Bilinear form and signature

Question

I have a bilinear form such that the associate matrix is

$$A = \left(\begin{matrix} 0&0&k\\ 0&k&0\\ k&0&0\end{matrix}\right)$$

For which $k$ the quadratic for has signature $(1,2)$?

Answer 1

This is all you need for Sylvester's Law of Inertia

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

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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \\ \end{array} \right) $$

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

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

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

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