Define $\ltdefn{S}{\complex{3}}{\complex{3}}$ by \begin{equation*} \lt{S}{\colvector{x_1\\x_2\\x_3}}=\colvector{4x_1+2x_2\\0\\x_1+3x_3-2} \end{equation*}
This function “looks” linear, but consider \begin{align*} 3\,\lt{S}{\colvector{1\\2\\3}}&=3\,\colvector{8\\0\\8}=\colvector{24\\0\\24} \end{align*} while \begin{align*} \lt{S}{3\,\colvector{1\\2\\3}}&=\lt{S}{\colvector{3\\6\\9}}=\colvector{24\\0\\28} \end{align*}
So the second required property fails for the choice of $\alpha=3$ and $\vect{x}=\colvector{1\\2\\3}$ and by Definition LT, $S$ is not a linear transformation. It is just about as easy to find an example where the first defining property fails (try it!). Notice that it is the “-2” in the third component of the definition of $S$ that prevents the function from being a linear transformation.