Is it possible to get less variables after Lagrange's method for quadratic form?

Question

For example, I have a quadratic form $ F = x_1^2 + x_2^2 + 4x_3^2 + x_4^2 + 2x_1x_2 + 4x_1x_3 - 2x_1x_4 + 4x_2x_3 - 6x_2x_4 $. After applying Lagrange's method of reduction to canonical form I get a result of $ y_1^2 - y_2^2 + y^3 $. Is this even possible in this case? How could the number of variables be lowered?

Answer 1

look up Sylvester's Law of Inertia. Here is a method that is largely the reverse order of Lagrange's. Results first

$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ - 2 & \frac{ 1 }{ 2 } & 1 & \frac{ 1 }{ 2 } \\ 3 & - 1 & - 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & - 2 & 3 \\ 0 & 1 & \frac{ 1 }{ 2 } & - 1 \\ 0 & 0 & 1 & - 1 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & \frac{ 1 }{ 2 } & - 1 & - 1 \\ 2 & - \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & \frac{ 1 }{ 2 } & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 & 1 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) $$

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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
$$ H = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 1 & 0 & 2 & - 1 \\ 0 & 0 & 0 & - 2 \\ 2 & 0 & 4 & 0 \\ - 1 & - 2 & 0 & 1 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & - 2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - 1 & - 2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & 1 & 2 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 1 & 0 & 0 & - 1 \\ 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 2 \\ - 1 & - 2 & 2 & 1 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - 1 & - 2 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & - 2 \\ 0 & 0 & 0 & 2 \\ 0 & - 2 & 2 & 0 \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & 0 & - 2 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 2 & - 2 \\ 0 & 2 & 0 & 2 \\ 0 & - 2 & 2 & 0 \\ \end{array} \right) $$

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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & 0 & - 2 & 1 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & - 2 \\ 0 & 0 & 1 & 1 \\ 0 & - 2 & 1 & 0 \\ \end{array} \right) $$

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$$ E_{6} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{6} = \left( \begin{array}{rrrr} 1 & 0 & - 2 & 1 \\ 0 & 1 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ \end{array} \right) , \; \; \; Q_{6} = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{6} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right) $$

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$$ E_{7} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - 1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{7} = \left( \begin{array}{rrrr} 1 & 0 & - 2 & 3 \\ 0 & 1 & \frac{ 1 }{ 2 } & - 1 \\ 0 & 0 & 1 & - 1 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ \end{array} \right) , \; \; \; Q_{7} = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 & 1 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{7} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ - 2 & \frac{ 1 }{ 2 } & 1 & \frac{ 1 }{ 2 } \\ 3 & - 1 & - 1 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & - 2 & 3 \\ 0 & 1 & \frac{ 1 }{ 2 } & - 1 \\ 0 & 0 & 1 & - 1 \\ 0 & 1 & \frac{ 1 }{ 2 } & 0 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 1 & \frac{ 1 }{ 2 } & - 1 & - 1 \\ 2 & - \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & \frac{ 1 }{ 2 } & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & - 4 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 0 & \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & - 1 & 1 & 1 \\ 0 & - 1 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 1 & 1 & 2 & - 1 \\ 1 & 1 & 2 & - 3 \\ 2 & 2 & 4 & 0 \\ - 1 & - 3 & 0 & 1 \\ \end{array} \right) $$

Answer 2

Yes, this can happen, but it is only possible when rank of the quadratic form is smaller than the number of variables that occur; thus, a minimal example occurs in dimension $2$.

For example, in the coordinates defined by the linear change of coordinates $$y_1 = x_1 + x_2, \qquad y_2 = x_2,$$ the quadratic form $$Q({\bf x}) := x_1^2 + 2 x_1 x_2 + x_2^2$$ is $$Q({\bf y}) = y_1^2 ,$$ which has rank $1$.

At least in case that the underlying field is algebraically closed, the minimum number of variables that can occur in the expression is the rank of the form; Sylvester's Law of Inertia implies that the same is true over $\Bbb R$.