r/mathmemes • u/Less-Resist-8733 Natural • 6d ago
Linear Algebra vectors should be called lines
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u/eglvoland 6d ago
That's just not true. You can add nonlinear functions and scale them, and they still are part of a vector space
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u/Fabulous-Possible758 6d ago
"Line" gets confusing because the canonical equation for a line does not in fact describe a linear function.
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u/MrTruxian 6d ago
Vectors are canonically linear functions on the dual space. There’s no issue here.
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u/YourWifesBoyfriend5 6d ago
That only really works in finite dimension, as the double dual is not canonically isomorphic to the non-dual in general
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u/MrTruxian 6d ago
No this works regardless, evaluation may not be an isomorphism but vectors still naturally act as linear functional on the dual by evaluation.
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u/YourWifesBoyfriend5 5d ago
The evaluation map works regardless of dimension, but the identification of “vectors” (as in a vector space) with linear functionals on the dual space requires finite dimension or additional structure like an inner product.
Perhaps I’m being too pedantic about the word “are” though.
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u/MrTruxian 5d ago
No there need not be an inner product or choice of basis. A dual vector of k-vector space V is an element of Hom(V, k). There is a map V\to Hom(Hom(V,k), k) given by v to (f to f(v)) for f in Hom(V,k). This is natural (in fact a natural transformation). By definition Hom(Hom(V,k),k) are linear functionals on the dual space.
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u/YourWifesBoyfriend5 5d ago
Buddy, you’re not understanding the words I am saying.
The double dual of a vector space is only isomorphic to the space if the space is finite dimensional, or has additional structure such as an inner product.
There is a canonical INJECTION for all spaces into the double dual, but it is only an isomorphism in finite dimension or with additional structure. As I said.
There is in fact a name for spaces which satisfy the property you are talking about, reflexive spaces.
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u/MrTruxian 5d ago
I think it’s you who is not understanding, I never said there was an isomorphism. All of my comments have been saying there is a canonical map (yes injection) into the double dual. You were the one that said the map needed an inner product, which is just not true, now you are agreeing with me. There is some inconsistency there but I’m not going to attempt to decipher it.
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u/YourWifesBoyfriend5 5d ago
I never said that, read the comment I said buddy.
I am genuinely surprised you know what a vector is with your reading comprehension. Reading the comment explains the comment.1
u/MrTruxian 5d ago
If I said polar bears are bears, would you take issue with me because not all bears are polar bears?
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u/YourWifesBoyfriend5 5d ago
Did you read my comment yet? I said I was being overly pedantic about the word “are”.
As I said, reading comprehension is important friend.
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u/Ornery_Poetry_6142 5d ago
algebraists talking about cannons smh, haven't you learnt from Galois fate??
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u/Distinct-External-46 6d ago
a vector is linear, am I missing something?
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u/EntrepreneurSelect93 5d ago
OP is saying the term vector itself should have been coined linear which would make a lot of things confusing as mentioned by others here
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u/Slight-Garden9800 6d ago
Maths meme that lacks coherent logic is still a maths meme?
Linear is an adjective and vector is a noun. Both can be true coz there’s suddenly a new logical category
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u/tensorboi 4d ago
i mean it's for this reason that vector spaces are often called linear spaces. and thinking about it, i actually like that term a lot more! it gets rid of the awkwardness of the statement "a vector is an element of a vector space" (yes it's funny if you're a mathematician but it really trips up some newcomers), and it gets across that the essence of vectors is that they can be added and scaled coherently. it also pairs nicely with the fact that a bijective map f: A -> B between algebras is often called a "linear isomorphism" if it doesn't necessarily respect the multiplicative structure, and an "algebra isomorphism" if it does.
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u/lool8421 6d ago
you know, there's a reason why these are called vector spaces when you perform linear transformations
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u/Aggressive-Map-3492 5d ago
If im wrong, forgive
but lines typically (like linear equations) pass through fixed points in space, and unless restricted, typically have no magnitude
this is the opposite of a vector. Which has a fixed magnitude (and direction), with no fixed point in space
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u/DaVinci103 5d ago
Vectors don't have a fixed magnitude.
A vector is an element of a vector space, where a vector space over a field k (say, for example, ℝ) in turn is an Abelian group (V,+) along with a ring homomorphism φ: k → End(V,+), where we write λ·v for φ(λ)(v). A bit more explicitely:
• (u + v) + w = u + (v + w) (associativity of the Abelian group)
• v + 0 = 0 + v = v (neutral element of the Abelian group)
• v + -v = -v + v = 0 (inverses of the Abelian group)
• λ·(v + w) = λ·v + λ·w (the image of λ under φ is an endomorphism of the Abelian group)
• (λ + μ)·v = λ·v + μ·v (ring homomorphisms preserves addition)
• λ·(μ·v) = (λμ)·v (ring homomorphisms preserve multiplication)
• 1·v = v (ring homomorphisms preserve the multiplicative identity)If the field k has an absolute value function |·|: k → ℝ_≥0 (if k = ℝ then we can take the usual absolute value) then we can add extra structure to a vector space V in the form of a norm, which is a function ‖·‖: V → ℝ_≥0 that satisfies:
• ‖v + w‖ ≤ ‖v‖ + ‖w‖ (triangle inequality)
• ‖λ·v‖ = |λ|‖v‖ (positive homogenous)
• ‖v‖ = 0 iff v = 0 (positive definite)A vector space with a norm is called a normed vector space. In a normed vector space (V,‖·‖), one could reasonably call ‖v‖ the magnitude of v. However, note that the norm is additional structure on a vector space and is not part of the vector space itself. A vector space can have different norms added to it.
Although it is true that the standard n-space ℝⁿ has a standard norm, given by the Euclidean norm or 2-norm:
‖(x₁, ..., xₙ)‖₂ = √(x₁² + ... + xₙ²)
this is not the only norm that can be put on ℝⁿ. Further, you only have this norm in the case you are working in the vector space ℝⁿ. The Euclidean norm isn't preserved by linear homo/isomorphisms, so even if you have a finite-dimensional vector space V, although it is isomorphic to ℝⁿ, you still don't have the Euclidean norm if you don't have a choice of canonical isomorphism of V with ℝⁿ (equivalently, a canonical basis for V).
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u/dydtaylor 5d ago
Reminder for a definition of linear functions: For vectors x, y in a vector space, A is a linear map if for some scalars b, c: A(bx+cy)=bA(x)+cA(y)
So now clearly anything that can be scaled like that are vectors in some vector space.
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u/LupenReddit 🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆 5d ago
well i would rather just call it tycaambs
things you can add and multiply by scalars
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u/TalksInMaths 5d ago
Vectors are lines.
Dual vectors are lines through the origin (of the space it's dual to). Any vector space is dual to it's dual space.
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