Canonical form of a Quadratic form.

Question

I've been given the following quadratic form to find the canonical form of: $$ Q(\bf{z})= z_1z_2 + 2z_2z_3 − 3z_3z_4 $$

through the method of forming perfect squares.

The method I've been taugh/show is to look at terms consisting of a specific variable, say $z_1$ and form the perfect square. We then set the canonical basis as the inside of the square (hopefully that makes sense...)

Until we get the quadratic form to look like:

$$ Q(z)=\alpha_1(\eta^1)^2+\alpha_2(\eta^2)^2+\alpha_3(\eta^3)^2+\alpha_4(\eta^4)^2 $$ where the alphas are the canonical coefficients.

Now I am stuck with the above problem as there is no square term. Typically my approach in these problems has been to start with a term that has a square term, and go from there. In this case, whenever I get to the final $\eta$ to find, I am left with two square terms, say $z_3$ and $z_4$

Is there something im missing?

Answer 1

When you don’t have any squared terms, a common trick is to pick one of the cross terms $z_iz_j$ and make the change of variables $z_i=\frac12(y_1+y_2)$, $z_j=\frac12(y_1-y_2)$. This change of variables comes from a polarization identity for quadratic forms. You then have a difference of squares with which you can continue.

Here, we can try $z_1=\frac12(y_1+y_2)$, $z_2=\frac12(y_1-y_2)$, obtaining $\frac14y_1^2-\frac14y_2^2+y_1z_3-y_2z_3-3z_3z_4$. After completing the squares a couple of times, you’ll once more be left with only a cross term, so apply another change of variables to it. When you’re all done, substitute for the $y_i$. (The factor of $\frac12$ in the change of variables is there to make this final substitution for the original variables “nicer.”)

Answer 2

The matrix identity below says your quadratic form can be written as $$\frac{1}{4} (x_1 + x_2 + 2 x_3)^2 - \frac{1}{4} (-x_1 + x_2 - 2 x_3)^2 - \frac{3}{4} ( x_3 + x_4)^2 +\frac{3}{4} (- x_3 + x_4)^2 $$

$$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 1 & - 2 & \frac{ 1 }{ 2 } & - 1 \\ 0 & 0 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 1 & - 2 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 0 & - 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & - 3 \\ 0 & 0 & - 3 & 0 \\ \end{array} \right) $$

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

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

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

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

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

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

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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 2 & 0 \\ 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 1 & - 2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & - 1 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & - 3 \\ 0 & 0 & - 3 & 0 \\ \end{array} \right) $$

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$$ E_{6} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{6} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 2 & 1 \\ 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) , \; \; \; Q_{6} = \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 1 & - 2 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 0 & - 1 & 1 \\ \end{array} \right) , \; \; \; D_{6} = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 & 0 \\ - 2 & 0 & 1 & 1 \\ 1 & 0 & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & - 3 \\ 0 & 0 & - 3 & 0 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - 2 & 1 \\ 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ \end{array} \right) = \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & - 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ 1 & - 2 & \frac{ 1 }{ 2 } & - 1 \\ 0 & 0 & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 2 & 0 & 0 & 0 \\ 0 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 0 & - 6 & 0 \\ 0 & 0 & 0 & \frac{ 3 }{ 2 } \\ \end{array} \right) \left( \begin{array}{rrrr} \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 & 0 \\ - 1 & 1 & - 2 & 0 \\ 0 & 0 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 0 & - 1 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 0 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 \\ 0 & 2 & 0 & - 3 \\ 0 & 0 & - 3 & 0 \\ \end{array} \right) $$

Answer 3

Gantmacher's version of Lagrange's method leads to $$ \left( \frac{1}{2} x_1 + \frac{1}{2} x_2 + x_3 \right)^2 - \left( -\frac{1}{2} x_1 + \frac{1}{2} x_2 - x_3 \right)^2 + \left( \frac{1}{2} x_2 + \frac{1}{2} x_3 -\frac{3}{2} x_4 \right)^2 - \left( -\frac{1}{2} x_2 + \frac{1}{2} x_3 +\frac{3}{2} x_4 \right)^2$$

Answer 4

Without spelling out all the computations; your quadratic form is given by the symmetric matrix $${\bf z}\begin{pmatrix} 0&\tfrac12&0&\hphantom{-}0\\ \tfrac12&0&1&\hphantom{-}0\\ 0&1&0&-\tfrac32\\ 0&0&-\tfrac32&\hphantom{-}0 \end{pmatrix}{\bf z}^{\top}=0,$$ and you want to find a linear transformation $T$ such that $$(T{\bf z})\begin{pmatrix} \alpha_1&0&0&0\\ 0&\alpha_2&0&0\\ 0&0&\alpha_3&0\\ 0&0&0&\alpha_4 \end{pmatrix}(T{\bf z})^{\top}=0,$$ so it suffices to diagonalize the symmetric matrix above.