
Re: Failing Linear Algebra:
Posted:
Apr 29, 2004 4:15 PM


Russell:
>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;
Oops. I didn't rotate in the wrong direction. I had a total brain fart and forgot which point comes first. Right...what I meant to say was that (1,0) rotates to (0,1).
>The vector (0,1) transforms to (1,0);
Right. Stupid mistake.
Still, I don't get how this result is written in matrix form. How would the mapping theta (pi/2) be represented in matrix and basis form?
>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.
OK, so this, I guess:
1 0 0 1 1 0 0 1
But, the second two are just multiples of the first two. So, they're unnecessary in a basis, right? So, the basis would just be: {(1,0), (0,1)} for a 90 degree rotation?
Question: if the rotation was by (pi/4) instead, would the basis have to be: {(1,0),(radical 2/2, radical 2/2), (0,1), (radical 2/2, radical 2/2)}?
Another area where I'm having some problems is with projection mappings and matrices.
>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.
Why would the first xvalue be negative though?
>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?
0 1 1 0
times (2, 1) =
(0 1 , 2  0) = (1, 2). Got it!
So this 90 degree rotation basis has to work for ANY unit vector then, no matter the direction it points in?
>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?
Yup, I can visualize that. They're 90 degrees apart. Wait...something else just hit me. (1, 2) isn't a unit vector. How come our 90 degree rotation basis works anyways? Would it work for a vector of any length?
>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 think I just learned how to do that yesterday. You convert the matrix into the identity matrix, and whatever has happened to the original identity matrix as a result of the conversions equals the inverse of the original matrix. I.E., you convert A into A^(1) using I. You can check that A*A^(1) = I, and then you know you're right.
With A^(1), you multiply one side of the standard basis by A^(1) and the other side by A, and that gives you basis alpha in terms of basis beta. Right? I'm a little hazy here, I think. But do I have the geneal idea down?

