end{bmatrix}=\begin{bmatrix}0&0\\3&1\end{bmatrix} $$ So $P+Q\ne I$ (note that it is still true that $(P+Q)K=K$, $H(P+Q)=H$).

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  • $\begingroup$ Thanks Martin. However, I think, the matrix $B$ that you defined is always singular since it will not have full rank. So, we don't have its inverse. $\endgroup$ – Jlamprong Nov 29 '13 at 23:10
  • $\begingroup$ You are right. Let me think about it. $\endgroup$ – Martin Argerami Nov 29 '13 at 23:27
  • $\begingroup$ I have edited to include what I think is a counterexample. $\endgroup$ – Martin Argerami Nov 29 '13 at 23:49
  • $\begingroup$ Thanks Martin. I am sorry, I forgot on the simmetricity of matrix $A$. $\endgroup$ – Jlamprong Nov 30 '13 at 3:15
  • $\begingroup$ It doesn't make a difference. I have changed $A$ to be symmetric. $\endgroup$ – Martin Argerami Nov 30 '13 at 3:17
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Note that $P^2=P$ and $Q^2=Q$. If $P+Q=I$, we must have $(P+Q)^2=P^2+Q^2$ and in turn $QP=0$. In other words, if $QP\ne0$, we obtain a counterexample. Now, $$ QP = \color{red}{K}\ (K^\top AK)^{-1}\ (K^\top AAH^\top)\ (HAH^\top)^{-1}\ \color{red}{H}. $$ If we put $n=2$ and $m=n-m=1$, then $\color{red}{K}$ is a column 2-vector, $\color{red}{H}$ is a row 2-vector and $(K^\top AK),\,(HAH^\top),\,(K^\top AAH^\top)$ are three scalars. So, if $\color{red}{K,H}$ and the three scalars are all nonzero, then $QP$ is necessarily nonzero and $P+Q\ne I$.

Hence we can construct a counterexample easily as follows. Let $$ H=(1,0),\ K=\pmatrix{0\\ 1},\ A=\pmatrix{2&1\\ 1&2},\ A^2=\pmatrix{5&4\\ 4&5}. $$ Then $HK=0$, $K,H$ are nonzero, $K^\top AK=HAH^\top=2$ and $K^\top AAH^\top=4$. One may verify that $QP=\pmatrix{0&0\\ 1&0}$ and $P+Q=\pmatrix{1&0\\ 1&1}$.

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  • $\begingroup$ Thank you very much for your help. However, I am sorry, I forgot on the simmetricity of matrix $A$. I have edited now. $\endgroup$ – Jlamprong Nov 30 '13 at 3:16
  • $\begingroup$ @Jlamprong See my new edit. Whether $A$ is symmetric or not is actually irrelevant. $\endgroup$ – user1551 Nov 30 '13 at 3:34
  • $\begingroup$ OK, thanks. I'll correct my formula... $\endgroup$ – Jlamprong Nov 30 '13 at 11:38

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