Find all vectors $(x,y)$ whose image under rotation $\frac\pi3$ is $(y,x)$ Find all vectors (x,y) whose image under rotation through the angle $\pi/3$ about the origin is (y,x). 
Using the appropriate rotation matrix I found that x=y. What do I need to do now to find the answer? 
 A: Let $(x’,y’)$ be the vector after the rotation. Then,
$$x’=y=x\cos\frac\pi3 -y\sin\frac\pi3$$
which leads to 
$$\frac yx=\frac{\cos\frac\pi3}{1+\sin\frac\pi3}
=\frac{\sin\frac\pi6}{1+\cos\frac\pi6} 
= \frac{2\sin\frac\pi{12}\cos\frac\pi{12}}{2\cos^2\frac\pi{12}} 
=\tan\frac\pi{12}$$
Thus, all the vectors that satisfy the requirement are
$(x,y)=t(1,\tan\frac\pi{12})$.
A: Algebraic approach: Rotating a vector by $\pi/3$ corresponds to multiplying by the matrix $$A=\begin{bmatrix}\cos\pi/3&-\sin\pi/3\\\sin\pi/3&\cos\pi/3\end{bmatrix}=\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}$$
And flipping the coordinates corresponds to multiplying by the matrix $$B=\begin{bmatrix}0&1\\1&0\end{bmatrix}$$So given a vector $v$, we want $Av=Bv$, which is to say $(A-B)v=0$. So that's what you have to do: Find the kernel of $A-B$.
Geometric approach: Flipping the coordinates means mirroring across the line $x=y$. So now you have to think: Which vectors end up in the same place under the reflection as they do under the rotation?
