representation of sum of squares of a globally positive quadratic function

Question

Consider the following quadratic function \begin{equation} f(y,z)=1697 y^2+57 y z+81 y+407 z^2-6 z+1 \end{equation} Using first and second derivatives test, we have shown that the global minimum of $f$ occurs at $(y_0,z_0)= (-\frac{66276}{2759467},\frac{24981}{2759467})$ and $f(y_0,z_0)=\frac{346}{2759467}>0$. This leads to $f(y,z)>0$ for all $y,z\in\mathbb{R}$. According to Hilbert's seventeenth problem (see https://en.wikipedia.org/wiki/Hilbert%27s_seventeenth_problem), $f(y,z)$ can be represented as sum of squares of other polynomials. How we write $f(y,z)$ as a sum of squares? Any reference, suggestion, idea, or comment is welcome. Thank you!

Answer 1

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 81 }{ 2 } & 1 & 0 \\ - 3 & \frac{ 600 }{ 227 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 227 }{ 2 } & 0 \\ 0 & 0 & \frac{ 692 }{ 227 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 81 }{ 2 } & - 3 \\ 0 & 1 & \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 81 & - 6 \\ 81 & 3394 & 57 \\ - 6 & 57 & 814 \\ \end{array} \right) $$

and so on...

Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 2 & 81 & - 6 \\ 81 & 3394 & 57 \\ - 6 & 57 & 814 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 2 & 81 & - 6 \\ 81 & 3394 & 57 \\ - 6 & 57 & 814 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 81 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 81 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & \frac{ 81 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 2 & 0 & - 6 \\ 0 & \frac{ 227 }{ 2 } & 300 \\ - 6 & 300 & 814 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 81 }{ 2 } & 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & \frac{ 81 }{ 2 } & - 3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 227 }{ 2 } & 300 \\ 0 & 300 & 796 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 81 }{ 2 } & \frac{ 24981 }{ 227 } \\ 0 & 1 & - \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & \frac{ 81 }{ 2 } & - 3 \\ 0 & 1 & \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 227 }{ 2 } & 0 \\ 0 & 0 & \frac{ 692 }{ 227 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 81 }{ 2 } & 1 & 0 \\ \frac{ 24981 }{ 227 } & - \frac{ 600 }{ 227 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 81 & - 6 \\ 81 & 3394 & 57 \\ - 6 & 57 & 814 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 81 }{ 2 } & \frac{ 24981 }{ 227 } \\ 0 & 1 & - \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 227 }{ 2 } & 0 \\ 0 & 0 & \frac{ 692 }{ 227 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 81 }{ 2 } & 1 & 0 \\ - 3 & \frac{ 600 }{ 227 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 227 }{ 2 } & 0 \\ 0 & 0 & \frac{ 692 }{ 227 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 81 }{ 2 } & - 3 \\ 0 & 1 & \frac{ 600 }{ 227 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 81 & - 6 \\ 81 & 3394 & 57 \\ - 6 & 57 & 814 \\ \end{array} \right) $$