Lagrange method

Question

I can't transform quadratic form to canonical using Lagrange method. $x_1^2 + 5x_2^2 - 4x_3^2 + 2x_1x_2 - 4x_1x_3$. I tried to solve it, but stopped at this moment: $x_1^2 + 2x_1x_2 + x_2^2 + 4x_2^2 - 4x_3^2- 4x_1x_3 = (x_1+x_2)^2 + 4x_2^2 - 4x_3^2- 4x_1x_3$.

Answer 1

In fact you want to get all occurrences of $x_1$ into the first bracket. So you start with $$x_1^2+5x_2^2+4x_2^2-4x_3^2+2x_1x_2-4x_1x_3=(x_1+x_2-2x_3)^2+4x_2^2+4x_2x_3-8x_3^2.$$

Can you continue from there?

Answer 2

The order you get to first with Lagrange's method is

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ - 2 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 1 & 5 & 0 \\ - 2 & 0 & - 4 \\ \end{array} \right) $$

or $$ (x+y-2z)^2 + 4 (y + \frac{1}{2}z)^2 - 9 z^2 $$ the same as

$$ (x+y-2z)^2 + (2y + z)^2 - 9 z^2 $$

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
$$ H = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 1 & 5 & 0 \\ - 2 & 0 & - 4 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 1 & 5 & 0 \\ - 2 & 0 & - 4 \\ \end{array} \right) $$

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

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 1 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 2 \\ 0 & 2 & - 8 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 9 \\ \end{array} \right) $$

==============================================

$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ \frac{ 5 }{ 2 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 1 & 5 & 0 \\ - 2 & 0 & - 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & \frac{ 5 }{ 2 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 9 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ - 2 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & - 9 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 1 & - 2 \\ 1 & 5 & 0 \\ - 2 & 0 & - 4 \\ \end{array} \right) $$