Let $T:\textbf{R}^{n}\to\textbf{R}^{m}$ be a linear transformation. Show that $T$ is continuously differentiable at every point, and in fact $T'(x) = T$ for every $x$.
My solution
According to this result, $T$ is continuous because it is Lipschitz. On the other hand, due to the linearity of $T$, for every $x_{0}\in\textbf{R}^{n}$, we have that \begin{align*} \lim_{x\to x_{0};x\neq x_{0}}\frac{\|T(x) - T(x_{0}) - T(x - x_{0})\|}{\|x - x_{0}\|} = \lim_{x\to x_{0};x\neq x_{0}}\frac{\|T(x) - T(x_{0}) - T(x) + T(x_{0})\|}{\|x-x_{0}\|} = 0 \end{align*}
Hence $T'(x) = T$. Once the derivative is unique, the result follows.
Is the wording of my proof right? Is there another way to approach it?