QUESIION 17
One-to-One Linear Operator
Choose all that apply $\cdot 5$ points
Select the linear operators that are one-to-one
$T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}y \\ x\end{array}\right)$
$T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}0 \\ x+y\end{array}\right)$
$T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}x+y \\ x-y\end{array}\right)$
$T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}2 x+y \\ 6 x+3 y\end{array}\right)$

Step 1 ：Given four linear operators, we need to identify which of them are one-to-one. A linear operator is one-to-one (or injective) if each element of the domain maps to a unique element of the codomain. In other words, no two different elements in the domain have the same image in the codomain.

Step 2 ：Let's analyze each operator one by one.

Step 3 ：First operator: $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}y \\ x\end{array}\right)$. This operator simply swaps the x and y coordinates. It is one-to-one because for every unique (x, y) in the domain, there is a unique (y, x) in the codomain.

Step 4 ：Second operator: $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}0 \\ x+y\end{array}\right)$. This operator is not one-to-one. The first coordinate in the codomain is always 0, regardless of the input. This means that different inputs can result in the same output. For example, (1, 2) and (3, -2) both map to (0, 3).

Step 5 ：Third operator: $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}x+y \\ x-y\end{array}\right)$. This operator is one-to-one. For every unique (x, y) in the domain, there is a unique (x+y, x-y) in the codomain.

Step 6 ：Fourth operator: $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{c}2 x+y \\ 6 x+3 y\end{array}\right)$. This operator is not one-to-one. The second coordinate is simply three times the first coordinate, so different inputs can result in the same output. For example, (1, 1) and (2, -1) both map to (3, 9).

Step 7 ：Final Answer: The one-to-one linear operators are $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}y \\ x\end{array}\right)$ and $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ such that $T\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}x+y \\ x-y\end{array}\right)$.