4 ms·

Why study finite-dimensional vector spaces in the abstract?

13 points by TriinT 17y ago

pbhjpbhj 17y ago
If R^3 is taken with vector (x1,x2,x3) defined using x1+x2+x3=0 then I can't quite picture that, I can only imagine the projections on the axes, not the actual plane ... anyone know of a simply 3D graphing tool online.
- TriinT 17y ago
  You can't picture a plane passing through the origin and whose normal vector is (1,1,1)?!? This is possibly the easiest plane to visualize, imho. What I find hard to visualize is the intersection of m hyperplanes in n-dimensional space. On 3-d graphing tools, try this out (a bit overkill, though): http://f-lohmueller.de/pov_tut/a_geo/a_geo__e.htm http://f-lohmueller.de/pov_tut/a_geo/a_geo__e.htm
  - pbhjpbhj 17y ago
    Following your first sentence, now I can. Easier than z=0? Just because I had a senior moment doesn't mean you need to go hyperbolic on me (nasty pun intended). Yes hypercubes and what not turn my brain inside out, like trying to catch Gas in a coarse net - surely you can never actually visualise them. As a note of personal history I've often found the simplest problems to be extremely difficult to grasp whilst still having the highest school exam results in my town as a child. FWIW, as an insight to how a broken brain works, I made the mistake of mentally choosing simple points (-1,-1,2), etc., and then connecting those points into a closed shape; in my minds eye excluding the region outside of the shape .. hence the problem.
- gjm11 17y ago
  Draw (or, better, imagine) a cube with corners at (+-1,+-1,+-1). The plane through the ++- vertices is x+y+z=1. The plane through the +-- vertices is x+y+z=-1. The plane you want is parallel to those and half-way between them. Does that help at all?
  - pbhjpbhj 17y ago
    Thanks to you and Jabavu, the "normal (1,1,1) vector through the origin" is the best description for me.
- JabavuAdams 17y ago
  You can write the equation of a plane in R^3 as A x1 + B x2 + C x3 + D = 0. So, in this case, we have (A,B,C) equal to (1,1,1), and D=0. (A,B,C) is the normal to the plane, and D is the perpendicular distance between the plane, and the origin. So, x1 + x2 + x3 = 0 defines a plane that passes through the origin (D=0), with normal vector (1,1,1) and of course (-1,-1,-1), and (2,2,2) etc. So it's a "diagonalish" plane that passes through the origin and "faces" (1,1,1).
tokenadult 17y ago
I'll note for the record that there are many other very interesting articles about mathematics living at that site, http://www.dpmms.cam.ac.uk/~wtg10/ http://www.dpmms.cam.ac.uk/~wtg10/ the academic home page for Timothy Gowers.

13 points by TriinT 17y ago

If R^3 is taken with vector (x1,x2,x3) defined using x1+x2+x3=0 then I can't quite picture that, I can only imagine the projections on the axes, not the actual plane ... anyone know of a simply 3D graphing tool online.

TriinT 17y ago
You can't picture a plane passing through the origin and whose normal vector is (1,1,1)?!? This is possibly the easiest plane to visualize, imho. What I find hard to visualize is the intersection of m hyperplanes in n-dimensional space. On 3-d graphing tools, try this out (a bit overkill, though): http://f-lohmueller.de/pov_tut/a_geo/a_geo__e.htm http://f-lohmueller.de/pov_tut/a_geo/a_geo__e.htm
- pbhjpbhj 17y ago
  Following your first sentence, now I can. Easier than z=0? Just because I had a senior moment doesn't mean you need to go hyperbolic on me (nasty pun intended). Yes hypercubes and what not turn my brain inside out, like trying to catch Gas in a coarse net - surely you can never actually visualise them. As a note of personal history I've often found the simplest problems to be extremely difficult to grasp whilst still having the highest school exam results in my town as a child. FWIW, as an insight to how a broken brain works, I made the mistake of mentally choosing simple points (-1,-1,2), etc., and then connecting those points into a closed shape; in my minds eye excluding the region outside of the shape .. hence the problem.
gjm11 17y ago
Draw (or, better, imagine) a cube with corners at (+-1,+-1,+-1). The plane through the ++- vertices is x+y+z=1. The plane through the +-- vertices is x+y+z=-1. The plane you want is parallel to those and half-way between them. Does that help at all?
- pbhjpbhj 17y ago
  Thanks to you and Jabavu, the "normal (1,1,1) vector through the origin" is the best description for me.
JabavuAdams 17y ago
You can write the equation of a plane in R^3 as A x1 + B x2 + C x3 + D = 0. So, in this case, we have (A,B,C) equal to (1,1,1), and D=0. (A,B,C) is the normal to the plane, and D is the perpendicular distance between the plane, and the origin. So, x1 + x2 + x3 = 0 defines a plane that passes through the origin (D=0), with normal vector (1,1,1) and of course (-1,-1,-1), and (2,2,2) etc. So it's a "diagonalish" plane that passes through the origin and "faces" (1,1,1).

tokenadult 17y ago

I'll note for the record that there are many other very interesting articles about mathematics living at that site, http://www.dpmms.cam.ac.uk/~wtg10/ http://www.dpmms.cam.ac.uk/~wtg10/ the academic home page for Timothy Gowers.