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Anonymous Wed 09 Sep 01:14:21 2020 No.12099876 View Reply Original Report

Quoted By: >>12099918 >>12100020 >>12100054 >>12102607

What exactly is a dual space V*? Is it a space of functions (transformations?). Like for example a vector space of 2D rotation matrices? Would it be dual to a vector space of 2D vectors? Would it consist of 2x2 matrices? No wait, then the result of this transformation should be a 2D vector. But the dual space consists of functions T such that T:V->R. So it maps a vector to a scalar? That resembles a dot product. Then that "function" T would be a 2D vector as well, same as the elements of a 2D vector space V it is the dual of?

Anonymous

Anonymous Wed 09 Sep 2020 01:20:38 No.12099895 Report

Quoted By:

Given a vector space V over the field F, the dual space V* is the set of all linear functionals from V to its underlying field F. With the operations of addition and scalar multiplication, V* is itself a vector space.

Anonymous

Anonymous Wed 09 Sep 2020 01:27:45 No.12099918 Report

Quoted By: >>12099928

>>12099876
>So it maps a vector to a scalar?
V* consists of the set of linear functionals from V into F. The functionals are just functions whose inputs are the vectors in V (elements of V), whose outputs are the scalars in F (elements of the field F), and who also satisfy the axioms of linearity.

I think you are overthinking it.

Anonymous

Anonymous Wed 09 Sep 2020 01:31:16 No.12099928 Report

Quoted By: >>12099982 >>12099990

>>12099918
Well I am trying to think of an example. If V is a vector space over R^n, then V* consists of n dimensional vectors that map V to R, so it is an inner product, or the dot product?

Anonymous

Anonymous Wed 09 Sep 2020 01:49:46 No.12099982 Report

Quoted By: >>12099990

>>12099928
Not even close. An inner product is a function $<,>: V \times V \mapsto \mathbb{R}$ that satisfies linearity in the first argument, conjugate symmetry, and positive definiteness. Whereas a linear functional is a function $f: V \mapsto \mathbb{F}$ such that for all $\alpha, \beta \in V, c \in \mathbb{F}$ we have the condition that $f(\alpha + c \beta) = f(\alpha) + c f (\beta)$

Anonymous

Anonymous Wed 09 Sep 2020 01:53:59 No.12099990 Report

Quoted By: >>12100026 >>12100054

>>12099928
>V* consists of n dimensional vectors that map V to R
No. V* consists of functions which map vectors in V to scalars in R, and satisfy this condition >>12099982
These functions themselves are vectors in the dual space V*.

Here's an example of a linear functional. Let $V = \mathbb{R}^3$ over the field $\mathbb{R}$ . Define $f(x,y,z) = x + 5y - 2z$ . As you can see, it takes these vectors in the vector space, and sends them to its field. Now these functions are themselves linear, so THEY satisfy the vector space axioms. Then V* is its own vector space, and these "linear functionals" are the vectors in V*.

Anonymous

Anonymous Wed 09 Sep 2020 02:01:35 No.12100020 Report

Quoted By: >>12100115

>>12099876
There are two notions of dual space-that of an algebraic one and a topological one.

The algebraic one is the simplest-it is the said of linear functionals on a vector space. If you wanted to describe this transformation with a matrix, it would be a row vector, so there is a nice interpretation in that respect. You are then right it does have a close relation to dot product. You could consider it a (0,2) tensor-by fixing an argument c and consider c x V -> R, have considered each c as specifying a linear functional, have found a similar representation as (1,1) tensor. v* corresponds to the linear transformation v x V -> R by the inner product <v, . > in this view. The dual space is isomorphic to the original vector space but not in a 'natural' way. The dual of the dual is isomorphic to the original vector space in a 'natural' way.

The topological dual is what happens when one considers between two topological vector spaces, continuous linear functionals. Then, it is in general not the original space, and the dual of the dual is not necessarily the original space either in a 'natural sense'.

Anonymous

Anonymous Wed 09 Sep 2020 02:02:56 No.12100026 Report

Quoted By: >>12100037

>>12099990
I think I understand that except I don't see how the statement you quoted is wrong. It seems to say the same thing. Since linear functionals are actually vectors of the same dimension as V.

Anonymous

Anonymous Wed 09 Sep 2020 02:08:54 No.12100037 Report

Quoted By:

>>12100026
It's not wrong but the terminology is a bit confusing, since while V* does consist of 'vectors', we also call those in V 'vectors'. Since they in fact have the same dimension, they are isomorphic-but, they aren't 'naturally', as it depends on a basis. Because of this, V* shouldn't be 'merely' thought of as another n dimensional vector space.

Anonymous

Anonymous Wed 09 Sep 2020 02:12:35 No.12100054 Report

Quoted By: >>12100079

>>12099990
>>12099876

it absolutely is close. fixing one element an inner product is linear. in fact for finite dimensional space you can parameterize the dual space by the vectors you take the dot product with. this is why dual is isomorphic in finite dimensional case to V

Anonymous

Anonymous Wed 09 Sep 2020 02:19:08 No.12100069 Report

Quoted By:

An interesting book on linear algebra that emphasizes these perspectives is Finite Dimensional Vector Spaces by Paul Halmos. Really though, this type of thinking lies in functional analysis.

Anonymous

Anonymous Wed 09 Sep 2020 02:24:45 No.12100079 Report

Quoted By:

>>12100054
I should mention that in the infinite dimensional case, we have functionals, or elements of the dual space, that are not simply a fixed inner product with some vector. but in this way we do have a natural injection of V \to V*, where each vector can give a linear functional. Its just not surjective in the infinte dim case

Anonymous

Anonymous Wed 09 Sep 2020 02:39:44 No.12100115 Report

Quoted By: >>12101892

>>12100020
>tensors
Yikes, that's another thing I can never wrap my head around, but can't we think of V and V* a vector and covector spaces, or covariant/contravariant vectors? For example if V are column vectors, wouldn't V* be row vectors (or the other way around)? So basically the basis of V transforms in the opposite way of the basis of V*. And I vaguely remember that the pair of covariant and contravariant vector spaces can be thought of the building blocks to construct tensors aka tensor product (maybe not all tensors but a particular types of tensors but I forget about the meaning of (m,n) in the tensor notation but I think it means covariance/contravariance. I feel like these are a bunch of related concepts and there is some sort of a circular dependency between them plus lots of definitions are ambiguous as they may mean different things to different people.

Anonymous

Anonymous Wed 09 Sep 2020 16:24:19 No.12101892 Report

Quoted By:

>>12100115
The way to think about it is to consider the difference between changing the components and the basis. If you change the basis, the basis components must change in the other direction. This is the concept of 'active' vs 'passive' transformation in physics. The 'circularity' is really duality, you can pick either direction and get the other.

For starting out, I would learn tensors in the notion of of space and dual spaces, and then the coordinate transformation interpretation. Only come to tensor product later. It seems to me to essentially specify a kind of multiplication of vector spaces-the best example I can think of this arising is in considering field extensions but that is a different topic.

Anonymous

Anonymous Wed 09 Sep 2020 20:20:41 No.12102607 Report

Quoted By: >>12103418 >>12103528 >>12107520

>>12099876
the babby way of thinking about dual space is that V is column vectors and V* is row vectors. the "evaluation" is matrix multiplication which gives a scalar in this case.

Anonymous

Anonymous Thu 10 Sep 2020 00:23:20 No.12103418 Report

Quoted By: >>12103733 >>12104502 >>12105317

>>12102607
Thats an interesting approach. There must be some kind of a map V to V* that converts column vectors (contravariant?) to row vectors (covariant?). Even though it is supposed to always exist by definition. What would you call that map?
> matrix multiplication
so the dot product of column and row vectors? But the dot product is a bilinear form V × V R. Whereas the vectors in V* are maps which are one forms VR. So confusing. There is some sort of a relationship between the dot product and the one form. It is kind of like saying 3 * 2 = 6 and * is the analogue of a bilnear form that takes 2 and 3 and returns 6. But one could also interpret 3 as an operator which takes 2 and spits out 6 (culling?) which would be similar to a one form, no?.

Anonymous

Anonymous Thu 10 Sep 2020 00:48:29 No.12103528 Report

Quoted By:

>>12102607
But what's known as "the evaluation homomorphism" would give the bidual space (which is btw isomorphic to the vector space)

Anonymous

Anonymous Thu 10 Sep 2020 01:47:40 No.12103733 Report

Quoted By:

>>12103418
>There must be some kind of a map V to V* that converts column vectors (contravariant?) to row vectors (covariant?). Even though it is supposed to always exist by definition. What would you call that map?
It's called the matrix transpose.

Anonymous

Anonymous Thu 10 Sep 2020 08:43:11 No.12104502 Report

Quoted By: >>12105317 >>12106066

>>12103418
The spaces $V$ and $V^*$ are isomorphic, but not naturally. There's no cannonical way of identification, they're isomorphic just because they have the same dimension. This is different when you equip $V$ with an inner product (actually any non-degenerate bilinear form suffices). Inner product gives you an isomorphism $V \to V^*$ given by $v \mapsto \langle v,- \rangle$ . You can also read it as "for any functional $T \in V^*$ there exists a unique vector $v \in V$ such that $T(x) = \langle v,x \rangle$ for all $x \in V$ ".

Look at how it works at $\mathbb{R}^n$ , assume I write general element $x \in \mathbb{R}^n$ as $x = (x^1,\dots,x^n)^T$ . Clearly any linear functional $T \colon \mathbb{R}^n \to \mathbb{R}$ is of the form $T(x) = a_1x^1 + \dots + a_n x^n$ for some uniquely determined scalars $a_i$ . But this is literally matrix multiplication of the row vector $(a_1,\dots,a_n)$ and the column vector $(x^1,\dots,x^n)^T$ . So we can understand every element of $\mathbb{R}^{n*}$ as a row vector. Now look at the inner product. For $x,y \in \mathbb{R}$ we have $\langle y,x \rangle = y^T x$ (matrix multiplication) and therefore the isomorphism $\mathbb{R^n} \to \mathbb{R}^{n*}$ described above sends $y$ to "matrix multiplication by $y^T$ from the left". So if we stick to the convention that $\mathbb{R}^{n*}$ are row vectors, the isomorphism is simply matrix transposition, $y \to y^T$ . (And conversely the linear functional $T$ with row vector $(a^1,\dots,a^n)$ described above corresponds to the dot product with the vector $(a^1,\dots,a^n)^T$ .

Anonymous

Anonymous Thu 10 Sep 2020 15:44:33 No.12105317 Report

Quoted By:

>>12103418
>(culling?)
lol i mean to say currying.

Thanks for this: >>12104502
it does look like the analogy with currying seems to make sense (that place holder, the dash:
>VV? given by v?v,??

Anonymous

Anonymous Thu 10 Sep 2020 19:50:30 No.12106066 Report

Quoted By: >>12106077

OK, I think I understand now. It may not be the "proper" way of thinking about this, but I think it makes sense. The linear functional in V* is a vector of the same dimension as a vector in V (except it is a row vs column) and it sends a vector in V to its field F. But how does it do it? Via the dot product of course. Thats the relationship between the bilinear form and the one-form, via currying! Nobody talks about currying in linear algebra but I think thats what it is. In laymans terms, the analog of the bilinear form is simple multiplication: 2*3=6. Now imagine 3 is a one form, that sends 2 to 6. Via the * which would be the place holder, for example 2-><3,*>. This pretty much corresponds to what this anon says: >>12104502 in proper linear algebra language.

Anonymous

Anonymous Thu 10 Sep 2020 19:53:47 No.12106077 Report

Quoted By:

>>12106066
should never teach CStards math more advanced than highschool algebra

Anonymous

Anonymous Fri 11 Sep 2020 04:53:33 No.12107520 Report

Quoted By: >>12107934 >>12108300

>>12102607
So V* is tensor, which contains two spatial information?

Anonymous

Anonymous Fri 11 Sep 2020 08:20:17 No.12107934 Report

Quoted By: >>12108292

>>12107520
I don't understand how you mean this

Anonymous

Anonymous Fri 11 Sep 2020 13:07:23 No.12108292 Report

Quoted By: >>12108589

>>12107934
Tensor of two spatial information

xx xy
xy yy

for 2D space.

Anonymous

Anonymous Fri 11 Sep 2020 13:12:15 No.12108300 Report

Quoted By: >>12108571

>>12107520
Wouldn't a rank-2 tensor be the result of the tensor product (bilinear form) between V and V*.
Anything is a tensor, so V* could also be thought to be a tensor which is a one-form in this case (0,1).

Anonymous

Anonymous Fri 11 Sep 2020 15:08:11 No.12108571 Report

Quoted By:

>>12108300
Sorry, I haven't focused, I just... Now it's something about linear transformation, for which you need vector and transformation vector, if it works like matrixes, therefore you get rank 2 tensor as result?

Anonymous

Anonymous Fri 11 Sep 2020 15:12:32 No.12108589 Report

Quoted By:

>>12108292
xx xy
yx yy
At least be proper about it

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