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Topic: Failing Linear Algebra:
Replies: 91   Last Post: Jan 10, 2007 12:56 PM

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 Guest
Re: Failing Linear Algebra:
Posted: Apr 30, 2004 9:31 PM

Russell:

>Another thing I didn't mention is that it's arbitrary which variables
>you choose to be free, and which "basic" as you call them. That is,
>you could rewrite the same equations with (say) the terms for x6 in
>the first column and those for x1 in the 6th column. Then x6 would be
>basic and x1 would be free; your results will look different
>numerically but they will turn out to work equally well in the
>equations.

Right. I understand why this is. Just as if you have x+y = 10, you can choose
to vary x OR y. The standard we've been using in class is to work left to
right, either in alphabetical order or numerical order. So, I think it's
obvious to most of us that x6 *could* be listed as a free variable while x1 is
listed as basic, but what's the point is rearranging the terms within the
equation? It's just something we've never bothered doing. But, like I said, I
think it's fairly obvious to everyone in my class that it doesn't matter which
variables are free and which are basic.

>Basis of the *system*? That is very confused terminology. Drum into
>your head the idea that a basis is something that applies to a *vector
>space*. And nothing else. (in linear algebra, that is.)

Maybe I meant to say *span* of the system?

>Get used to thinking about matrices etc as abstract objects
>in their own right, not just some weird technique for solving systems
>of equations.

Right. I think I understand this. My professor explained that the matrix
notation is really just a handy way of organizing the notation and keeping the
terms all alligned.

>Hint: in your example below, which very important vector is not in the
>set of solutions?

The 0 vector?

>Anyhow, your vector doesn't work in a span; try multiplying it by
>2 and see if it satisfies your system. You need to subtract the
>constants out (make it homogeneous) so I think you should have
> (-6, 0, 1, -3, 1, 0, 0).

OK, right. I think the technique I used can only be used in a homogeneous
system.

>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.

OK, mine don't work because my system wasn't homogeneous. I'm still not really
sure *why* your three vectors would be a basis for the kernel though. Their
span would produce all equations that equal 0? Is that where it comes from?

>The solution set is the set of all vectors of form (5,8,1,10,0,0,0)+V
>where V is a member of the span of my three vectors.

OK, and that solution is the entire vector space, right? So, in this case, dim
Im = 4 (the values 5, 8, 1, 10) and dim Ker = the 3 zero values, right? 4 + 3
= 7, which makes sense because we're in R^7? Is that, in essence, the
Dimension Theorem?

>
>With three unconstrained variables that you can set to anything, the
>point is that you have a whole 3-dimensional space of vectors that
>will "hold" in your homogeneous equation. That's a lot more than what
>you say, with the free variables all set to zero (but yes that does
>happen to be a solution too).

So, no matter what those last three free variables are, the system of equations
still holds. But, in algebraic structures, the kernel was the set of all
things that map to zero. Doesn't the dimension of three imply that there are
three things, or three variables, that map to 0?

>You tell me. How many equations do you have in how many unknowns?

One equation, two unknowns.

>How many rows will there be in your matrix if you write this as a
>matrix equation, and how many columns?

One row, two columns.

>Not sure what to say here, are you bothered by abstract notation like
>{e_1, e_2, e_3} being a basis for some 3-dimensional vector space?

I think I'm OK with that notation, because I understand that to be the
{(1,0,0), (0,1,0), (0,01)} form, so that e1 is (1,0,0), e2 is (0,1,0), and e3
is (0,0,1), and together they form the "standard" basis for R^3. If I'm right
about everything I've just said, then I'm fine with that "e" notation for the
standard bases.

>Or
>is the problem that you can't seem to grasp bases in R^n other than
>the standard one?

Right. That's the problem. I'm just getting how to change from one basis to
another, but I really don't understand how bases other than the standard ones
are useful.

Date Subject Author
4/24/04 Daniel Grubb
4/24/04 Marc Olschok
4/24/04 Daniel Grubb
4/24/04 Marc Olschok
4/24/04 Daniel Grubb
4/24/04 Thomas Nordhaus
4/24/04 Dave Rusin
4/25/04 Jonathan Miller
4/25/04 Felix Goldberg
4/24/04 Daniel Grubb
4/28/04 Tim Mellor
4/28/04 James Dolan
4/28/04 Daniel Grubb
4/28/04 James Dolan
4/28/04 Daniel Grubb
4/28/04 gersh@bialer.com
4/29/04 Daniel Grubb
4/29/04 Dave Rusin
4/28/04 Guest
4/29/04 Guest
4/28/04 Guest
1/10/07 David C. Ullrich
4/29/04 Dave Rusin
4/28/04 Guest
1/10/07 Law Hiu Chung
1/10/07 Dave Seaman
1/10/07 Marc Olschok
1/10/07 George Cox
4/28/04 Guest
1/10/07 Dave Rusin
4/28/04 Lee Rudolph
4/28/04 Guest
4/28/04 Guest
1/10/07 Marc Olschok
1/10/07 Toni Lassila
4/29/04 Guest
1/10/07 M L
1/10/07 Thomas Nordhaus
4/30/04 Guest
1/10/07 David C. Ullrich
1/10/07 Toni Lassila
4/30/04 Guest
1/10/07 George Cox
1/10/07 Marc Olschok
4/30/04 Guest
4/30/04 Guest
4/27/04 Guest
1/10/07 Thomas Nordhaus
1/10/07 David C. Ullrich
1/10/07 Dave Rusin
1/10/07 David C. Ullrich
5/9/04 James Dolan
5/10/04 David C. Ullrich
5/10/04 James Dolan
5/10/04 David C. Ullrich
5/10/04 Marc Olschok
5/10/04 David C. Ullrich
4/27/04 Guest
1/10/07 Thomas Nordhaus
4/27/04 Guest
1/10/07 magidin@math.berkeley.edu
1/10/07 David C. Ullrich
1/10/07 Marc Olschok
1/10/07 David C. Ullrich
1/10/07 Tim Mellor
4/28/04 Daniel Grubb
4/28/04 Daniel Grubb
4/27/04 Guest
1/10/07 David C. Ullrich
4/28/04 Dave Rusin
4/28/04 Daniel Grubb
4/27/04 Guest
1/10/07 Marc Olschok
4/24/04 Wayne Brown
4/24/04 Thomas Nordhaus
4/24/04 David Ames