
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 >>>added together in any way, >> >>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 addition to* those above.
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 7th9th 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 90degree 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 yvector rotates counterclockwise to a unit vector in the x direction. Since this yvector 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 xvector 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 2tuple. 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/2rotation 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 what they are talking about, when you read the relevant passages.
> >? > > >>(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 >>your prof does it too, > >Yes.
Some books do it the other way; don't worry about that now, stick to the convention used in your class.

