For the first one, x+ y= 1, C is the line segment between A and B. For the second, C is the line segment from (0, 0) to (3, 4).
If x+y = 1, then we are talking about the points (x,1-x) = (0,1) + x(1,-1). For each such point, we get the point in C:
x(2,1) + (1-x)(1,3) = (2x,x) + (1-x,3-3x) = (x+1,3-2x) = (1,3) + x(1,-2), which may be more familiar to you as the line y = 5 - 2x, which runs through (2,1) and (1,3).
This should make sense: C is a linear map, and thus should map lines to lines.
In a similar vein, C maps the unit square [0,1] x [0,1] to the parallelogram bounded on one side by the vector that runs from 0 to A (the image of the vector from (0,0) to (1,0) on the x-axis), on another by the vector from 0 to B (the image of the vector from (0,0) to (0,1) on the y-axis) on the third side by the vector that runs from A to A+B, and on the fourth side by the vector that runs from B to A+B.
Intutively, C takes the vector from (0,0) to (x,0) to the vector from (0,0) to (2x,x), and the vector from (0,0) to (0,y) to the vector from (0,0) to (y,3y) mapping a rectangular grid to a skewed one of tiled parallelograms.
thanks very much!!!
b)in what sense are these affine combinations since x+y=1?
c)how do you obtain the original square [0,1]X[0,1]? so if x is fixed at x=0, C ranges from (0,0) to (1,3)=0B~is mapped from segment (0,0) to (1,0), if x=1 then C ranges from (2,1) to (3,4)~(1,0) to (1,1), AD, if y fixed at y=0, then C ranges (0,0) to (2,1) 0A~(0,0) to (1,0), y fixed at y=1, then C ranges from (1,3) (3,4) BD~ (0,1) to (1,1). and then x or y can take values like 1/2 which result in parallel line segments like x=1/2 results in a segment parallel to 0B and joining midpoints 0A to BD. so the solution is the whole area of the parallelogram with sides included?
"Intutively, C takes the vector from (0,0) to (x,0) to the vector from (0,0) to (2x,x), and the vector from (0,0) to (0,y) to the vector from (0,0) to (y,3y) mapping a rectangular grid to a skewed one of tiled parallelograms."
I'm sorry I don't follow the mapping.... I can see that if y=0 is fixed then C ranges from (0,0) to (2x,x) or if x=0 fixed then C ranges from (0,0) (y,3y)
the problem statement implies that holding x fixed y ranges from 0 to 1 over the reals so the original map is the whole square [0,1]X[0,1] which is a mapped linearly to the parallelogram kind of like a simple linear mapping of a Jacobian integral, higher powers of the Jacobian are used for physics. Soooo you can also map surfaces and transform them in R3 R4, Rn? I wonder if you can map spheres to wrinkled spheres for instance and create other transformations. another problem: c) A=(1,1,1,0), B=(0,1,1,1), C=(1,1,0,0), D=xA+yB+zC, find xyz such that D=(1,5,3,4). y=4, x=-1, z=2. d) prove that no choice of x,y,z can equal D=(1,2,3,4), y=4, x=-1, z=2, -1+4=2=5 does not=2