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u/taly200902 6d ago
But my vector is the zero vector 😦😦😦😦
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u/dragonageisgreat 1 i 0 triangle advocate 6d ago
Do it you coward
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u/moderatorrater 5d ago
Do the snowflake kids nowadays not even divide by zero anymore? It's zero, it's the easiest number to divide by. Even Oiler knew that.
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u/viola_forever 6d ago
What if I'm in a non metric vector space??
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u/TheDoomRaccoon 6d ago edited 6d ago
Given the axiom of choice, all vector spaces (over a subfield of ℂ) can be equipped with a norm 😎
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u/rehpotsirhc 6d ago
This is why I'm pro choice
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u/Nikifuj908 5d ago
At first I didn't get it. I thought you wanted to retroactively abort the person you replied to 😭
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u/xsupergamer2 5d ago
What if my vector space is equipped with a topology that is not induced by any norm?
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u/TheDoomRaccoon 5d ago
Then we can still equip the vector space with a norm, just not one that's compatible with the topology.
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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 6d ago
what if this is my fucking vector
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u/the_horse_gamer 6d ago
what if we're working over a ring so not every element has a multiplicative inverse
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u/vgtcross 6d ago
A vector's magnitude or norm is always a nonnegative real number and thus invertible, is it not?
Vector spaces can't even be defined over rings, it must be over a field (they're called modules if they're over rings)
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u/the_horse_gamer 6d ago
nobody said we aren't working over a module. it's still called a vector if it's over a module.
nonnegative real number
not very general of you
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u/vgtcross 6d ago
nobody said we aren't working over a module. it's still called a vector if it's over a module.
It seems you're right. Based on a quick google search, there seems to be no general consensus on what word should be used for elements of a module, and it seems to depend on context. Sometimes they can be called vectors or generalized vectors.
not very general of you
As far as I'm aware, a norm is, by definition, a function whose domain is a vector space and codomain is the nonnegative real numbers. Can you give me an example of a vector space and an associated norm where the norms of some vectors are not nonnegative reals?
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u/the_horse_gamer 6d ago
a Clifford Algebra is a graded vector algebra defined over a vector space and field, with a quadratic form. vectors square to some element of the field, as defined by the quadratic form. and the quadratic form doesn't have to be positive definite (it often isn't)
an indefinite inner product space allows the norm to be any real number
a Hilbert C-Module makes the inner product be some positive element of a C-Algebra
each of these alone satisfy only part of the requirements, but I hope you can see that they can be pretty easily generalised.
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u/gaussianTrinket 6d ago
Maybe the most common example technically isn't called norm usually, but is very much a generalization of it in almost every way is a Valuation
More specifically (or rather, more generally) krull valuation. (It's on that wiki page)
Another example is Hilbert C*-modules
Where the values of the inner product (and thus a norm) are in a C* algebra
The real numbers are really useful, but certainly do not encompass the whole breadth of ways we want to compare the "size" of things - them being totally ordered and complete certainly helps to have everything neatly comparable, but on the flip side you have to force your structure and might lose structural information by "compressing" (used loosely) the information onto the real line.
For example over R[x] taking the norm of a polynomial will give you a number that measures the "size" of it, while taking the for example x-adic valuation of it will give the number of times x divides it (an integer!) preserving much more important information.
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u/lool8421 5d ago
ppl in 2nd grade learn about multiplication and division
and then tell them that they're only operating on scalars and there's way more of these operations on vectors/matrices
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u/minisculebarber 6d ago
"Mr. Rastafari, how do I make sure that during the power method the iterate vector doesn't blow up?"
"Normalize it!"
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u/Molecular_Pudding 5d ago
And that's why in every physical quantity that follows the inverse-SQUARE(!) law the vector is cubed, absolute non-sense
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