Re: [Algorithms] decompose onto non-orthogonal vectors

[Klaus H. <k_h...@os...>]

----- Original Message -----
From: Scott Justin Shumaker <sjs...@um...>
To: <gda...@li...>
Sent: Saturday, July 15, 2000 9:15 AM
Subject: Re: [Algorithms] decompose onto non-orthogonal vectors


> Okay, let me try and explain this as simply as possible.  You want to find
> scalars u,v such that u*a + v*b = p.  Since these are 3-dimensional
> vectors, you have 3 equations and 2 unknowns:
>
> u*a1 + v*b1 = p1
> u*a2 + v*b2 = p2
> u*a3 + v*b3 = p3
>
> where p1,p2,p3 are the coordinates of the point p, likewise for a1,a2,a3
> and a, and b1,b2,b3 and b.
>
> How do you solve a system of 3 linear equations with 2 unknowns?  If you
> don't know Gaussian elimination, this can be done with some simple
> algebraic rules and substituion.

Okay, this may become very embarrassing for me now... :) How can you solve a
system of 3 linear equations in two unknows, u, v, if they cannot be reduced
to two linear equations (for example, a2 = a3, b2 = b3, and p2 = p3)???

Let's assume that we change the 3 linear equations as follows:

u*a1 + v*b1 + w*c1 = p1
u*a2 + v*b2 + w*c2 = p2
u*a3 + v*b3 + w*c3 = p3

Then there's a unique solution, if

      | a1 b1 c1 |
det | a2 b2 c2 | != 0
      | a3 b3 c3 |

If we set c1 = c2 = c3 = 0, then we have the same system as above (yours),
but with 3 unknowns:

u*a1 + v*b1 + w*0 = p1
u*a2 + v*b2 + w*0 = p2
u*a3 + v*b3 + w*0 = p3

Then
               | a1 b1 0 |
Dn = det | a2 b2 0  | = 0
               | a3 b3 0 |

Since Dn = 0, we get a division by zero, and thus there's an infinite number
of solutions.

... or did I just miss something???

Niki