0
$\begingroup$

I just learn how to complete square of a quadratic form and then I met a problem without square term: we work over the field of read numbers. Reduce the quadratic form $$q(x_1,x_2,x_3,x_4)=x_1x_2+x_2x_3+x_3x_4$$ to the diagonal form and calculate its signature.

I do it in this way and I don't know whether it is right or wrong:

Since there is no square terms, I set $y_1=x_1,y_2=x_2-x_1,y_3=x_3,y_4=x_4$. So I have $q=y_1(y_2+y_1)+(y_2+y_1)y_3+y_3y_4$. Then I complete the square $q=(y_1+\frac{y_2+y_3}{2})^2-\frac{1}{4}(y_3-(y_2+2y_4))^2+\frac{1}{4}(y_2+2y_4)^2-\frac{1}{4}y_2^2$.

Is it the right way to do the diagonalization and conclude that the signature is $n_+-n_-=2-2=0$?

Thank you very much!

| cite | improve this question | | | | |
$\endgroup$
2
$\begingroup$

The standard way (Gauß's reduction) consists, when there are no square terms, in using repeatedly the identity: $$ab=\frac14(a+b)^2-\frac14(a-b)^2$$ so as to eliminate the variables one after the other.

In detail: \begin{align*} x_1x_2+x_2x_3+x_3x_4&=(x_1+x_3)x_2+x_3x_4\\ &=\frac14(x_1+x_2+x_3)^2-\frac14(x_1-x_2+x_3)^2+x_3x_4\\ &=\frac14(x_1+x_2+x_3)^2-\frac14(x_1-x_2+x_3)^2+\frac14(x_3+x_4)^2-\frac14(x_3-x_4)^2 \end{align*} The signature of the quadratic form is $(2,2)$ (the signature is a pair of natural numbers).

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.