# Thread: y=2x mapped onto itself under transformation T

1. ## y=2x mapped onto itself under transformation T

$\displaystyle \begin{pmatrix} k & -2\\1-k & k \end{pmatrix}$, where k is constant
A transformation T:R^2 arrow R^2 is represented by the matrix A.
Find the value of k for which the line y=2x is mapped onto itself under T.

What I did:
$\displaystyle \begin{pmatrix} k & -2\\1-k & k \end{pmatrix}$$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$=$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$
implies $\displaystyle \begin{pmatrix} x(k-4) \\x(1+k) \end{pmatrix}$=$\displaystyle \begin{pmatrix} x\\2x \end{pmatrix}$
Therefore k=5 and k=1 !!
What I should do:
x(1+k)=2x(k-4)
k=9
Where is my principle wrong? Please point it out, because i fear its something fundamental and i may have to suffer direly for it later.
edit:k=5, k=1

2. Hi,
your principle is wrong because by = you're saying that a point (x,2x) must map to itself. This condition is too strong and as you showed it has no solution for k. What you have to consider is $\displaystyle \begin{pmatrix} k & -2\\1-k & k \end{pmatrix} \begin{pmatrix} x\\2x \end{pmatrix}=\begin{pmatrix} w\\2w \end{pmatrix}$ (because (x,2x) can be mapped to arbitrary point (w,2w))
and after eliminating w from two simultaneous equations you get the right result (or you get necessary condition for k, and by linearity you see that it is also sufficient)