I'm new here, I would like some help in solving this particular Quadratic form with matrices using the method described:
Quadratic forms - Completing squares
$$A = \begin{bmatrix} 1 & 1 & 2\\ 1 & -1 & 1\\ 2 & 1 & 3\end{bmatrix}$$
I've managed to solve the example given on the link provided and I understand why things are as they are pretty well without much trouble but I can't seem to solve my own problem, why? What I mostly want to know is how to find Q (Coefficient matrix)
I show the algorithm below. It is helpful in some ways. However, it need not give the "simplest" expression; here we could also write $$ (x+y+2z)^2 - (y+z)^2 - y^2 = x^2 - y^2 + 3 z^2 + 2 yz + 4 zx + 2xy $$
The method used to go in this direction is usually called repeated "completing the square," and hardly requires matrices to do by hand.
$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 2 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & - 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 1 & 2 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \\ \end{array} \right) $$
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As there are just three elementary matrices $E_j,$ we end up with $$ P = E_1 E_2 E_3 $$ and $$ P^{-1} = Q = E_3^{-1} E_2^{-1} E_1^{-1} $$
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
$$ H = \left(
\begin{array}{rrr}
1 & 1 & 2 \\
1 & - 1 & 1 \\
2 & 1 & 3 \\
\end{array}
\right)
$$
$$ D_0 = H $$
$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 1 & 1 & 2 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \\ \end{array} \right) $$
==============================================
$$ E_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 2 \\ 0 & - 2 & - 1 \\ 2 & - 1 & 3 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 1 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & - 1 \\ 0 & - 1 & - 1 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 3 }{ 2 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - \frac{ 1 }{ 2 } \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - \frac{ 3 }{ 2 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 2 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - \frac{ 3 }{ 2 } \\ 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - \frac{ 1 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 2 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 2 & 0 \\ 0 & 0 & - \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 2 \\ 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 1 & 2 \\ 1 & - 1 & 1 \\ 2 & 1 & 3 \\ \end{array} \right) $$
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