Esta es una prueba de mi primera fórmula copiada de Scientific WorkPlace:
Sean los vectores en \(\mathbb{R}^3\):
\[\left\{ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}, \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix}, \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \right\}\]
Demostraremos que son linealmente independientes y forman una base:
\[\begin{vmatrix} 1 & 2 & 3 \\ 1 & 0 & -1 \\ -1 & -2 & 3 \end{vmatrix} = -12 \neq 0\]
Por lo que los vectores son linealmente independientes y forman una base.
Sea \(\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\)
Entonces su norma es:
\[\|\vec{v}_1\| = \left\| \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \right\| = \sqrt{3}\]
Así que el primer vector unitario es:
\[\hat{u}_1 = \frac{1}{\sqrt{3}}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix}\]
\[\vec{v}'_2 = \vec{v}_2 - (\vec{v}_2 \cdot \hat{u}_1)\hat{u}_1\]
Sustituimos:
\[\vec{v}'_2 = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} - \left( \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} \right) \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix}\]
Realizamos primero el producto interno:
\[\begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} = \frac{4}{3}\sqrt{3}\]
Entonces:
\[\vec{v}'_2 = \begin{pmatrix} 2 \\ 0 \\ -2 \end{pmatrix} - \left( \frac{4}{3}\sqrt{3} \right) \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ -\frac{4}{3} \\ -\frac{2}{3} \end{pmatrix}\]
Calculamos su norma:
\[\|\vec{v}'_2\| = \left\| \begin{pmatrix} \frac{2}{3} \\ -\frac{4}{3} \\ -\frac{2}{3} \end{pmatrix} \right\| = \frac{2}{3}\sqrt{2}\sqrt{3}\]
\[\hat{u}_2 = \frac{1}{\frac{2}{3}\sqrt{2}\sqrt{3}} \begin{pmatrix} \frac{2}{3} \\ -\frac{4}{3} \\ -\frac{2}{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix}\]
\[\vec{v}'_3 = \vec{v}_3 - (\vec{v}_3 \cdot \hat{u}_1)\hat{u}_1 - (\vec{v}_3 \cdot \hat{u}_2)\hat{u}_2\]
Sustituimos:
\[\vec{v}'_3 = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} - \left( \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} \right) \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} - \left( \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} \right) \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix}\]
Productos internos:
\[\begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} = -\frac{1}{3}\sqrt{3}\]
\[\begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} = \frac{1}{3}\sqrt{2}\sqrt{3}\]
Operando los términos:
\[\left( -\frac{1}{3}\sqrt{3} \right) \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} = \begin{pmatrix} -\frac{1}{3} \\ -\frac{1}{3} \\ \frac{1}{3} \end{pmatrix}\]
\[\left( \frac{1}{3}\sqrt{2}\sqrt{3} \right) \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ -\frac{1}{3} \end{pmatrix}\]
\[\vec{v}'_3 = \begin{pmatrix} 3 \\ -1 \\ 3 \end{pmatrix} - \begin{pmatrix} -\frac{1}{3} \\ -\frac{1}{3} \\ \frac{1}{3} \end{pmatrix} - \begin{pmatrix} \frac{1}{3} \\ -\frac{2}{3} \\ -\frac{1}{3} \end{pmatrix} = \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix}\]
Norma y vector unitario:
\[\|\vec{v}'_3\| = \left\| \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix} \right\| = 3\sqrt{2}\]
\[\hat{u}_3 = \frac{1}{3\sqrt{2}} \begin{pmatrix} 3 \\ 0 \\ 3 \end{pmatrix} = \begin{pmatrix} \frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2} \end{pmatrix}\]
\[\|\hat{u}_1\| = \left\| \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} \right\| = 1\]
\[\|\hat{u}_2\| = \left\| \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} \right\| = 1\]
\[\|\hat{u}_3\| = \left\| \begin{pmatrix} \frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2} \end{pmatrix} \right\| = 1\]
\[\hat{u}_1 \cdot \hat{u}_2 = \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} = 0\]
\[\hat{u}_1 \cdot \hat{u}_3 = \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2} \end{pmatrix} = 0\]
\[\hat{u}_2 \cdot \hat{u}_3 = \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix} \cdot \begin{pmatrix} \frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2} \end{pmatrix} = 0\]
Ya que son normales y ortogonales se concluye que son ORTONORMALES.
\[\left\{ \begin{pmatrix} \frac{1}{3}\sqrt{3} \\ \frac{1}{3}\sqrt{3} \\ -\frac{1}{3}\sqrt{3} \end{pmatrix}, \begin{pmatrix} \frac{1}{6}\sqrt{2}\sqrt{3} \\ -\frac{1}{3}\sqrt{2}\sqrt{3} \\ -\frac{1}{6}\sqrt{2}\sqrt{3} \end{pmatrix}, \begin{pmatrix} \frac{1}{2}\sqrt{2} \\ 0 \\ \frac{1}{2}\sqrt{2} \end{pmatrix} \right\}\]