
Vector Transformations.
Hi everyone. I'm a little confused by this math problem:
The line transformation T fixes the line y = 4x pointwise and maps every point on the line x + 4y = 0 to the origin. What are the eigenvalues and corresponding eigenvectors of the matrix A of T?
I'm pretty sure I know how to calculate eigenvalues and their corresponding eigenvectors using the characteristic equation, but I don't know how I'm suppose to find "matrix A of T" or even what the resulting transformation is suppose to look like.
Any help would be greatly appreciated.

It is not necessary to find the matrix representing T. "T fixes the line y= 4x pointwise" means that if T does not change points on that line; if y= 4x, then T(x, y)= (x, y) or T(x, 4x)= (x, 4x)= 1(x, 4x). That Tells you that 1 is an eigenvalue with eigenvector (1, 4). Saying that T " maps every point on the line x + 4y = 0 to the orign" means that T(4y, y)= (0, 0)= 0(4y, y) and that tells you that 0 is an eigenvalue with eigen vector (4, 1).
But, in fact, it is not hard to find the the matrix representing T. The two lines, y= 4x, and x= 4y are perpendicular (their slopes are 4 and 1/4) so we could use (1, 4) and (4, 1) as basis vectors every vector, (x, y), can be written as a linear combination of (1, 4) and (4, 1). (x, y)= a(1, 4)+ b(4, 1)= (a 4b, 4a+ b) so a 4b= x and 4a+ b= y. Solve for a and b: 17a= x+ 4y so a= (x+ 4y)/17 and 17b= y 16x so b= (y 16x)/17. In particular, (1, 0)= 1/17(1, 4) 16/17(4, 1) and (0, 1)= 4/17(1, 4)+ 1/17(4, 1).
T(1, 0)= 1/17 T(1, 4) 16/17(4, 1)= 1/17 (1, 4)= 1/17(1, 0)+ 4/17(0, 1) and T(0, 1)= 4/17 T(1, 4)+ 1/16 T(4, 1)= 4/17 T(1, 4)= 4/17(1, 0)+ 16/17(0, 1). The coefficients in those equations give the columns of the matrix representation of T:
$\displaystyle T= \begin{pmatrix}\frac{1}{17} & \frac{4}{17} \\ \frac{4}{17} & \frac{16}{17}\end{pmatrix}$
But, as I said, you don't need to find that matrix representation the information given tells you directly what the eigenvalues and eigenvectors are.

Ah, I see now. Thank you very much! I can see why you are rated MathHelp Forum's most helpful member of 2009!