Base diagonal para la forma bilineal en $\mathbb{R}[x]_2$ .

Question

Para $f,g \in V =\mathbb{R}[x]_2$ tenemos la forma bilineal \begin{eqnarray*} \phi : &V \times V &\rightarrow \mathbb{R} \\ &(f,g) &\mapsto \int_{-1}^1 xf(x)g(x)dx. \end{eqnarray*} Demostré que es degenerada observando que ker $(\phi_L) = \left\{a_0\left(1 - \frac{5}{3}x^2\right) \in V \mid a_0 \in \mathbb{R}\right\} \neq \{0\}$ . (Tal vez esto ya es donde se equivoca). Ahora voy a dar una base de $V$ para la cual la matriz asociada a $\phi$ es diagonal. Ahora bien, aunque tengo un teorema que garantiza la existencia de dicha base, un método para construir tal base se me escapa. $$ H = \left( \begin{array}{rrr} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

Answer 1

Multiplicando las columnas de $P$ por $( 1,-2, -5)$ sugiere la base $$ \{ \; 1+x, \; \; 1-x, \; \; 3 - 5 x^2 \; \} $$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - \frac{ 3 }{ 5 } & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & \frac{ 2 }{ 3 } & 0 \\ \frac{ 2 }{ 3 } & 0 & \frac{ 2 }{ 5 } \\ 0 & \frac{ 2 }{ 5 } & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} \frac{ 4 }{ 3 } & 0 & 0 \\ 0 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$ \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 4 }{ 3 } & 0 & 0 \\ 0 & - \frac{ 1 }{ 3 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - \frac{ 3 }{ 5 } & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & \frac{ 2 }{ 3 } & 0 \\ \frac{ 2 }{ 3 } & 0 & \frac{ 2 }{ 5 } \\ 0 & \frac{ 2 }{ 5 } & 0 \\ \end{array} \right) $$

ORIGINAL: Multiplico su matriz por $15/2$ para obtener números enteros. Para aplicarlo a su matriz específica, multiplique ambos $H$ y $D$ por $2/15.$

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - \frac{ 3 }{ 5 } & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} \frac{ 1 }{ 2 } & - 1 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ \frac{ 3 }{ 10 } & - \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 10 } \\ - 1 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) $$

Algoritmo discutido en http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
$$ H = \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) $$ $$ D_0 = H $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 10 & 5 & 3 \\ 5 & 0 & 3 \\ 3 & 3 & 0 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 10 & 0 & 3 \\ 0 & - \frac{ 5 }{ 2 } & \frac{ 3 }{ 2 } \\ 3 & \frac{ 3 }{ 2 } & 0 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 3 }{ 10 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 10 } \\ 1 & \frac{ 1 }{ 2 } & - \frac{ 3 }{ 10 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 10 } \\ - 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & \frac{ 3 }{ 2 } \\ 0 & \frac{ 3 }{ 2 } & - \frac{ 9 }{ 10 } \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 10 } \\ - 1 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 1 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ - \frac{ 3 }{ 5 } & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 3 }{ 5 } \\ 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} \frac{ 1 }{ 2 } & - 1 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ \frac{ 3 }{ 10 } & - \frac{ 3 }{ 5 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 10 & 0 & 0 \\ 0 & - \frac{ 5 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 10 } \\ - 1 & 1 & - \frac{ 3 }{ 5 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 0 & 5 & 0 \\ 5 & 0 & 3 \\ 0 & 3 & 0 \\ \end{array} \right) $$