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

 Messages: [ Previous | Next ]
 Daniel Grubb Posts: 526 Registered: 12/10/04
Re: Failing Linear Algebra:
Posted: Apr 28, 2004 10:15 AM

>>Actually from what you say here it seems pretty likely that
>>you do know what linear independence _is_, but the
>>way you're stating the definition is totally wrong.

>Right. That's my point. I *do* know the definition, but the phrasing always
>gets me. It's like this, probably, for a bunch of the main terms. Still, you
>say the number one problem is that students don't understand all the
>definitions. Is my understanding of "independence" then not good enough? Or
>is it?

As others have said, if the phrasing is a problem, then you don't
understand the definition. That can be because you don't have the concept,
or because you can't say things coherently. The problem is that, to
do math, you *have* to say things coherently. Precise language is
crucial. Fuzzy concepts are simply not good enough. Let's look at your
definition.

>Let me try that one...independence means a group of vectors (in
>homogenous form???) such that if they all equal the zero vector, then
>the only possible way for that is the each coefficient of every vector
>has to equal 0 too.

I'm going to pick this apart so you can see how language gets used.

>independence means a group of vectors

No. 'We say that a set of vectors is independent when....'

As you said it, you imply that independence *is* a set of vectors, but
it isn't. A set of vectors can have the property of independence. We will
be defining what it takes for a collection of vectors to have that
property.

>(in homogenous form???)

This certainly suggests some lack of understanding. The set of vectors
is in homogeneous form? A *set* cannot be in homogeneous form. An
algebraic expression can be. Be careful exactly what it is that has a form.

>such that if they all equal the zero vector

All the vectors are equal to 0? Are you sure you want each and every one
of the vectors to be 0? In other words, there is only one vector in the
set? As you have said it, you mean that the only independent set of
vectors is the set with only the zero vector. Is that what you want?
I suspect not.

>then
>the only possible way for that is the each coefficient of every vector
>has to equal 0 too.

What coefficient? You didn't mention or introduce any coefficients! You
only have some vectors which are all 0!

Let's look at a real definition:

A set of vectors {v_1 , ...v_n} is independent if the only way for
a linear combination c_1 v_1 +...c_n v_n to be the 0 vector is when
c_1 = ...=c_n=0.

Some things to notice:

*If* c_1 = ...=c_n =0, then for *any* collection of vectors {v_1, ..v_n},
independent or not, c_1 v_1 +...c_n v_n=0. Do you see why? So it is
possible for a linear combination to be 0, yet the vectors in
that linear combination not to be independent.

The definition is precise and checkable: If you can find c_1,...,c_n
that are not all 0 such that c_1 v_1 +...c_n v_n =0, then the set
{v_1 ,...v_n} is *not* independent. Do you see why? If you can show
that no such collection of c_i exists, then the set *is* independent.
Do you see why?

More importantly, do you see the difference between your definition and
the one I gave?

--Dan Grubb

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