I have a quadratic form: $x_{1}^{2} + x_{2}^{2} + 3x_{3}^{2} + 4x_{1}x_{2} + 2x_{1}x_{3} + 2x_{2}x_{3}$
How could I find canonical form of that? Trying to use Lagrange method, I came to this:
$(x_{1} + x_{2} + x_{3})^{2} - 2x_{1}x_{2} + 2(\sqrt{3} - 1)x_{1}x_{3} + 2(\sqrt{3} - 1)x_{2}x_{3}$
But as you see, replacements here do not work.
This exercise is from task book, so I know the answer, but not a solution, it’s $y_{1}^{2} + y_{2}^{2} - y_{3}^{2}$
I'm guessing what canonical form might mean...
$$ (x+2y+z)^2 - 3 (y + \frac{z}{3})^2 + \frac{7}{3} z^2 $$ If it is required to have only $\pm 1$ as coefficients, $$ (x+2y+z)^2 - \left(y \sqrt 3 + \frac{z}{\sqrt3}\right)^2 + \left( \frac{\sqrt 7}{\sqrt3} z\right)^2 $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & \frac{ 1 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & \frac{ 7 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 3 \\ \end{array} \right) $$
how to do it
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 3 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$
$$ H = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 3 \\ \end{array} \right) $$
==============================================
$$ E_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & - 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 1 & 0 & 1 \\ 0 & - 3 & - 1 \\ 1 & - 1 & 3 \\ \end{array} \right) $$
==============================================
$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & - 2 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 3 & - 1 \\ 0 & - 1 & 2 \\ \end{array} \right) $$
==============================================
$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & - 2 & - \frac{ 1 }{ 3 } \\ 0 & 1 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & \frac{ 7 }{ 3 } \\ \end{array} \right) $$
==============================================
$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - 2 & 1 & 0 \\ - \frac{ 1 }{ 3 } & - \frac{ 1 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 3 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - 2 & - \frac{ 1 }{ 3 } \\ 0 & 1 & - \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & \frac{ 7 }{ 3 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 1 & \frac{ 1 }{ 3 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 3 & 0 \\ 0 & 0 & \frac{ 7 }{ 3 } \\ \end{array} \right) \left( \begin{array}{rrr} 1 & 2 & 1 \\ 0 & 1 & \frac{ 1 }{ 3 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 1 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 3 \\ \end{array} \right) $$