Find the canonical form of quadratic form $$x_1x_2-4x_1x_4+3x_2x_3+5x_2x_4-x_3x_4$$
I got the following expression: $$x_1x_2-4x_1x_4+3x_2x_3+5x_2x_4-x_3x_4 =$$$$= (x_1+\frac{1}{2}x_2-2x_4)^2 + \frac{1}{4}(x_2-3x_3+5x_4)^2+\frac{9}{4}(x_3-\frac{2}{9}x_4)^2-x_1^2-\frac{328}{81}x_4^2$$ But might one have more square terms than the number of the variables?
If you must have $\pm 1$ or zero on the diagonal, multiply my $D$ on both sides by
which changes the relation to $ (PR)^T H(PR) = D_1 $
$$ R = \left( \begin{array}{rrrr} \frac{1}{\sqrt 2} & 0 & 0 & 0 \\ 0 & \sqrt 2 & 0 & 0 \\ 0 & 0 & \frac{1}{\sqrt{40}} & 0 \\ 0 & 0 & 0 & \frac{ \sqrt{40} }{ 11 } \\ \end{array} \right) $$
$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 5 & 4 & 0 & 1 \\ - \frac{ 13 }{ 8 } & - \frac{ 11 }{ 10 } & 1 & - \frac{ 11 }{ 40 } \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 0 & - 4 \\ 1 & 0 & 3 & 5 \\ 0 & 3 & 0 & - 1 \\ - 4 & 5 & - 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 5 & - \frac{ 13 }{ 8 } \\ 1 & \frac{ 1 }{ 2 } & 4 & - \frac{ 11 }{ 10 } \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & - \frac{ 11 }{ 40 } \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 40 & 0 \\ 0 & 0 & 0 & - \frac{ 121 }{ 40 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 3 }{ 2 } & - 3 & \frac{ 11 }{ 40 } & 1 \\ \frac{ 1 }{ 2 } & - 9 & 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 40 & 0 \\ 0 & 0 & 0 & - \frac{ 121 }{ 40 } \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 3 }{ 2 } & \frac{ 1 }{ 2 } \\ - 1 & 1 & - 3 & - 9 \\ 0 & 0 & \frac{ 11 }{ 40 } & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 1 & 0 & - 4 \\ 1 & 0 & 3 & 5 \\ 0 & 3 & 0 & - 1 \\ - 4 & 5 & - 1 & 0 \\ \end{array} \right) $$