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

 Messages: [ Previous | Next ]
 Russell Blackadar Posts: 586 Registered: 12/12/04
Re: Failing Linear Algebra:
Posted: Apr 28, 2004 9:30 PM

On 28 Apr 2004 22:23:59 GMT, Anonymous wrote:

>Russell:
>

>>>It's a group of vectors that can be multiplied by any scalar and/or

>>
>>Not just any way -- there are precise conditions that the sum must
>>satisfy. Look for the axiomatic definition in your book(s). It's not
>>at all difficult to memorize. Do it.

>
>I already have memorized it. The space has to be closed under addition and
>scalar mult, and contain the zero vector. Therefore,

Not therefore -- some of the properties you list below are *in

it satisfies those 8
>properties:
>
>*v* + *w* = *x* (in the space)
>c*v* = c*v* (in the space)
>0*v* = 0
>1*v* = *v*
>
>*v* + *w* = *w* + *v*
>(*v* + *w*) + *x* = *v* + (*w* + *x*)
>*v* - *v* = 0
>c(s*v*) = (cs)(*v*)

Good. But you also need c(v+w) = cv + cw.

>
>>The algebra is the important thing, not the picture. And your current
>>understanding of the algebra is insufficient to keep you from getting
>>confused on the upcoming test.

>
>But, I did so well in algebra in 7th-9th grades. How much different can this
>be?

By that I meant only that you need to focus more on the equations,
symbols, etc. and worry less about the geometrical interpretations;
you seem to have an insufficiently abstract notion of what a vector
is. I see you've made a good start by listing axioms instead of
talking vaguely about planes and solids.

Regarding why this is harder, who really knows. But as (I think)
Prof. Ullrich said, this is typically the course where students who
are "good at math" first start having real difficulty. The habits
that you learn in passing this course will stand you in good stead in
the future. Even though (as Acid Pooh said) it won't get easier, one
of the things you are struggling with now is attitude, and once you
win that battle, you won't have as much difficulty with *that* as you
are having now. Everyone has to get over the hurdle sooner or later;
it's called mathematical maturity.

> I've got the definitions well enough: I got a B- on the first exam. It's
>the *concepts* since image/kernel/basis that have confused me, like I said.

Well, what folks are telling you here is that if you find yourself
confused, and you can't quickly write out a precise definition of the
thing you are working on, then *that* is the place to start getting
yourself unconfused. You flunked the initial test here in sci.math,
but you are starting to make a good comeback.

>
>>>I guess I could also really use some help with understanding how a
>>>mapping gets converted into a matrix, and then how to solve it.

>>
>>I like the term "linear transformation" and I think you should use it
>>too;

>
>OK. See? I never knew what that term meant. Now I do.
>

>>Do you know
>>what restrictions I'm talking about? f(a+b) = f(a)+f(b) and
>>f(ca) = cf(a) of course. That, by definition, is what makes the
>>mapping *linear*.

>
>Right. And also the 0 vector.

No, I think you aren't getting it here. The zero vector is a member
of a *linear space* -- i.e. a vector space. It has nothing really to
do with a *linear transformation*, which is a mapping between spaces.
The word "linear" appears in both phrases, and for similar reasons,
but they are not the same.

>
>>Try working it out for the 90-degree rotation I mentioned
>>above, and then try your matrix out on some 2D column vectors to see
>>if they really do turn 90 degrees when you multiply by the matrix.

>
>If something is rotated 90 degrees, the first point, cosine, goes from 1 to 0.
>And sine goes from 0 to 1. I still don't really get how to represent that in a
>matrix.

First of all, sine and cosine aren't points; instead, they are (or
rather they give) the two components of a *single* point (aka vector)
on the unit circle. They are indeed relevant to this problem, but
let's not think in terms of sines and cosines just yet.

Think basis vectors. What is the basis we are using here?

>
>Theta(pi/2) of (0,1) becomes (1,0).

I think you've rotated in the wrong direction; positive pi/2 is a
counterclockwise rotation. The vector (0,1) transforms to (-1,0);
that is, a unit y-vector rotates counterclockwise to a unit vector in
the -x direction. Since this y-vector was our second basis vector, we
now have the *second* column of our matrix. But we still need to find
the first column. Hint: how does a unit x-vector transform in a pi/2
(counterclockwise) rotation?

> Does that mean the matrix is just:
>
>/ 1>| 0|
>\ /
>2X1

Our vector space is R^2, so there are *two* basis vectors, and each of
them is a 2-tuple. So the matrix should be 2x2. Hopefully you have
worked the matrix out in your head by now, so I'll write it down:

0 -1
1 0

That is, the first column is a unit in the y direction, and the second
column is a unit in the -x direction.

We're done; but it never hurts to check our result. (And hopefully,
the following will give you some insight.) Let's see if this matrix
works for some arbitrary vector, say, (2,1). Express that vector as a
column vector, multiply by our matrix, and you get the answer (-1,2)
expressed as a column vector, right? Plot the two points (2,1) and
(-1,2) on some graph paper and draw in the lines from the origin; what
angle do you see? We have just demonstrated that multiplying by our
matrix has the effect of rotating our (arbitrary) vector by pi/2.

In other words, our matrix expresses the pi/2-rotation linear
transformation in the basis {(1,0),(0,1)}. A different basis would
give a different matrix for the same linear transformation. In fact
one of the things you may be asked to do is, given a matrix in one
basis, compute the matrix for the same linear transformation in a
different basis. I won't go into that here; it's in all of your books
and hopefully you have enough insight now to be able to understand

>
>?
>
>

>>(I am assuming you multiply with matrix on the left and column vector
>>on the right, as is done in Schaum's outline; hopefully that is how

>
>Yes.

Some books do it the other way; don't worry about that now, stick to
the convention used in your class.

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