DemostraciĆ³n: Sean $x, y, x_1, x_2, y_1, y_2 \in \mathbb{R}^n$ y $\lambda \in \mathbb{R}$
- $$x \cdot y =\sum_{i=1}^{n} \xi_i \eta_i = \sum_{i=1}^{n} \eta_i \xi_i = y \cdot x \ \blacksquare $$
- $$ \left(\lambda x\right) \cdot y = \sum_{i=1}^{n} \left( \lambda \xi_i\right)\eta_i = \sum_{i=1}^{n} \lambda \left(\xi_i \eta_i\right) = \lambda \sum_{i=1}^{n} \xi_i \eta_i = \sum_{i=1}^{n} \xi_i\left( \lambda \eta_i\right)$$ $$ \therefore \left( \lambda x\right) \cdot y = \lambda \left( x \cdot y\right) = x \cdot \left( \lambda y \right) \blacksquare $$
- Si $x_1= \left[ \begin{matrix} \xi_{i1} \end{matrix}\right],\ x_2= \left[\begin{matrix} \xi_{i2}\end{matrix}\right]$ entonces \begin{align} (x_1 + x_2) \cdot y &= \sum_{i=1}^{n} \left(\xi_{i1} + \xi_{i2}\right) \eta_i = \sum_{i=1}^{n} \left( \xi_{i1} \eta_{i} + \xi_{i2} \eta_{i}\right) \\ &= \sum_{i=1}^n \xi_{i1} \eta_i + \sum_{i=1}^n \xi_{i2} \eta_i = x_1 \cdot y + x_2 \cdot y \blacksquare \end{align}
- \begin{align} x \cdot(y_1+y_2) &= (y_1 + y_2) \cdot x & \text{(por 1)}\\ &= (y_1 \cdot x ) + (y_2 \cdot x) & \text{(por 3)} \\ &= x \cdot y_1 + x \cdot y_2 & \text{(por 1)} \blacksquare \end{align}
- $x \cdot x = {\xi_1}^2 + {\xi_2}^2 + ... + {\xi_n}^2$; como ${\xi_i}^2 \geq 0 \ \ \ \ \ \ \ \ i = 1,2,...,n$ $$\Rightarrow {\xi_1}^2 + {\xi_2}^2 + ... + {\xi_n}^2 \geq 0$$ \begin{align} &\Rightarrow ) \ x = 0 \Rightarrow x \cdot x = \sum_{i=1}^{n} (0) ^2 = 0 & \\ &\Leftarrow ) \ x \cdot x = 0 \Rightarrow \sum_{i=1}^n {\xi_i}^2 = 0 & \\ & \Rightarrow \xi_i = 0 \ \ \ \ i = 1,2,...n \\ & \Rightarrow x = 0 \blacksquare \end{align}