My goal is to factor $$6h^2 + 8k^2 + 12l^2 − 12hk − 12hl + 16kl$$ into $$6 (h - k - l)^2 + 4 l^2 + 2 (k + l)^2$$ I don't seem to understand how to come up with those factors without guessing my way forward. I have tried factoring the ≥1 variable terms, for instance trying to group squares of $ h,l $ or $k,l$ etc.
What would be a proper strategy when factoring this?
On
$$6h^2 + 8k^2 + 12l^2 \underbrace{− 12hk − 12hl} + 16kl= 6h^2-12h(k+l)+8k^2 + 12l^2+ 16kl$$ $$ = 6h^2-12h(k+l)+\underbrace{6(k+l)^2-6(k+l)^2}+8k^2 + 12l^2+ 16kl$$ $$ = 6(h-k-l)^2+6l^2+4kl+2k^2 $$
$$ = 6(h-k-l)^2+4l^2+\underbrace{2l^2+4kl+2k^2} $$
$$ = 6(h-k-l)^2+4l^2+2(l+k)^2 $$
On
it's not that bad if you pick up pen and paper and just start from something. For example, all the quadratic terms have coefficients $6, 8, 12.$ So it would be stupid not to separate a square term starting with $$6(h\pm k\pm l)^2$$ first. Then you look at the signs of the mixed terms, and should be easily deduce the remaining signs and come up with $6(h-k-l)^2.$
After you complete this square, the remaining term is $2k^2+6l^2+4kl$ and you can easily decompose this into sum of squares either using the same idea or just straight up using the quadratic formula from high school.
On
We have $$\begin{align}6h^2 + 8k^2 + 12l^2 − 12hk − 12hl + 16kl&=\color{red}{6h^2-12hk}+\color{blue}{8k^2+16kl+8l^2}+\color{green}{4l^2}\\&=\color{red}{6(h^2-2hk)}+\color{blue}{6(k^2+2kl+l^2)}+\color{blue}{2(k^2+2kl+l^2)}+\color{green}{4l^2}\\&=\color{purple}6(\color{red}{h^2-2hk}+\color{blue}{k^2+2kl+l^2})+\color{blue}{2(k+l)^2}+\color{green}{4l^2}\\&=\color{purple}{6(h^2-2h(k+l)+(k+l)^2)}+\color{blue}{2(k+l)^2}+\color{green}{4l^2}\\&=6(h-(k+l))^2+4l^2+2(k+l)^2\end{align}$$ as required.
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
$$ H = \left( \begin{array}{rrr} 6 & - 6 & - 6 \\ - 6 & 8 & 8 \\ - 6 & 8 & 12 \\ \end{array} \right) $$ $$ D_0 = H $$ Each step is a choice of "elementary" matrix $E_j$ and then $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ so that we always have these three: $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$ $$ $$ $$ H = \left( \begin{array}{rrr} 6 & - 6 & - 6 \\ - 6 & 8 & 8 \\ - 6 & 8 & 12 \\ \end{array} \right) $$
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$$ E_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 6 & 0 & - 6 \\ 0 & 2 & 2 \\ - 6 & 2 & 12 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - 1 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & 2 & 2 \\ 0 & 2 & 6 \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - 1 & - 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 6 & - 6 & - 6 \\ - 6 & 8 & 8 \\ - 6 & 8 & 12 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 1 & 0 \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 1 & 1 & 0 \\ - 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 6 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 1 & - 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 6 & - 6 & - 6 \\ - 6 & 8 & 8 \\ - 6 & 8 & 12 \\ \end{array} \right) $$
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