A is $2×2$ matrix s.t. $\forall x \in R^2, Ax$ is vector obtained from projecting $x$ to $2x_1-x_2=0$, give A satisfying the condition as stated above
What I've done so far:
$A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$, $x=\begin{bmatrix}y_1\\y_2\end{bmatrix}$ which means we can write the following:$\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}y_1\\y_2\end{bmatrix}=\begin{bmatrix}ay_1+by_2\\cy_1+dy_2\end{bmatrix}$
projection=$\frac{2y_1-y_2}{y_1^2+y_2^2}<y_1,y_2>$
However, at this point, I feel like I'm understanding the direction in which I should go incorrectly.
If $u$ is a unit vector, then $f(x) := (u^\top x) u = (uu^\top) x$ is the projection onto the span of $u$ (a one-dimensional subspace). So all you need to do is find a unit vector that lies on the line $\{(x_1, x_2) : 2x_1 - x_2=0\}$.