Math 464/514-Hotline
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Please look here for answers to questions on homework and other 
class materials, exams etc.
I will post questions (without any identifying info!) and answers
in the reverse order they are received.
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Wed Sep 27 21:00:48 MDT 2006
Class will be cancelled tomorrow due to illness. Please turn your homeworks 
into my mailbox for the grader. I have modified the syllabus accordingly.
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Mon Sep 25 21:42:50 MDT 2006
> on problem #18, p.124 (sect 2.5): who is matrix A?
it is the incidence matrix of some graph (abstract)
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Wed Sep 13 17:04:39 MDT 2006
> I had a couple of questions about the homework that is due tomorrow.
> On Sect. 2.1: On #7(d) when you are showing/not showing that the set is a
> subspace, can you assume that when you muliply by a scalar, that scalar is
> finite (< infinity)?
if you allow infinity as a point of the space, then infinity*convergent
sequence could be defined; but that could lead to the following problem:
imagine the sequence (converging to zero)
a_k:= { 0, k even; 1/k, k odd}
then  the sequence
b_k := inf * a_k = { 0, k even; inf, k odd}
has two limit points (0, inf) and so it is not convergent.
 That is, if you set inf*0 = 0. So you get into problems, since there is
really no meaningful way to define inf * 0 (except in the sense of limits,
but that means you can not allow infinity as a legitimate scalar).
 That is no different of how infinity would play in a finite dimensional
vector space setting:
 if V is a subspace, then, for any scalar a,
   a*V = { x = a*y | y element of V} = V
(except when a = 0), that a*V is a subspace. But if a = inf,
a*V is not a subspace without serious redefinition.
So, for the scalars, we want to stick to the definition of R (or C)
that does not allow infinity as a point, and only eccepts it in the sense
of limits.

> On Sect. 2.2: For the second matrix in #5, the second row is inconsistent, so
> is it necessary to solve the system?
 No need, as there is no solution.
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