I think 3D geometry has a lot of quirks and has so many results that un_intuitively don’t hold up. In the link I share a discussion with ChatGPT where I asked the following:
assume a plane defined by a point A=(x_0,y_0,z_0), and normal vector n=(a,b,c) which doesn’t matter here, suppose a point P=(x,y,z) also sitting on the space R^3. Question is:
If H is a point on the plane such that (AH) is perpendicular to (PH), does it follow immediately that H is the projection of P on the plane ?
I suspected the answer is no before asking, but GPT gives the wrong answer “yes”, then corrects it afterwards.
So Don’t we need more education about the 3D space in highschools really? It shouldn’t be that hard to recall such simple properties on the fly, even for the best knowledge retrieving tool at the moment.
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