
Re: Failing Linear Algebra:
Posted:
Apr 28, 2004 6:32 PM


Russell:
> If the kernel >is {0}, i.e. the set of rows (or the set of columns) of the matrix is >linearly independent, and the matrix is square, then the matrix is >invertible and you have an isomorphism. Otherwise, one or more rows >or columns is a linear combination of the others; and that means there >are nonzero vectors which, when multiplied by your matrix, give zero. >In other words, those vectors are members of the kernel of your linear >transformation. See how it all fits together?
Right. So, if a matrix is homogenous and in echelon form, the number of free variables is the same as the kernel, because if any free variable is set equal to 0, then the system holds. Therefore, the number of basic variables is the same as the image, right?
>Btw you asked elsewhere what an eigenvalue is. The definition is >simplicity itself, learn it!
Yeah, I know the definition. I think I stated it in a post already.
>Btw I found nothing objectionable >about Lipschutz from an abstract point of view; if he made it any >simpler he would mislead, but he treads that fine line very nicely >IMHO.
OK, thanks. I'll be sure to use it to study.
>Usenet will probably be too slow for your needs this week; who wants >to type in those matrices in ASCII anyway.
Right, Usenet can be a little slow, but I thank you guys for your speed in responding. It's my fault that I left this thread (and all its great responses) hanging for a while. I forgot about it.... Anyways, even if your suggestions are too slow to help me on Friday's test, they may come in handy on my final. So, please, don't hesitate to fill me in whenever I've made a mistake.
BTW, I'm good enough with matrices that I can understand them if they're just typed like:
2 1 4 6 3 5 8 1 1 5 2 2
etc.

