Let $T:\textbf{R}^{n}\to\textbf{R}^{m}$ be a linear transformation. Show that there exists a number $M > 0$ such that $\|Tx\|\leq M\|x\|$. Conclude in particular that every linear transformation from $\textbf{R}^{n}$ to $\textbf{R}^{m}$ is continuous.
My solution
Let $[T]_{\mathcal{B}}^{\mathcal{B}'} = [T(e_{1})^{T},T(e_{2})^{T},\ldots,T(e_{n})^{T}]$, where $\mathcal{B} = \{e_{1},e_{2},\ldots,e_{n}\}$ and $\mathcal{B}' = \{f_{1},f_{2},\ldots,f_{m}\}$.
Thus, according to the triangle inequality as well as the Cauchy-Schwarz inequality, we have that \begin{align*} \|Tx\| & = \|x_{1}T(e_{1})^{T} + x_{2}T(e_{2})^{T} + \ldots + x_{n}T(e_{n})^{T}\|\\\\ & \leq|x_{1}|\|T(e_{1})^{T}\| + |x_{2}|\|T(e_{2})^{T}\| + \ldots + |x_{n}|\|T(e_{n})^{T}\|\\\\ & = \langle(|x_{1}|,|x_{2}|,\ldots,|x_{n}|),(\|T(e_{1})^{T}\|,\|T(e_{2})^{t}\|,\ldots,\|T(e_{n})^{T}\|)\rangle\\\\ & \leq \sqrt{|x_{1}|^{2} + |x_{2}|^{2} + \ldots + |x_{n}|^{2}}\sqrt{\|T(e_{1})^{T}\|^{2} + \|T(e_{2})^{T}\|^{2} + \ldots + \|T(e_{n})^{T}\|^{2}} = M\|x\| \end{align*}
where $M = \sqrt{\|T(e_{1})^{T}\|^{2} + \|T(e_{2})^{T}\|^{2} + \ldots + \|T(e_{n})^{T}\|^{2}}$ and we assume that $T\neq 0$.
If $T = 0$, then any $M + 1$ does the job.
Consequently, $T$ is continuous because it is Lipschitz.
I would like to know if I am reasoning correctly. Any other solution is equally welcome.
Yes, it correct! Well done! Basically this shows that every linear map between finite dimensional vector spaces is continuous.
One suggestion: I would write the basis $\mathcal{B}'= \{f_1, \dots, f_m\}$ because you already used $e_1, \dots, e_n$.
Maybe one nitpick: the question asks for $M > 0$. If $T= 0$, your proof gives $M=0$, but you can simply add $1$ to your choice of $M$.