Let $V=\mathbb R^3$. Consider the quadratic form $$q:V\to\mathbb R, (x_1,x_2,x_3)^T\rightarrow 5x_1^2+5x_2^2-4x_3^2-14x_1x_2+4x_1x_3+4x_2x_3$$
And let $\beta:V \times V\to \mathbb R$ the symmetric form which has $q$ as quadratic form.
With polarisation, i.e. $\beta(v,w)=\frac12(q(v+w)-q(v)-q(w))$ I calculated the transformation matrix of $\beta$ wrt to the standard basis B, i.e. $[\beta]_{B\times B}=$ $
\begin{bmatrix}
5 & -7 & 2 \\
-7 & 5 & 2 \\
2 & 2 & -4
\end{bmatrix}
$.
Now I want to determine a basis $\cal B$ such that $[\beta]_{\cal B \times \cal B}=\begin{bmatrix}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & 0
\end{bmatrix}$.
I tried to consider $\beta(b_1,b_1)=1 $ and $\beta(b_2,b_2)=-1$, but I do not see how to proceed, it is getting really messy. How should one handle this task?
Best Answer
Added: the cleanest expression before reaching the $\pm 1$ stage is $$ 8\left(2x - y - z \right)^2 - 3\left(3x - y - 2z \right)^2 = 5 x^2 + 5 y^2 - 4 z^2 + 4 yz + 4 zx - 14 xy.$$ This becomes $$ \left( \sqrt {32} \;x - \sqrt 8 \; y - \sqrt 8 \; z \right)^2 - \left( \sqrt {27} \;x - \sqrt 3 \; y - \sqrt {12} \;z \right)^2 = 5 x^2 + 5 y^2 - 4 z^2 + 4 yz + 4 zx - 14 xy.$$
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 7 }{ 5 } & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 7 }{ 5 } & 1 & 0 \\ \frac{ 2 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 7 }{ 5 } & \frac{ 2 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) $$
At this stage, to force diagonal elements to be $\pm 1,$ we will need to introduce some square roots, but we can throw these in a diagonal matrix that is its own transpose $S^T = S, \; \; $ let us call it $S$ for square root, $$ S = \left( \begin{array}{rrr} \frac{1}{\sqrt 5} & 0 & 0 \\ 0 & \frac{\sqrt 5}{ \sqrt{24} } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ so that $$ S^T D S = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
Finally $$ S^T P^T H P S = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
and your basis matrix is $R = PS$
The algorithm for the first part, all rational entries, is:
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
$$ H = \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) $$
==============================================
$$ E_{1} = \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 7 }{ 5 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 5 & 0 & 2 \\ 0 & - \frac{ 24 }{ 5 } & \frac{ 24 }{ 5 } \\ 2 & \frac{ 24 }{ 5 } & - 4 \\ \end{array} \right) $$
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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - \frac{ 2 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & - \frac{ 2 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 7 }{ 5 } & \frac{ 2 }{ 5 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & \frac{ 24 }{ 5 } \\ 0 & \frac{ 24 }{ 5 } & - \frac{ 24 }{ 5 } \\ \end{array} \right) $$
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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 7 }{ 5 } & \frac{ 2 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$
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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 7 }{ 5 } & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 7 }{ 5 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 7 }{ 5 } & 1 & 0 \\ \frac{ 2 }{ 5 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 5 & 0 & 0 \\ 0 & - \frac{ 24 }{ 5 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 7 }{ 5 } & \frac{ 2 }{ 5 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 5 & - 7 & 2 \\ - 7 & 5 & 2 \\ 2 & 2 & - 4 \\ \end{array} \right) $$