*Question 1* $T(\mathbf{v})=\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix}\mathbf{v}+\begin{pmatrix}0 \\ 0\end{pmatrix}$ $T(\mathbf{v})=\begin{pmatrix}-1 & 0\\0 & -1\end{pmatrix}\mathbf{v}+\begin{pmatrix}4 \\ 4\end{pmatrix}$ $T(\mathbf{v})=\begin{pmatrix}-1 & 0\\0 & 1\end{pmatrix}\mathbf{v}+\begin{pmatrix}4 \\ 0\end{pmatrix}$ $T(\mathbf{v})=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}\mathbf{v}+\begin{pmatrix}0 \\ 4\end{pmatrix}$ *Question 2* For all $x,y\in X,$ $\begin{align} f(x,x)&\leq f(x,y)+f(y,x) & \text{since $f$ satisfies triangle inequality} \\ & \leq 2f(x,y) & \text{since $f$ is symmetric} \\ 0 &\leq 2f(x,y) & \text{since $f$ is non-degenerate} \end{align}$therefore $f(x,y)\geq 0.$ $\square$ Let $x,y \in \mathbb{R}.$ Define $f(x,y)=\min(|x-y|,1).$ Then $f$ is a metric on $\mathbb{R}$ since: 1. $f(x,y)=0\implies |x-y|=0\implies x=y$ hence $f$ is **non-degenerate**. 2. $f(x,y=\min(|x-y|,1)=\min(|y-x|,1)=f(y,x)$ hence $f$ is **symmetric**. 3. Since for all $u,v\in \mathbb{R},$ $|u-v|\leq|u|+|v|,$ $\begin{align} \min(|u-v|,1) &\leq \min(|u|+|v|, 1) \\ &\leq \min(|u|,1) + \min(|v|,1) \tag{1} \end{align}$*(Note here that I use the fact that $\min(a+b,1)\leq \min(a,1)+\min(b,1)$ where $a,b\geq{0}$ which you can justify by a simple case by case analysis)*. Let $z\in \mathbb{R}.$ Substituting $u$ and $v$ for $x-z$ and $y-z$ respectively in $(1)$ gives $\min(|x-y|,1)\leq \min(|x-z|,1)+\min(|y-z|,1)$That is, $f$ satisfies the **triangle inequality**. Since $f(x,y) \leq {1},$ $f$ is not surjective onto $[0,\infty).$ $\square$ Suppose $X$ contains at least two distinct elements. Choose distinct $x,y\in X,$ then $(x,y)\neq (y,x)$ but $f(x,y)=f(y,x)$ since $f$ is symmetric. Hence, it is not possible for $f$ to be injective on any such $X.$ However, if $|X|\leq 1$ then $f$ is certainly injective. $\square$ *Question 3* Note that $\delta((x_{1}, y_{1}), (x_{2}, y_{2})) = d((x_{1}, \sqrt{2} y_{1}), (x_{2}, \sqrt{2} y_{2}))$where $d$ is the standard Euclidean metric on $\mathbb{R}^2.$ We need to show that $\delta$ satisfies the three properties of metrics: 1. **Non-degenerate**: $\delta((x_{1}, y_{1}), (x_{2}, y_{2})) = 0 \implies d((x_{1}, \sqrt{2} y_{1}), (x_{2}, \sqrt{2} y_{2}))=0.$ Since $d$ is non-degenerate, this implies $(x_{1},\sqrt{ 2 }y_{1})=(x_{2},\sqrt{ 2 }y_{2})\implies(x_{1},y_{1})=(x_{2},y_{2}).$ 2. **Symmetry**: $\begin{align}\delta((x_{1}, y_{1}), (x_{2}, y_{2})) &= d((x_{1}, \sqrt{2} y_{1}), (x_{2}, \sqrt{2} y_{2})) \\ &= d( (x_{2}, \sqrt{2} y_{2}),(x_{1}, \sqrt{2} y_{1})) & \text{by symmetry of $d$}\\&= \delta((x_{2},y_{2}), (x_{1},y_{1})) \end{align}$ 3. **Triangle inequality**: Let $(x_{3},y_{3})\in \mathbb{R}^2$ $\begin{align}\delta((x_{1}, y_{1}), (x_{2}, y_{2})) &= d((x_{1}, \sqrt{2} y_{1}), (x_{2}, \sqrt{2} y_{2})) \\ &\leq d((x_{1},\sqrt{ 2 }y_{1}),(x_{3},\sqrt{ 2 }y_{3}))+d((x_{3},\sqrt{ 2 }y_{3}),(x_{2},\sqrt{ 2 }y_{2})) & \text{since $d$ satisfies T.I.} \\ &= \delta((x_{1},y_{1}),(x_{3},y_{3}))+\delta((x_{3},y_{3}),(x_{2},y_{2})) \end{align} $ Hence $(\mathbb{R}^2,\delta)$ is a metric space. $\square$ *Question 4* Define $f:L\to \mathbb{R}$ by $f(x,y)=\frac{\sqrt{ 5 }}{5}x+\frac{2\sqrt{ 5 }}{5}y$ (rotate $L$ about $O$ by $\tan^{-1}(2)$ clockwise and take its $x$ coordinate). Define $g:\mathbb{R}\to L$ by $g(r)=(x,1+2x)$ where $x$ satisfies $\frac{\sqrt{ 5 }}{5}x+\frac{2\sqrt{ 5 }}{5}(1+2x)=r.$ It is straightforward to check that $g$ and $f$ inverses of each other and so $f$ is bijective. Let $(x_{1},y_{1}),(x_{2},y_{2})\in L$ then $\begin{align} d_{1}(f(x_{1},y_{1}),f(x_{2},y_{2}))&= \left\lvert \frac{\sqrt{ 5 }}{5}x_{1}+\frac{2\sqrt{ 5 }}{5}y_{1} - \frac{\sqrt{ 5 }}{5}x_{2}-\frac{2\sqrt{ 5 }}{5}y_{2} \right\rvert \\ &= \left\lvert \frac{\sqrt{ 5 }}{5}x_{1}+\frac{2\sqrt{ 5 }}{5}(1+2x_{1}) - \frac{\sqrt{ 5 }}{5}x_{2}-\frac{2\sqrt{ 5 }}{5}(1+2x_{2}) \right\rvert \\ &= \sqrt{ 5 }\; \lvert x_{1}-x_{2} \rvert \\ &= \sqrt{ 5(x_{1}-x_{2})^2 } \\ &= \sqrt{ 5(x_{1}^2 -2x_{1}x_{2}+x_{2}^2) }\\ &= \sqrt{ x_{1}^2 - 2x_{1}x_{2}+x_{2}^2 + 4x_{1}-8x_{1}x_{2} + 4x_{2}^2 } \\ &= \sqrt{(x_{1}-x_{2})^2 + (2x_{1}-2x_{2})^2 } \\ &= \sqrt{ (x_{1}-x_{2})^2 + (y_{1}-y_{2})^2} = d((x_{1},y_{1}),(x_{2},y_{2})) \end{align}$Hence $f$ is indeed an isometry. $\square$