Exercise about finding when a quadratic form has isotropic vectors and then finding them.

geometryquadratic-formsvectors

I found this exercise among those given in last year's geometry exam and I can't understand the solution given by the professor. The solution also seems incomplete because the isotropic vectors are not found. Could someone kindly explain the solution and clarify how to find isotropic vectors?

Find for which a and b the quadratic form:

$x^2+2axy+by^2$

has isotropic vectors and in that case find them.

Solution:
$x^2+2axy+by^2$=$x^2+2axy+a^2y^2+(b-a^2)y^2=(x+ay)^2+(b-a^2)y^2$

Then if $ b \le a^2 $ there are isotropic vectors,otherwise not.
In other words, the determinant of the associated matrix ́is $\le0$ if and only if we have isotropic vectors.

Best Answer

Welcome!

To find isotropic vectors, you have to find vectors $(x,y)$ such that $f(x,y)=0$ ($f$ is the quadratic form). To find them, they write the quadratic form as a linear combination of squares of linear forms , using Gauß' method and obtaining $$f(x,y)=(x+ay)^2+(b-a^2)y^2.$$

The coefficients are, respectively $1$ and $b-a^2$. Now

  • If $b>a^2$, both coefficients are positive, so the quadratic form can be $0$ only if each linear form is $0$, i.e. $\; y=0$, $x+ay=0$, whence $(x,y)=(0,0)$. Thus, in this case the quadratic form is definite, and as the coefficients are positive, it is definite positive.
  • If $b\le a^2$, we have \begin{align} f(x,y)=0&\iff (x+ay)^2=(a^2-b)y^2\iff x+ay=\pm\sqrt{a^2-b}y\\&\iff x=\bigl(-a\pm\sqrt{a^2-b}\bigr)y. \end{align}

The latter equations are the equations of the isotropic cone.

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