Timeline for I want a smooth orthogonalization process
Current License: CC BY-SA 4.0
33 events
| when toggle format | what | by | license | comment | |
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| Oct 2 at 6:16 | comment | added | Ryan Budney | I don't know any place that describes this particular procedure in detail. | |
| Sep 27 at 0:56 | comment | added | Jabby | Are you able to provide a link/reference for this procedure where I can look into the details? | |
| Sep 26 at 23:24 | comment | added | Ryan Budney | Yes, you would have a fixed orthonormal basis for every vector space. The orthonormal basis would largely vary continuously with the vector space, but when you make the transitions through the linear ordering (i.e. when you vector space transitions from not projecting onto the (y,z) plane to only projecting onto either the (x,z) or (x,y) plane, there would be a discontinuity. | |
| Sep 26 at 23:06 | comment | added | Jabby | Thanks @RyanBudney. Are you saying that we can project a pre-determined set of reference vectors onto the subspace that the basis vectors span, and then perform Gram-Schmidt on those projected vectors, which would make the resulting orthonormal basis the same for all basis vectors that span the same subspace. | |
| Sep 26 at 16:45 | comment | added | Mihail | I'm surprised nobody mentioned it, but this is a linear algebra problem which would suit math.stackexchange better then mathoverflow. | |
| Sep 26 at 16:11 | comment | added | Ryan Budney | Lexicographical ordering is an easy one to code. So $(x,y) < (x,z) < (y,z)$. | |
| Sep 26 at 16:06 | comment | added | Ryan Budney | on the coordinate axis planes so that you choose the "preferred" coordinate axis among all the available ones. That would be one device to set the piecewise domains for your decomposition. | |
| Sep 26 at 16:05 | comment | added | Ryan Budney | I see Sammy answered your question on the other forum. If you want a piecewise-smooth section, i.e. with discontinuities, one standard approach is to use the Schubert cells for the Grassmannians. For this, you take your subspace, and consider its projections on to the various coordinate subspaces. So if the orthogonal projection of your subspace to the $x,y$-plane is an isomorphism, you would then transport the $x,y$ coordinates onto your plane, apply Gram-Schmidt to them, and that would be your canonical basis for the plane. That kind of thing. You would have to choose a linear ordering.. | |
| Sep 26 at 15:52 | comment | added | Ryan Budney | What you are asking for does not exist. There is no section of the projection map from the Stiefel manifold to the Grassman manifold. | |
| S Sep 26 at 6:40 | history | suggested | Mihail | CC BY-SA 4.0 |
corrected the set brackets
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| Sep 26 at 1:12 | review | Suggested edits | |||
| S Sep 26 at 6:40 | |||||
| Sep 25 at 20:36 | comment | added | Abdelmalek Abdesselam | What could possibly help is the fact that there are explicit determinantal formulas for the Gram-Schmidt process en.wikipedia.org/wiki/… | |
| Sep 25 at 20:36 | comment | added | Jabby | If different sets of basis vectors span the same subspace, I ideally want the same orthonormal basis (I have asked this question in more general terms here bit.ly/3XXq6rE). This is a problem for me because I am separately learning a distribution over coordinates on the manifold defined by the orthonormalized basis vectors, and so if the orthonormalized basis vectors flip their direction, the coordinates get mapped to the wrong points on the manifold. | |
| Sep 25 at 20:28 | comment | added | Pedro Lauridsen Ribeiro | If $\epsilon$ is large, then both bases are not close to each other and then there is no reason to expect that their GS orthonormalizations should be close to each other either. More importantly: why is this a problem to you? In either case the subspace generated by it is always the same. | |
| Sep 25 at 20:15 | comment | added | Jabby | I don't think it's a rounding issue. The aspect of my initial example that is potentially misleading is the suggestion that epsilon needs to be very small or infinitesimal. Even if epsilon is large and the matrix is never close to degenerate, the problem remains: the orthonormalized basis vectors for the same subspace are very different depending on the sign of epsilon. | |
| Sep 25 at 19:59 | comment | added | Ryan Budney | Your problem sounds closer to a rounding error issue than an issue with Gram-Schmidt. For matrices that are near-degenerate, i.e. $det(A)$ small but non-zero, it can be a subtle problem to decide if $det(A)>0$ or $det(A)<0$, in part because the determinant is a high-degree polynomial so it intensifies the ambiguity. Presumably the best solution is to consider how your matrix $A$ is created, and how that process may lead it to be near-degenerate. You could then maybe fix the problem before it happens. | |
| Sep 25 at 19:54 | comment | added | Jabby | @RyanBudney To be honest, I'm not sure if the kind of axis flipping in my example occurs in practice, or how I would even know. If I knew that one of my orthonormalized axes was pointing in a direction that I didn't want it to (according to some convention), I could flip its sign and solve my problem. It seems to me that once I am able to detect if the problem exists, I am able to solve it... | |
| Sep 25 at 19:54 | comment | added | მამუკა ჯიბლაძე | In spirit of reinforcement learning methods - is it possible to use the gradient descent for something like $\sum_i(1-\langle b_i,b_i\rangle)^2+\sum_{i<j}\langle b_i,b_j\rangle^2$? | |
| Sep 25 at 19:48 | comment | added | Pedro Lauridsen Ribeiro | Your edit seems to be superfluous. By definition, basis vectors must be linearly independent and thus are nonzero. The GS procedure always takes bases to bases, but if you pick a finite set of linearly dependent vectors (particularly, if the set contains the zero vector) the GS procedure is no longer defined there because the orthogonalization part of the GS algorithm will then necessarily produce the zero vector, which cannot be normalized. Maybe this is at the root of some of your concerns in the OP...? | |
| Sep 25 at 17:13 | comment | added | Ryan Budney | If your example is (as you say) misleading and all you are asking about is whether or not Gram-Schmit is a smooth process, then the answer is yes, if you consider Gram-Schmidt to be defined on the space of linearly independent vectors (of whatever number) then yes it is a smooth mapping. All the standard variations of Gram-Schmidt are smooth mappings, including the ones you find in undergraduate linear algebra texts. | |
| Sep 25 at 14:28 | answer | added | Vít Tuček | timeline score: 8 | |
| Sep 25 at 9:08 | history | edited | Jabby | CC BY-SA 4.0 |
added 61 characters in body
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| Sep 25 at 8:47 | history | edited | Jabby | CC BY-SA 4.0 |
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| Sep 25 at 7:41 | answer | added | Robert Wegner | timeline score: 1 | |
| Sep 25 at 1:24 | history | became hot network question | |||
| Sep 25 at 0:14 | answer | added | Tom Goodwillie | timeline score: 11 | |
| Sep 24 at 23:05 | answer | added | Will Sawin | timeline score: 12 | |
| Sep 24 at 20:21 | review | Close votes | |||
| Oct 3 at 12:17 | |||||
| Sep 24 at 20:03 | comment | added | Ryan Budney | Your question appears mis-formulated. You say 𝐵 is a basis and you are changing it slightly, but your $\pm \epsilon$ perturbations are not a basis when $\epsilon=0$. So it's unclear what you are looking for, as your formulation is contradicting the language you are using to discuss the problem. | |
| Sep 24 at 18:48 | comment | added | Andrei Smolensky | In your example, there is no continuous path between these two bases in the space of bases of a given 2-dim space (it must go through a a spanning set of a 1-dim subspace). If you allow extra dimensions (say, rotate the plane around x-axis), it works (GS gives an ONB for each of the intermediate planes, preserving the orientation - the latter changes along the path). | |
| Sep 24 at 18:19 | comment | added | Jabby | Are you saying that no method exists that does what I am looking for? | |
| Sep 24 at 17:55 | comment | added | Pedro Lauridsen Ribeiro | The Gram-Schmidt process is smooth (you can tell that just by looking at the formulae) - what is happening in your example is that the two bases in $\mathbb{R}^2$ provided in the second paragraph of the OP have different orientations and thus lie in two different connected components of the space of all bases in $\mathbb{R}^2$. In that case, the corresponding GS orthonormalizations will have opposite orientations as well (recall that the GS procedure preserves orientation) and thus don't need to be close to each other. The same is true for any finite-dimensional vector space. | |
| Sep 24 at 17:19 | history | asked | Jabby | CC BY-SA 4.0 |