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So given a hyperplane $x_0 + ... x_n = C$, we know that the normal $n = (1, ..., 1)^T$. How can we find a projection matrix $P$ for this hyperplane that can be used to get orthogonal projections of vectors?

There is one other similar question Projection on the hyperplane $H: \sum x_i=0$, but I can't follow how they derived the projection matrix itself for the case when $x_0 + ... + x_n = 0$.

wednesdaymiko
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1 Answers1

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There is no projection matrix if $C\neq 0$. This is because the projection $P$ onto your hyperplane is not linear. If it were, then we'd have $P(0) = 0$, but $0$ isn't even in the hyperplane. So what we are looking for is an affine operator $P$.

Actually, I derived this in a former question of yours. Look at the answer and observe that the operator you look for is $$ P(v) = (A^\top A)^{-1}A^\top v - (A^\top A)^{-1}A^\top y. $$ If $C$ were zero then the projection matrix is $(A^\top A)^{-1}A^\top$. $A$ just contains the basis vectors of $x_0 + x_1 + ... x_n = 0$ which is in fact a subspace.

Hyperbolic PDE friend
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  • So for example, let's take the plane $x + y + z = 3$. If I take the basis vectors $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ and project them onto the plane, and then take those projections and put them into a matrix, that's then a "projection matrix" for $x + y + z = 3$? – wednesdaymiko Jan 15 '22 at 15:05
  • No, not quite, because the projection onto $x+y+z=3$ is not linear, it is affine. I deleted this in my answer. This would only work for $x+y+z=0$. But you can make the ansatz $P(x) = Mx+b$ where $P$ is the projection onto $x+y+z = 3$ and the matrix $M$ and the vector $b$ are to be found. Now compute $P(0)$ using e.g. the normal vector. This gives you $b$. Then, to find the first row $M_1$ of $M$ calculate $P(e_1)$ - again using the normal vector. $P(e_1)$ gives you $M_1+b$, so $M_1 = P(e_1) - b$. Same for $M_2$, $M_3$. – Hyperbolic PDE friend Jan 15 '22 at 15:14
  • Thanks. I think I understand why there's no projection matrix - or can visualize it in $R^3$ why it's the case - because of the offset.

    Instead, I guess what I'm looking for is the closest point on plane between a vector and the plane.

     – wednesdaymiko Jan 15 '22 at 16:29
  • And this is then subtracting the normal vector from a vector, which will then land on the plane, and it will also be of the shortest distance. – wednesdaymiko Jan 15 '22 at 16:30
  • Thats exactly it. If there were a matrix it would be linear. So $P(0)=0$. But this would mean that $0$ fulfils $x_1+x_2+x_3=3$ which is does not – Hyperbolic PDE friend Jan 15 '22 at 16:30
  • Yes, it will be of the shortest distance. You just have to calculate which $\lambda$ fulfils $v-\lambda(1, 1, 1) \in \text{Hyperplane}$ – Hyperbolic PDE friend Jan 15 '22 at 16:31
  • I think I'm finally getting it. Thank you for the help and have an amazing weekend. – wednesdaymiko Jan 15 '22 at 16:36
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    You, too. No problem, feel fre to ask if something is unclear – Hyperbolic PDE friend Jan 15 '22 at 16:37