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this post was submitted on 16 Jun 2025
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Asklemmy
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I couldn't make sense of the first paragraph, are you sure it is right ?
Pretty sure, yes. I'm probably just explaining badly.
There's a full 360 degrees of rays perpendicular to (AH) starting at H. That would be true of line to a point in 3D. In 2D there would be exactly 2 possibilities (left and right), while in 4D they would correspond to an ordinary sphere, and hyperspheres in higher dimensions yet.
Together, they take up a plane. Only points on a certain (infinite) line going through this new plane and H will actually orthogonally map to H, and it's the same one that's normal to to original plane. Let's call the line L.
If point P wasn't in this plane, (PH) couldn't be perpendicular to (AH). It is in the new plane, but we still don't know for sure it's on line L, so it's not true that that implies it projects to H.
~~I tried again, I don't find mistakes in your statements, I just don't see how they make up for "instant in-mind proofs" for the problem~~ I think I see it now, nevermind. Your got a very good visualization for 3D CanadPlus. It seems so intuitive that "the set of points that map to H with orthogonal projection is a straight line", but do you happen to have a pocket proof for that ?
Uhh, that the preimage of a point like H is a line? Off the top of my head, I'd use the fact it's a shifted copy of the kernel. Well, assuming without loss of generality that we're in a vector space and not just an affine space.
Using basic rules and notions from linear algebra and the theory of functions:
For a projection O in space V, your preimage L is defined as {l∈V | O(l) = H}. Using the linearity of O you can turn that into {l∈V | O(l-H) = 0}, which is equivalent to {y∈V | O(y) = 0} by setting l=y+H. Definitionally, an affine subspace is constructed from the members of a subspace added to a constant like that. The kernel, {y∈V | O(y) = 0}, is a subspace because any linear combination of vectors within it will, once you apply and distribute the operator using linearity again, turn into a sum of 0s, meaning the result must always be another member of the kernel.All that's left is to prove it's a 1D affine subspace, AKA an infinite line. Every point w in the domain V is in some preimage, by the definition of a function, and so using the same math you can construct it as O(w)+k for some k in the kernel. O(r)=r for all r in the range by the definition of a projection, which you can use to both show it's a subspace and can't contain any basis of the kernel (expanding that out I'll leave as an exercise). So, the dimension of the range and the kernel have to add to that of the whole domain. This actually holds for all other linear operators as well.
Our space is 3D and the provided plane is 2D. 3-2=1, QED.
Probably there's a proof from the axioms of Euclidean geometry that doesn't need linear algebra, but I was never good at that sort of thing. It's also worth noting that any set defined purely by linear operators and affine linear subspaces will again be (describable as) affine linear. It's like a closure property.
How first reading felt:
How the second reading felt at the beginning:
How it ended up:
What is
{y∈V | O(y) = 0}? If the plane doesn't pass through $0_V$ then how would that 0 be the image of some point ? Most likely you're using something from linear algebra that I didn't learn in my course (I didn't learn projection I think, only examples when learning matrices).Answer, at risk of making it worse:
I was assuming this is a linear projection in a (non-affine) vector space, from the beginning. All linear operators have to to map the origin (which I've just called 0; the identity of vector addition) to itself, at least, because it's the only vector that's constant under scalar multiplication. Otherwise, O(0)*s=O(0*s) would somehow have a different value from O(0). That means it's guaranteed to be in the (plane-shaped) range.I can make this assumption, because geometry stays the same regardless of where you place the origin. We can simply choose a new one so this is a linear projection if we were working in an affine space.
Can I ask why you wanted a proof, exactly? It sounds like you're just beginning you journey in higher maths, and perfect rigour might not actually be what you need to understand. I can try and give an intuitive explanation instead.
Does "all dimensions that aren't in the range must be mapped to a point/nullified" help? That doesn't prove anything, and it's not even precise, but that's how I'd routinely think about this. And then, yeah, 3-2=1.
Hmm. Where did the question in OP come from?
They're abstractly defined by idempotence: Once applied, applying them again will result in no change.
There's other ways of squishing everything to a smaller space. Composing your projection to a plane with an increase in scale to get a new operator gives one example - applied again, scale increases again, so it's not a projection.
I'm actually old and lurked in university stuff for a long time and dropped out of engineering in university and started with math all anew, yet at the same time I'm still a beginner.
I don't exactly remember How I started thinking about the "distance between plane and a point formula", I think I stumbled upon it while organizing my old bookmarks. Tried to make a proof, and in the process that question came, and when I couldn't solve it on the fly I though like "it's so over for me". Then ChatGPT also got it wrong and was like "It's so over for mankind". And I ended up making this post to share my despair. Actually many answers were eye opening.
Haha, I thought it was a homework question. It would be a pretty good one; it's not hard to answer, but the a proof touches on a lot of things. I probably would have gone about this differently if I hadn't thought I was addressing someone who's actively studying these things. Hopefully you still knew most of the terms I was using.
The missing part, because including an exercise is low-key a dick move if you were just curious:
Any basis vector k can't be 0 (that would be dumb), so if O(k)=0 it fails idempotence and can't be in the range. Therefore, all kernel bases are not in the range.For the range being a subspace, O(a+b)=O(a)+O(b)=a+b, and you can extend that to any linear combination of range vectors.
I guess you'd need to include the proof that vector (sub)spaces must have a basis to make it airtight, so we know the kernel has any dimensional at all. But, then it's just the pigeonhole principle, since you can choose a basis for the whole space made up from bases of the two subspaces.
Best of luck.
hhhhhhh homework in the summer ?
Although I know in Japan they give them such horrors