For proving the linear transformation, use that the following properties:

\(T(\alpha x + \beta y) = \alpha T (x) + \beta T (y)\)

and \(T(\alpha x) = \alpha T(x)\)

where alpha and beta are the scalars.

(d)Given that,

\(x = r \cos \theta\)

\(y = r \sin \theta\)

and \(\overrightarrow{T}(x,y)=(r \cos(\theta+\varphi), r \sin(\theta+\varphi))\)

for some constant \(\varphi\)

Now showing in below T is linear transformation,

\(T(\alpha x_{1}(x,y)+\beta x_{2}(x,y))=[\alpha\{r_{1} \cos(\theta= \varphi), r_{1} \sin (\theta=\varphi)\}+\beta\{r_{2} \cos(\theta=\varphi), r_{2} \sin (\theta=\varphi\}]\)

\(= \alpha T (r_{1} \cos(\theta = \varphi), r_{1} \sin(\theta = \varphi)) + \beta T (r_{2} \cos(\theta = \varphi), r_{2} \sin(\theta = \varphi))\)

\(= \alpha T (x_{1} (x, y)) + \beta T (x_{2}(x, y))\)

and \(T(\alpha(x, y)) = {\alpha r \cos(\theta = \varphi), \alpha r \sin(\theta = \varphi)}\)

\(= \alpha (r \cos(\theta = \varphi), r \sin(\theta = \varphi))\)

\(= \alpha T (x, y)\)

Hence, rotation is linear transformation. (f) Given that a fixed vector r and T maps each point to its reflection with respect to vector r, \(\overrightarrow{T}(\overrightarrow{x})=\overrightarrow{x}-2 \overrightarrow{x} r\)

\(=2\overrightarrow{x}_{r}-\overrightarrow{x}\)

Now showing in below T is linear transformation,

\(T(\alpha x + \beta y) = (\alpha x + \beta y) - 2 (\alpha x + \beta y)_{\varphi}

\(= \alpha x + \beta y - 2 \alpha x_{\varphi} - brta y_{\varphi}\)

\(=(\alpha x - 2 \alpha x_{\varphi}) + (\beta y - \beta y_{\varphi})\)

\(= \alpha (x - 2x_{\varphi}) + \beta (y - y_{\varphi})\)

\(= \alpha T (x) + \beta T (y)\)

and \(T (\alpha x) = \alpha x - 2(\alpha x)_{\varphi}\)

\(= \alpha x - 2 \alpha x_{\varphi}\)

\(= \alpha (x- 2x_{\varphi})\)

\(= \alpha T (x)\)

Hence, reflection is linear transformation.

Therefore, above satisfies both properties of linear transformation.

Hence, both rotation and reflection are linear transformation.

\(T(\alpha x + \beta y) = \alpha T (x) + \beta T (y)\)

and \(T(\alpha x) = \alpha T(x)\)

where alpha and beta are the scalars.

(d)Given that,

\(x = r \cos \theta\)

\(y = r \sin \theta\)

and \(\overrightarrow{T}(x,y)=(r \cos(\theta+\varphi), r \sin(\theta+\varphi))\)

for some constant \(\varphi\)

Now showing in below T is linear transformation,

\(T(\alpha x_{1}(x,y)+\beta x_{2}(x,y))=[\alpha\{r_{1} \cos(\theta= \varphi), r_{1} \sin (\theta=\varphi)\}+\beta\{r_{2} \cos(\theta=\varphi), r_{2} \sin (\theta=\varphi\}]\)

\(= \alpha T (r_{1} \cos(\theta = \varphi), r_{1} \sin(\theta = \varphi)) + \beta T (r_{2} \cos(\theta = \varphi), r_{2} \sin(\theta = \varphi))\)

\(= \alpha T (x_{1} (x, y)) + \beta T (x_{2}(x, y))\)

and \(T(\alpha(x, y)) = {\alpha r \cos(\theta = \varphi), \alpha r \sin(\theta = \varphi)}\)

\(= \alpha (r \cos(\theta = \varphi), r \sin(\theta = \varphi))\)

\(= \alpha T (x, y)\)

Hence, rotation is linear transformation. (f) Given that a fixed vector r and T maps each point to its reflection with respect to vector r, \(\overrightarrow{T}(\overrightarrow{x})=\overrightarrow{x}-2 \overrightarrow{x} r\)

\(=2\overrightarrow{x}_{r}-\overrightarrow{x}\)

Now showing in below T is linear transformation,

\(T(\alpha x + \beta y) = (\alpha x + \beta y) - 2 (\alpha x + \beta y)_{\varphi}

\(= \alpha x + \beta y - 2 \alpha x_{\varphi} - brta y_{\varphi}\)

\(=(\alpha x - 2 \alpha x_{\varphi}) + (\beta y - \beta y_{\varphi})\)

\(= \alpha (x - 2x_{\varphi}) + \beta (y - y_{\varphi})\)

\(= \alpha T (x) + \beta T (y)\)

and \(T (\alpha x) = \alpha x - 2(\alpha x)_{\varphi}\)

\(= \alpha x - 2 \alpha x_{\varphi}\)

\(= \alpha (x- 2x_{\varphi})\)

\(= \alpha T (x)\)

Hence, reflection is linear transformation.

Therefore, above satisfies both properties of linear transformation.

Hence, both rotation and reflection are linear transformation.