Ejercicios resueltos de proceso de Gram-Schmidt

$\vec{u_1} = \vec{v_1} = (1{,}0{,}1)$

$\text{proj}_{\vec{u_1}}\vec{v_2} = \frac{\vec{v_2} \cdot \vec{u_1}}{\vec{u_1} \cdot \vec{u_1}}\vec{u_1}$

$\vec{v_2} \cdot \vec{u_1} = (1{,}1{,}0) \cdot (1{,}0{,}1) = 1$

$\vec{u_1} \cdot \vec{u_1} = (1{,}0{,}1) \cdot (1{,}0{,}1) = 2$

$\text{proj}_{\vec{u_1}}\vec{v_2} = \frac{1}{2}(1{,}0{,}1) = (\frac{1}{2},0,\frac{1}{2})$

$\vec{u_2} = \vec{v_2} - \text{proj}_{\vec{u_1}}\vec{v_2} = (1{,}1{,}0) - (\frac{1}{2},0,\frac{1}{2}) = (\frac{1}{2},1,-\frac{1}{2})$

$\text{proj}_{\vec{u_1}}\vec{v_3} = \frac{(0{,}1{,}1) \cdot (1{,}0{,}1)}{2}(1{,}0{,}1) = \frac{1}{2}(1{,}0{,}1) = (\frac{1}{2},0,\frac{1}{2})$

$\text{proj}_{\vec{u_2}}\vec{v_3} = \frac{(0{,}1{,}1) \cdot (\frac{1}{2},1,-\frac{1}{2})}{\|(\frac{1}{2},1,-\frac{1}{2})\|^2}(\frac{1}{2},1,-\frac{1}{2})$

$= \frac{1 - \frac{1}{2}}{\frac{1}{4} + 1 + \frac{1}{4}}(\frac{1}{2},1,-\frac{1}{2}) = \frac{\frac{1}{2}}{\frac{3}{2}}(\frac{1}{2},1,-\frac{1}{2}) = \frac{1}{3}(\frac{1}{2},1,-\frac{1}{2})$

$\vec{u_3} = (0{,}1{,}1) - (\frac{1}{2},0,\frac{1}{2}) - (\frac{1}{6},\frac{1}{3},-\frac{1}{6}) = (-\frac{2}{3},\frac{2}{3},\frac{2}{3})$

Base ortogonal: $\{(1{,}0{,}1), (\frac{1}{2},1,-\frac{1}{2}), (-\frac{2}{3},\frac{2}{3},\frac{2}{3})\}$

$\vec{u_1} = (2{,}0{,}0)$

$\text{proj}_{\vec{u_1}}\vec{v_2} = \frac{(1{,}1{,}0) \cdot (2{,}0{,}0)}{(2{,}0{,}0) \cdot (2{,}0{,}0)}(2{,}0{,}0) = \frac{2}{4}(2{,}0{,}0) = (1{,}0{,}0)$

$\vec{u_2} = (1{,}1{,}0) - (1{,}0{,}0) = (0{,}1{,}0)$

$\text{proj}_{\vec{u_1}}\vec{v_3} = \frac{(1{,}1{,}1) \cdot (2{,}0{,}0)}{4}(2{,}0{,}0) = \frac{1}{2}(2{,}0{,}0) = (1{,}0{,}0)$

$\text{proj}_{\vec{u_2}}\vec{v_3} = \frac{(1{,}1{,}1) \cdot (0{,}1{,}0)}{1}(0{,}1{,}0) = (0{,}1{,}0)$

$\vec{u_3} = (1{,}1{,}1) - (1{,}0{,}0) - (0{,}1{,}0) = (0{,}0{,}1)$

$\vec{e_1} = \frac{\vec{u_1}}{\|\vec{u_1}\|} = \frac{(2{,}0{,}0)}{2} = (1{,}0{,}0)$

$\vec{e_2} = \frac{\vec{u_2}}{\|\vec{u_2}\|} = \frac{(0{,}1{,}0)}{1} = (0{,}1{,}0)$

$\vec{e_3} = \frac{\vec{u_3}}{\|\vec{u_3}\|} = \frac{(0{,}0{,}1)}{1} = (0{,}0{,}1)$

Base ortonormal: $\{(1{,}0{,}0), (0{,}1{,}0), (0{,}0{,}1)\}$ (base canónica)

$f_1(x) = 1$

$\langle x, 1 \rangle = \int_0^1 x \cdot 1 dx = \frac{x^2}{2}\bigg|_0^1 = \frac{1}{2}$

$\langle 1, 1 \rangle = \int_0^1 1 \cdot 1 dx = 1$

$\text{proj}_{f_1}x = \frac{1/2}{1} \cdot 1 = \frac{1}{2}$

$f_2(x) = x - \frac{1}{2}$

$\langle x^2, 1 \rangle = \int_0^1 x^2 dx = \frac{x^3}{3}\bigg|_0^1 = \frac{1}{3}$

$\text{proj}_{f_1}x^2 = \frac{1/3}{1} \cdot 1 = \frac{1}{3}$

$\langle x^2, x-\frac{1}{2} \rangle = \int_0^1 x^2(x-\frac{1}{2})dx = \int_0^1 (x^3-\frac{x^2}{2})dx = \frac{x^4}{4} - \frac{x^3}{6}\bigg|_0^1 = \frac{1}{4} - \frac{1}{6} = \frac{1}{12}$

$\langle x-\frac{1}{2}, x-\frac{1}{2} \rangle = \int_0^1 (x-\frac{1}{2})^2 dx = \int_0^1 (x^2-x+\frac{1}{4})dx = \frac{1}{12}$

$\text{proj}_{f_2}x^2 = \frac{1/12}{1/12}(x-\frac{1}{2}) = x-\frac{1}{2}$

$f_3(x) = x^2 - \frac{1}{3} - (x-\frac{1}{2}) = x^2 - x + \frac{1}{6}$

$g_1(x) = \frac{1}{\sqrt{1}} = 1$

$g_2(x) = \frac{x-\frac{1}{2}}{\sqrt{1/12}} = 2\sqrt{3}(x-\frac{1}{2})$

$g_3(x) = \frac{x^2-x+\frac{1}{6}}{\sqrt{1/180}} = 6\sqrt{5}(x^2-x+\frac{1}{6})$

Base ortonormal: $\{1, 2\sqrt{3}(x-\frac{1}{2}), 6\sqrt{5}(x^2-x+\frac{1}{6})\}$

Supongamos que existen escalares $c_1, c_2, \ldots, c_n$ tales que:

$c_1u_1 + c_2u_2 + \cdots + c_nu_n = 0$

Para cualquier índice $i$, tomamos el producto interno de ambos lados con $u_i$:

$\langle c_1u_1 + c_2u_2 + \cdots + c_nu_n, u_i \rangle = \langle 0, u_i \rangle = 0$

$c_1\langle u_1, u_i \rangle + c_2\langle u_2, u_i \rangle + \cdots + c_n\langle u_n, u_i \rangle = 0$

Como el conjunto es ortogonal: $\langle u_j, u_i \rangle = 0$ para $j \neq i$

Por tanto, todos los términos se anulan excepto:

$c_i\langle u_i, u_i \rangle = 0$

Como $u_i \neq 0$ por hipótesis: $\langle u_i, u_i \rangle = \|u_i\|^2 > 0$

Por tanto: $c_i = 0$ para todo $i = 1, 2, \ldots, n$

Conclusión: La única solución de la ecuación de dependencia lineal es la trivial, por tanto el conjunto es linealmente independiente.