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?
Best Answer
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.”)