Computing the matrix for a linear transformation

Question

For a point $(x, y)$ in the plane $\mathbb{R}^{2}$, define $T(x, y)$ to be the point on the line $\ell = \{(x, y) \in \mathbb{R}^{2} \mid y = 2x\}$ that is closest to $(x, y)$. Show that the mapping is linear, and find the $2\times 2$ matrix associated with this mapping.

I know how to prove a function is linear, but I'm not sure about how to find this matrix. I tried plugging in "test points" to determine values of the function. For example, $(0, 0)$ should map to $(0, 0)$. Can someone help me please?

Answer 1

Hint: You are given a point $(x,y)$ and you want to find the point $(w,2w)$ closest to it. In other words, you want to pick $w$ such that $$ \left\Vert (w,2w)-(x,y)\right\Vert ^{2} =\left(w-x\right)^{2}+\left(2w-y\right)^{2} $$ is minimized. To do so, take the derivative of the above with respect to $w$, set it to zero, and solve for $w$. What do you see?

Answer 2

From your description, I think every point $\left( x, y \right)$ in $\mathbb{R}^2$ is mapped to the point $\left( \dfrac{x + 2y}{5}, \dfrac{2x + 4y}{5} \right)$ on the line $l$. So, $\left( 1, 0 \right)$ is mapped to $\left( \dfrac{1}{5}, \dfrac{2}{5} \right)$ and the point $\left( 0, 1 \right)$ is mapped to $\left( \dfrac{2}{5}, \dfrac{4}{5} \right)$.

Therefore, the required matrix would be $\left[ \begin{matrix} \frac{1}{5} & \frac{2}{5} \\ \frac{2}{5} & \frac{4}{5} \end{matrix} \right]$.

To obtain the formula for $T$

Let $\left( x_1, y_1 \right) \in \mathbb{R}^2$ be given. We wish to find a point $\left( x, 2x \right)$ on the line $l$ such that the distance between $\left( x_1, y_1 \right)$ and $\left( x, 2x \right)$ is minimum. It is same as saying that we have to minimize the function $f \left( x \right) = \left( x - x_1 \right)^2 + \left( 2x - y_1 \right)^2$. To find a stationary point, we use $f' \left( x \right) = 0$. This gives, $x = \dfrac{x_1 + 2y_1}{5}$. Now, $f'' \left( x \right) = 10 > 0$. Hence, $f$ is minimum at $\dfrac{x_1 + 2y_1}{5}$. Thus, the point $\left( x_1, y_1 \right)$ in the plane $\mathbb{R}^2$ is mapped to the point $\left( \dfrac{x_1 + 2y_1}{5}, \dfrac{2x_1 + 4y_1}{5} \right)$. Hence,

$$T \left( x, y \right) = \left( \dfrac{x + 2y}{5}, \dfrac{2x + 4y}{5} \right)$$