2
$\begingroup$

Let $Ax = y$ be a linear system with $A \in \Bbb R^{n,m}, n<m$. Consider $H_i := \{x \in \Bbb R^{m}: A_ix = y_i\}, i \in \{1,\dots,n\}, y_i \in \Bbb R$ where $A_i$ denotes the $i$-th row of $A$. Now, regard $\mathcal{P}_{H_i}z := z - {A_i}^{\intercal} {y_i - A_i z \over \|A_i\|^2} $. This is apparently an orthogonal projection onto the hyperplane $H_i$ but I fail to show it. Any help is appreciated.

$\endgroup$
1
$\begingroup$

$Px$ is the orthogonal projection onto $H$ if and only if $Px\in H$ and $\langle x-Px,z-Px\rangle = 0$ for all $z\in H$.

If $Px = x + \frac{y_i-A_ix}{\|A_i\|^2}A_i^T$ (there must be a plus!), then $A_iPx = A_ix + \frac{y_i-A_ix}{\|A_i\|^2}A_iA_i^T = A_ix + (y_i-A_ix) = y_i$, hence $Px\in H_i$. Also, if $z\in H_i$, \begin{align} \langle x-Px,y-Px\rangle &= -\frac{y_i-A_ix}{\|A_i\|^2}\left\langle A_i^T,z-x-\frac{y_i-A_ix}{\|A_i\|^2}A_i^T\right\rangle\\ &= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - A_ix - \frac{y_i-A_ix}{\|A_i\|^2}A_iA_i^T\right)\\ &= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - A_ix - y_i + A_ix\right)\\ &= -\frac{y_i-A_ix}{\|A_i\|^2}\left(A_iz - y_i\right) = 0, \end{align} since $A_iz=y_i$. So, $P$ is indeed the orthogonal projection onto $H_i$.

$\endgroup$
  • $\begingroup$ $\mathcal{P}_{H_i}z \in H_i$ if and only if $A_i\mathcal{P}_{H_i}z = y_i$ but I get $A_i\mathcal{P}_{H_i}z = -y_i + 2A_iz$ $\endgroup$ – Pazu Oct 9 at 16:31
  • $\begingroup$ I have edited the answer... $\endgroup$ – amsmath Oct 9 at 16:39
  • $\begingroup$ Well in the concerning paper there is a minus but I guess it must be a typo. $\endgroup$ – Pazu Oct 9 at 16:46
  • $\begingroup$ Yes, it's definitely a typo. $\endgroup$ – amsmath Oct 9 at 16:47
-2
$\begingroup$

Hint: To show it, it's enough to prove $\mathcal P_{H_i}(z)$ is $z$ if $z\in H_i$ and is $0$ if $z\perp H_i$ (i.e. if $z=\lambda A_i^T$).

$\endgroup$
  • 1
    $\begingroup$ This is false. We are considering the projection onto an affine subspace here. $\endgroup$ – amsmath Oct 9 at 16:19
  • $\begingroup$ Ah, indeed.. Sorry $\endgroup$ – Berci Oct 9 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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