DemostraciĆ³n.
- $$ $$ $\begin{align*} A (x_1 + x_2) &= \begin{pmatrix} {r_1}^t (x_1 + x_2) \\ {r_2}^t (x_1 + x_2) \\ \vdots \\ {r_m}^t (x_1 + x_2) \\ \end{pmatrix} = \begin{pmatrix} {r_1}^t x_1 + {r_1}^t x_2 \\ {r_2}^t x_1 + {r_2}^t x_2 \\ \vdots \\ {r_m}^t x_1 + {r_m}^t x_2 \\ \end{pmatrix} \\ \\ & = \begin{pmatrix} {r_1}^t x_1 \\ {r_2}^t x_1 \\ \vdots \\ {r_m}^t x_1 \\ \end{pmatrix} + \begin{pmatrix} {r_1}^t x_2 \\ {r_2}^t x_2 \\ \vdots \\ {r_m}^t x_2 \\ \end{pmatrix} = A x_1 + A x_2 \end{align*} $ $$\blacksquare $$
- $$ A (\lambda x) = \begin{pmatrix} {r_1}^t (\lambda x) \\ {r_2}^t (\lambda x)\\ \vdots \\ {r_m}^t (\lambda x) \\ \end{pmatrix} = \begin{pmatrix} \lambda ({r_1}^t x) \\ \lambda ({r_2}^t x) \\ \vdots \\ \lambda ({r_m}^t x) \\ \end{pmatrix} = \lambda \begin{pmatrix} {r_1}^t x \\ {r_2}^t x \\ \vdots \\ {r_m}^t x \\ \end{pmatrix} = \lambda (A x) \blacksquare $$
- Si $ B = \begin{pmatrix} {s_1}^t \\ {s_2}^t \\ \vdots \\ {s_m}^t \\ \end{pmatrix}$ entonces: $ \begin{align*} (A+B)x &= \begin{pmatrix} ({r_1}^t + {s_1}^t) x \\ ({r_2}^t + {s_2}^t) x \\ \vdots \\ ({r_m}^t + {s_m}^t) x\\ \end{pmatrix} = \begin{pmatrix} {r_1}^t x + {s_1}^t x \\ {r_2}^t x + {s_2}^t x \\ \vdots \\ {r_m}^t x + {s_m}^t x\\ \end{pmatrix} \\ \\ &= \begin{pmatrix} {r_1}^t x \\ {r_2}^t x \\ \vdots \\ {r_m}^t x \\ \end{pmatrix} + \begin{pmatrix} {s_1}^t x \\ {s_2}^t x \\ \vdots \\ {s_m}^t x \\ \end{pmatrix} = A x + B x \end{align*}$ $$\blacksquare$$