forumbasta Posted March 24, 2011 Share Posted March 24, 2011 Hi everyone, Given two different reference frames in a vector space; say left and right. v is a vector defined in the left frame and u is a vector defined in the right frame. What is the nature of a matrix A that can satisfy the equality u= A.v? Thank you Link to comment Share on other sites More sharing options...

DrRocket Posted March 24, 2011 Share Posted March 24, 2011 Hi everyone, Given two different reference frames in a vector space; say left and right. v is a vector defined in the left frame and u is a vector defined in the right frame. What is the nature of a matrix A that can satisfy the equality u= A.v? Thank you 1. There is no meaning in mathematics to the term "reference frame". It is a concept from physics. I think you are referring to what in linear algebra is called a set of basis vectors. 2. Given two sets of basis vectors x1, x2, .., xn and y1,y2,...yn a change of basis is a matrix T for which the ith column is the coordinates of xi in terms of y1, y2, .., yn 3. So to solve the vector equation u= A.v in terms of matrices you need to express everything in terms of a single basis. The transformation T is actually the identity linear transformation, though not the identity matrix. If it takes the "left frame" to the "right frame" then your equation in matrix terms, referred to the "right frame" is u= ATv This reduces the problem to solving u= B.v in a single reference frame. Unless the vector space is 1-dimensional there are infinitely many such matrices B for any given u and v. 4. If with respect to a fixed basis, and given u and v you want to find a linear transformation such that u=Bv you can simply send v to u and extend linearly. There are infinitely many ways to do this in dimension 2 and above. One simple way is this: If u has coordinates (u1, u2, ..., un) and v has coordinates (v1, v2, ..., vn) B can be a diagonal matrix with diagonal elements u1/v1, u2/v2, ... , un/vn Link to comment Share on other sites More sharing options...

Bignose Posted March 25, 2011 Share Posted March 25, 2011 [math]A_{ij} = \frac{\partial x_i}{\partial y_j}[/math] where x and y are components of the basis vectors in the two different coordinate systems. This is basically the definition of a tensor. I recommend Synge and Schild Tensor Calculus for much more info. Link to comment Share on other sites More sharing options...

ajb Posted March 25, 2011 Share Posted March 25, 2011 The most general thing is that the matrix belongs to [math]GL(n, \mathbb{R})[/math] if the vector space is over the reals and has dimension n. In practice you may be interested in subgroups of this. Link to comment Share on other sites More sharing options...

forumbasta Posted March 25, 2011 Author Share Posted March 25, 2011 (edited) Thank you DrRocket, I meant by "reference frame" coordinate system. Your reply is nearly what I am looking for. My accurate request is: The vectors u and v are defined in two different coordinate systems. "The matrix B have to be the product of two matrices, one of them is a transformation matrix from one of the coordinate system to the other". As you specified B = A.T I am very grateful if you could support the statement between quotations by some references. Thank you again, have a nice weekend Edited March 25, 2011 by forumbasta Link to comment Share on other sites More sharing options...

DrRocket Posted March 26, 2011 Share Posted March 26, 2011 Thank you DrRocket, I meant by "reference frame" coordinate system. Your reply is nearly what I am looking for. My accurate request is: The vectors u and v are defined in two different coordinate systems. "The matrix B have to be the product of two matrices, one of them is a transformation matrix from one of the coordinate system to the other". As you specified B = A.T I am very grateful if you could support the statement between quotations by some references. Thank you again, have a nice weekend Any text on linear algebra. Hoffman and Kunze would do. Link to comment Share on other sites More sharing options...

ajb Posted March 28, 2011 Share Posted March 28, 2011 (edited) 1. There is no meaning in mathematics to the term "reference frame". As a geometer I would probably read that as picking a basis of sections for the tangent bundle over some smooth manifold. The word "frame" is sometimes used to refer to a basis, but as you state, this really comes from physical language. Using physical language I approve of, but it can be confusing for mathematicians not exposed to theoretical physics. Edited March 28, 2011 by ajb Link to comment Share on other sites More sharing options...

DrRocket Posted March 28, 2011 Share Posted March 28, 2011 As a geometer I would probably read that as picking a basis of sections for the tangent bundle over some smooth manifold. I hope this is not part of your strategy for writing popularizations of mathematics. Link to comment Share on other sites More sharing options...

Xittenn Posted March 28, 2011 Share Posted March 28, 2011 (edited) My two cents? I say this way too often but, some more details would better help with the question. I think the most common reason, and I say this because you still haven't given one, is in graphics programming when one wishes to switch from DirectX to OpenGL or they have a mesh that was done in left hand and it is required to put it in right and vice verci; bones, texture coordinates and all. If I am off topic I apologize now. The problem with finding one matrix to do this transformation in this case is the two matrices that you are trying to transform between are compositions of linear matrices themselves. They are comprised of a 3x3 scaling, a 3x3 rotation and a concatenated translation vector. There is no proper way to do this in this scenario, as such. You would have to decompose the matrices and apply the relevant transformations independently. This has a lot to do with the properties of affine matrices and where the composed matrix is not. Please correct me where I'm wrong on this I still have a hard time wrapping my brain around the whole of it unless I have been working on it myself, and I haven't! Edited March 28, 2011 by Xittenn Link to comment Share on other sites More sharing options...

ajb Posted March 29, 2011 Share Posted March 29, 2011 (edited) I hope this is not part of your strategy for writing popularizations of mathematics. Only when wishing to bamboozle people. More generally, a frame usually means a collection of sections that forms a basis of the typical fibre of a vector bundle. So, over a point we can understand this to be synonymous with a basis of a vector space. But I don't think this is common language in linear algebra. Edited March 29, 2011 by ajb Link to comment Share on other sites More sharing options...

DrRocket Posted March 29, 2011 Share Posted March 29, 2011 Only when wishing to bamboozle people. More generally, a frame usually means a collection of sections that forms a basis of the typical fibre of a vector bundle. So, over a point we can understand this to be synonymous with a basis of a vector space. But I don't think this is common language in linear algebra. Yes. You understand. I understand. All the non-mathematicians are scratching their heads and pointing at the crazy mathematicians, with some justification. Link to comment Share on other sites More sharing options...

imatfaal Posted March 29, 2011 Share Posted March 29, 2011 Yes. You understand. I understand. All the non-mathematicians are scratching their heads and pointing at the crazy mathematicians, with some justification. Yep. I just noticed that your picture and ajb's both show a face at exactly the same angle/aspect and both with teeth showing - admittedly Tom Baker is slightly less scary than the bear - you're not two avatars of the same crazy mixed-up hyper-mathematician are you? Link to comment Share on other sites More sharing options...

DrRocket Posted March 29, 2011 Share Posted March 29, 2011 Yep. I just noticed that your picture and ajb's both show a face at exactly the same angle/aspect and both with teeth showing - admittedly Tom Baker is slightly less scary than the bear - you're not two avatars of the same crazy mixed-up hyper-mathematician are you? No we are two different crazy mixed-up hyper-mathematicians. Link to comment Share on other sites More sharing options...

Xerxes Posted April 1, 2011 Share Posted April 1, 2011 Only when wishing to bamboozle people. Well you are certainly bamboozling me!Like... More generally, a frame usually means a collection of sections that forms a basis of the typical fibre of a vector bundle. So, over a point we can understand this to be synonymous with a basis of a vector space. So. I am temporarily divorced from my texts and tutorial notes, accordingly lemme try this: At each and every point on a manifold, we will have a tangent vector space. Each of these spaces is entitled to an arbitrary basis (let's assume these spaces are finite-dimensional), then there may be a set-theoretic union of all such possible bases over all points in the "underlying" manifold which is called, if I recollect correctly, the frame bundle. OK. Of course the basis vectors for each and every tangent space comprises of vectors (they are a subspace-space of the tangent space), so that a section of the frame bundle is, essentially by definition, a vector field. Is this what you mean? And what you call the "collection" of all such sections you call a "frame"? Hmm, something seems a bit wrong here, for surely, if our tangent spaces are finite-dimensional, this need not imply that elements in the section are? So how can the "collection of sections" be synonymous with a basis? Moreover, the seems no limit to the number of sections I may take of my so-called frame bundle Not feeling very bright today, sorry (too much travelling!) Link to comment Share on other sites More sharing options...

ajb Posted April 1, 2011 Share Posted April 1, 2011 (edited) At each and every point on a manifold, we will have a tangent vector space. Each of these spaces is entitled to an arbitrary basis (let's assume these spaces are finite-dimensional), then there may be a set-theoretic union of all such possible bases over all points in the "underlying" manifold which is called, if I recollect correctly, the frame bundle. This forms a principle bundle, rather than a vector bundle. The tangent bundle is an associated vector bundle with respect to the frame bundle. OK. Of course the basis vectors for each and every tangent space comprises of vectors (they are a subspace-space of the tangent space), so that a section of the frame bundle is, essentially by definition, a vector field. Is this what you mean? A section of the frame bundle is not a vector field. It is a frame. There is the notion a parallelisable manifold. That is a manifold such that the frame bundle has a global section. This means that we have a finite collection of vector fields that restricted to each point defines a basis for the tangent space at that point. Generally you won't be able to do this. And what you call the "collection" of all such sections you call a "frame"? Hmm, something seems a bit wrong here, for surely, if our tangent spaces are finite-dimensional, this need not imply that elements in the section are? So how can the "collection of sections" be synonymous with a basis? Moreover, the seems no limit to the number of sections I may take of my so-called frame bundle Yes, so I have been quite slack with all this. Sections of a vector bundle form a vector space. There maybe no sections other than the zero section. However, what I am really talking about are local sections over some open subset of the manifold. As the vector bundles are always locally trivial, by frame I really mean a choice of local trivialisation. (Or we explicitly assume we are only talking about trivial vector bundles) So, a frame of E over U is a smooth choice of basis [math](s^{1}(x), s^{2}(x)\cdots ,s^{n}(x))[/math] for all [math]x \in U[/math]. In particular, if a vector bundle of dimension n (dimension of the vector spaces) and there are n linear independent (global sections) then the vector bundle is trivial. A better way to think about vector bundles is as locally free modules (over the structure sheaf). The locally free part in essence means that locally we can always trivialise, or in other words we can always find a basis locally. Edited April 1, 2011 by ajb Link to comment Share on other sites More sharing options...

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