Question

3. Let L be the linear transformation on R2 that reflects each point P across the line y kx (k>0) a) (2 marks) Show that v and v2 - 1 are eigenvectors of L. b) (1 mark) What is the eigenvalue corresponding to each eigenvector? (Hint: No need to calculate the characteristic polynomial or solve a matrix equation. Geometric reasoning should suffice to solve this problem. Drawing a diagram is recommended!)

Answer 1

The point lies on $y=kx$ so it remains unaffected by the transformation

That is, $T(1,k)=(1,k)$ so it is an eigenvector with eigenvalue 1

The point doesn't lie on our graph

Equation of the line perpendicular to the given line and passing through the given point is:

$y-1=-\frac{1}{k}(x-(-k))\Rightarrow y-1=-\frac{1}{k}(x+k)$

Solving $y-1=-\frac{1}{k}(x+k)$ and $y=kx$ we get

$kx-1=-\frac{1}{k}(x+k)\Rightarrow k^2x-k=-(x+k)\Rightarrow (k^2+1)x=0\Rightarrow x=0$

And

y=0

So the intersection point is

(0,0)

The reflected point is one which lies on $y-1=-\frac{1}{k}(x+k)$ and which has same distance from
(0,0)
as the point

Which is to say it is the point
A, B)
where $|B-kA|=|b-ka|\Rightarrow B-kA=-(1-k(-k))=-(1+k^2)$ (since we want a different point)

Along with $B-1=-\frac{1}{k}(A+k)$ so that

$B=1-\frac{1}{k}(A+k)=-\frac{A}{k}$

Thus, $B-kA=-\frac{A}{k}-kA=-(k^2+1)A/k=-(1+k^2)\Rightarrow A=k\Rightarrow B=-1$

So the reflected point is $(A,B)=(k,-1)$ which is note is

Thus, is also an eigenvector which eigenvalue