How do I determine the rank and signature of a quadratic form

Question

The quadratic form is given: $f=x_1x_2-x_2x_3$ I have to determine a rank and signature of it. What is the way to do it? ( I know that usually I have to transform a matrix into upper triangular form to determine the rank, but in this case on diagonal it is all zeros and there is no way I can transform it to be a upper diagonal matrix)

$$\left[\begin{matrix}0 & \frac{1}{2} & 0 \\ \frac{1}{2} & 0 & -\frac{1}{2} \\ 0 & -\frac{1}{2} & 0 \end{matrix}\right]$$

The only thing I could think of was to add to the first column the third one and the first row to the third row, then the first column will be all zeros and I willl change to columns to look it as follows:

$$\left[\begin{matrix} \frac{1}{2} & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & 0\end{matrix}\right]$$ I now that by switching rows or columns you change determinan, but is it important in this case? Any case, the rank for this form is $r=2$ am I right? ( I just don't know for sure how to determine it)

Answer 1

Write the quadratic form on its reduced form i.e. with squared terms:

$$x_1x_2-x_2x_3=x_2(x_1-x_3)=\frac14(x_2+x_1-x_3)^2-\frac14(x_2-x_1+x_3)^2.$$ hence the signature is $(1,1)$ and the rank is $r=2$.

Answer 2

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

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

$$ H = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 1 & 0 & - 1 \\ 0 & - 1 & 0 \\ \end{array} \right) $$

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

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

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

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

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