I need some help regarding this example
$$ q(x)= x_1^2 + 7x_2^2 + 2x_3^2 +6x_1x_2 - 4x_1x_3 - 4x_2x_3 $$ $$ =x_1^2 + (6x_2 - 4x_3)x_1 + 7x_2^2 - 4x_2x_3 + 2x_3^2 $$ $$ = (x_1 + (3x_2 - 2x_3))^2 - (3x_2 - 2x_3)^2 + 7x_2^2 - 4x_2x_3 + 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2x_2^2 + 8x_2x_3 - 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2(x_2 - 2x_3)^2 - 8x_3^2 - 2x_3^2 $$ $$ = (x_1 + 3x_2 - 2x_3)^2 - 2(x_2 - 2x_3)^2 - 10x_3^2$$
So I get the eigenvalues -2, 1 and -10 while my teacher gets -2, 1 and 6 and I'm wondering how does he get 6?
And also, I sort of understand the process but not completely so if anyone could explain how we complete squares on quadratic forms then i'd be grateful, I know it has to do with Sylvester's Law of Inertia but I didn't understand that well from the book.
method discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
Note that the matrix $H$ has characteristic polynomial $x^3 - 10x^2 + 6x + 12$ and (irrational) eigenvalues
$$-0.8119658377479428698001875737, \; \; \; \; 1.605232411407321548171091093, \; \; \;9.206733426340621321629096481 $$
$$ H = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$ $$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & - 2 & 4 \\ - 2 & 4 & 2 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 4 \\ 0 & 4 & - 2 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 4 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$
$$ P^T H P = D $$ $$ Q^T D Q = H $$ $$ H = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$
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$$\left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & - 2 & 4 \\ - 2 & 4 & 2 \\ \end{array} \right) $$
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$$\left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 4 \\ 0 & 4 & - 2 \\ \end{array} \right) $$
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$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P = \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 3 & 1 & 0 \\ - 4 & 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 3 & - 4 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & - 2 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & 6 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 0 & 1 & - 2 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 3 & - 2 \\ 3 & 7 & - 2 \\ - 2 & - 2 & 2 \\ \end{array} \right) $$