Have you found the eigenvalues and eigenvectors for A? Remember that if is an eigenvalue and v a corresponding eigenvector, then so that if , any line in the v direction is "stretched" by a factor of . If , it is "squeezed" by that factor. If 0 is an eigenvalue (which is true if and only if det(A)= 0), the entire line in the direction of v is collapsed to 0 and, in fact, the entire plane is collapsed to the line perpendicular to v.

A linear function has the nice property that it maps lines to lines. That is, determine what T(1, 2), T(2, 0), and T(0, 0) are and draw the lines between them.2.a)Given is linear and T(1,2) = (2,4) and T(2,0) = (0,0), graph T(F) (the image of F under T), where F is the set of points in located on the boundary of the triangle determined by the points (1,2), (2,0), and (0,0).

(I don't know how one would go about graphing so maybe some guidance as to how I'd do so for this problem would be nice)

I don't open ".doc" attachments.b)Graph T(G), where G is the set of points strictly inside the triangle F. (In other words, G is the set of points not in F, but bounded by F.) [u](same applies for this problem)

.doc attached with those 2 questions along with one more.

Thanks again.