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August 1

Is there a name for this specific type of matrix?

I am looking for information regarding n-by-n matrices of this type:

1      1/a   1/ab  1/abc  1/abcd  1/abcde
a      1     1/b   1/bc   1/bcd   1/bcde
ab     b     1     1/c    1/cd    1/cde
abc    bc    c     1      1/d     1/de
abcd   bcd   cd    d      1       1/e
abcde  bcde  cde   de     e       1

namely: all 1's on the principle diagonal with n-1 arbitrary non-zero entries (a,b,c,...) directly below the diagonal; each row is the scalar product with each of a,b,c... (in that order) of the one above it; and every element is the inverse of its transposed location. This kind of matrix crops up frequently (either in full or in part) in the form of units-of-measurement or currency conversion tables. I was wondering if it had a mathematically-recognized name, since I cannot find it anywhere in Wikipedia's list of matrices, and my web searches using the seemingly obvious terms is just dredging up lots of irrelevancy. DWIII (talk) 13:09, 1 August 2015 (UTC)[reply]

I doubt it has a name as such. It's a rank one matrix, which is the outer product if the vectors (1, a, ab, abc, abcd, abcde) and (1, 1/a, 1/ab, 1/abc, 1/abcd, 1/abcde) Sławomir
Biały 13:22, 1 August 2015 (UTC)[reply]

Geometry question

Hello, I have an arbitrary 3D vector (u,v,w) through the origin. Perpendicular to this, centred on the origin, I have a circle. I need to choose one point (any point) on this circle to use as a base point for further operations. Obviously I can see various practical ways to do this, but they all seem ugly and ad hoc. I believe (someone correct me if I am wrong) that there is no way to choose the point such that there are no discontinuities as (u,v,w) moves (hairy ball theorem?). Even so, is there a "nice" way to do it? Something that has some symmetry or elegance to it? 109.153.231.34 (talk) 22:58, 1 August 2015 (UTC)[reply]

What do you mean by "further operations"? Also, the hairy ball theorem is a statement about the tangent vector field, not the normal, as you appear to describe, and one can easily define a continuous normal vector field on the circle (or any other orientable surface).

Additionally, I'm pretty sure that no point on the circle in particular is "better" than another with respect to the origin, because every point (by definition) is the same distance away from the origin. There always exists an orthogonal basis in which your vector (u,v,w) has all components but one zero. For example, if u = 1, v = 2, w = 2, then with the basis (1/3, 2/3, 2/3), (2/3, 1/3, -2/3), (-2/3, 2/3, -1/3) your vector is expressed simply as (3, 0, 0). --Jasper Deng (talk) 00:37, 2 August 2015 (UTC)[reply]

By the way, I also assume you meant a sphere not a circle. A circle is a two-dimensional figure and the hairy ball theorem does not apply to it.--Jasper Deng (talk) 00:49, 2 August 2015 (UTC)[reply]

The nature of the "further operations" is not relevant. All that matters is that I have arbitrary (u,v,w) and I want a pleasing and symmetrical way to choose one point on the circle that I described, in terms of u, v and w. I do not see that changing the basis has any relevance to what I am trying to achieve. I mean circle, not sphere. I am not sure that the hairy ball theorem applies or is relevant, hence the question mark. If it is relevant, the sphere is locus of the endpoints of normalised (u,v,w). 109.153.231.34 (talk) 01:39, 2 August 2015 (UTC)[reply]

The hairy ball theorem applies in the following way. Let U = (u,v,w). A point V on the circle of radius 1 centered at the original perpendicular to U lives on the unit sphere. There is a unique tangent unit vector W to this circle such that U,V,W is a positively oriented basis. So if it were possible to choose the point V continuously in U, then it would also be possible to choose W continuously, but W is then a non-vanishing vector field on the sphere, which contradicts the hairy ball theorem.

In response to the original post, what is effectively being asked is whether there is a "nice" lift from the 2-sphere into the space of orthonormal frames. There is not one, but what is sometimes helpful is to work with the frames directly, which can be described using unit quaternions. Sławomir
Biały 01:45, 2 August 2015 (UTC)[reply]

Thanks very much for your reply. After playing with this problem for some little while, I was kind of reaching the conclusion that there was no way of doing what I wanted, so it's good (or bad!) to have it confirmed. I am not familiar with orthonormal frames or quaternions, but I assume from your initial statement that machinations with these will not give me the "coordinates of point on circle = nice function of (u,v,w)" that I seek, right? 109.153.231.34 (talk) 02:04, 2 August 2015 (UTC)[reply]

(edit conflict)Pick a fixed vector Z. Then W=U×Z will be perpendicular to U, and W/|W| will be on the circle described. So if you pick Z=(0, 0, 1) then you get (v, -u, 0)/√(u²+v²), and this is probably about as 'nice' an expression as you can get. This fails, as the hairy ball theorem says it must somewhere, when U is parallel to Z, so you have to make an additional rule for that exception. If |U|=1 then it represents a point on the earth and the problem amounts to assigning a direction to each point on the earth in a 'nice' way. One way would be to pick one of the four directions of the compass, North, South, East or West. This works everywhere except at the poles and N, S, E, W are basically undefined there. --RDBury (talk) 02:47, 2 August 2015 (UTC)[reply]

Thanks, yeah, this is the sort of thing that I have been coming up with, but of course (v, -u, 0)/√(u²+v²) does not meet my standards of "nice" or "symmetrical" (as it fails on u = v = 0 and does not treat u, v and w equally). 109.153.231.34 (talk) 02:58, 2 August 2015 (UTC)[reply]

If you want symmetrical then take Z = (1, 1, 1). That gives you (v-w, w-u, u-v)/√(2(u²+v²+w²-uv-uw-vw)). Still fails when u=v=w but that's the hairy ball theorem at work. --RDBury (talk) 23:29, 2 August 2015 (UTC)[reply]

Oh yes, that's interesting, thanks! 109.153.231.34 (talk) 00:14, 3 August 2015 (UTC)[reply]

Here's how I would do it, although perhaps this is the ugly ad hoc way you already had in mind:

1) Pick some arbitrary point on the circle (highest X, Y or Z coord would do). That's you first point.

2) After moving the circle, pick the point on the moved circle which is closest to the previous point. This becomes your new point.

So, just continue finding the new point closest to the previous one. That's easy to do with numerical methods (just test the distance, or better yet the square of the distance, on sample points evenly spaced around the circle). StuRat (talk) 23:58, 2 August 2015 (UTC)[reply]