Demostración: Sean $x, y \in \mathbb{R}^n$ y $\alpha \in \mathbb{R}$
- Si $ \ x \in \mathbb{R}^n, \ \ \varphi(x)=x^t$, si $ \ x^t \in {\mathbb{R}^n}^*, \ \ \varphi^{-1}(x^t)=\left( {x^t}\right) ^t$ Entonces tendremos que $$\varphi^{-1}\left(x^t\right) = \varphi^{-1}(\varphi x) = x $$ Pero $ \varphi^{-1}\left(x^t\right) = \left( x^t\right) ^t$ $$ \therefore \ x = \left( x^t\right) ^t \blacksquare $$
- $$ x+y = \begin{pmatrix} \xi_1+\eta_1\\ \xi_2+\eta_2\\ \vdots \\ \xi_n+\eta_n \end{pmatrix} $$ $$\therefore \left( x+y\right)^t = \left[ \begin{matrix} \xi_1+\eta_1 \ \xi_2+\eta_2 \ ... \ \xi_n+\eta_n \end{matrix} \right] = \left[ \begin{matrix} \xi_1 \ \xi_2 \ ... \ \xi_n \end{matrix} \right ] + \left[ \begin{matrix} \eta_1 \ \eta_2 \ ... \ \eta_n \end{matrix} \right] = x^t + y^t \ \blacksquare$$
- $$ \alpha x = \begin{pmatrix} \alpha \xi_1\\ \alpha \xi_2\\ \vdots \\ \alpha \xi_n \end{pmatrix}$$ $$\therefore \left(\alpha x\right)^t = \left[ \begin{matrix} \alpha \xi_1 \ \alpha \xi_2 \ ... \ \alpha \xi_n \end{matrix} \right] = \alpha \left[ \begin{matrix} \xi_1 \ \xi_2 \ ... \ \xi_n \end{matrix} \right ] = \alpha x^t\ \blacksquare $$