
Re: Failing Linear Algebra:
Posted:
May 1, 2004 9:23 AM


Another correction! This is good, I'm learning stuff. I think this will be the last one or I risk looking a bit too much like JSH.
On Thu, 29 Apr 2004 19:40:06 0700, russell@mdli.com wrote:
[basis for kernel  mine  vs. bogus basis for solution set  his]
>My vectors are a basis  of the kernel, not of the solution set >(which has no basis). Yours are not a basis of any space relevant to >the problem, AFAICS.
They're not a basis of any relevant *subspace* of our domain. But just for the record, I realize now that it *is* a basis for a 3D vector space that has + and * redefined to transform first into the kernel space, perform the operation normally there, and transform back; and that *is* a relevant vector space (namely, it is a vector space of all solution vectors). And that isn't the only way one could define + and * to generate the same solution set from that basis; but I think I've digressed far enough here... If Anonymous has any clue what I mean with all this, he will be doing very well indeed, otherwise not to worry.
Lesson: in math, one should never assert the nonexistence of something just because one has failed to think of an example.
[snip to R^2 example]
>You see the trouble your method got you into. Forget trying to get a >basis for your solution set  it isn't a vector space!
Not a vector space with the usual +,* operations of R^2, that is.

