Find the axis of symmetry of $A = \{(x,y) \in \mathbb{R^2}\vert f(x,y)=0\}$ for this linear operator.

Question

Say $A = \{(x,y) \in \mathbb{R^2}\vert f(x,y)=0\}$ for the $f(x,y) = 4x^2 - 4xy+y^2-{\sqrt5} x - 2{\sqrt5}y+20$

Linear operator $T : \mathbb{R}^2 \to \mathbb{R^2}$, $T(x) = x- 2\frac{<x,u>}{<u,u>}u$

Find the unit vector $u\in \mathbb{R^2}$ in the 2nd quadrant satisfying $A=\{T(x) \vert x\in A\}$.

From here, my solution starts. I plot the graph of the $A$ having new coordinates $x'=\frac{1}{\sqrt5}\begin{bmatrix} 1 \\2 \\\end{bmatrix}$ and $y'=\frac{1}{\sqrt5}\begin{bmatrix} -2 \\ 1\end{bmatrix}$

The $x'$ and $y'$ are eigenvectors of the $5$ and $0$ respectively. I got the $(y')^2=x'-4$ (The green axis is $x'$ and blue axis is $y'$)

Let me consider the vector $v$ perpendicular to the $u$. Then $T(u) = -u$ and $T(v) =v$. So the question requires the vector axis of symmetry lied on the 2nd quadrant. But in my thought, this is "nonsense" because the graph is symmetric for the green(in the 1st quadrant) not the blue(in the 2nd quadrant). Moreover the answer is $\frac{1}{\sqrt5}\begin{bmatrix} -2 \\ 1\end{bmatrix}$. I can't understand why the answer should be like that.

Thanks and regards.

Answer 1

Note that the linear transformation $T$ is not a reflection. If you compare this transformation with e.g. this, you will notice that there's a sign difference. In other words, the map $T$ not only reflects along the green line but also changes the sign. It is this change of sign that makes $y'$ the right answer.

As a small aside, it is a bit confusing to use $x$ and $y$ first as coordinates of a point in $\mathbb{R}^{2}$, so as elements of $\mathbb{R}$, and then use $x$ as a vector in $\mathbb{R}^{2}$.