1. ## 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.

2. 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:

$T= \begin{pmatrix}\frac{1}{17} & \frac{4}{17} \\ \frac{4}{17} & \frac{16}{17}\end{pmatrix}$

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.

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