Classify a quadratic form depending on a parameter

by parishilton   Last Updated November 08, 2018 23:20 PM

Let $Q:\mathbb{R^3}\to\mathbb{R}$ be a quadratic form, $Q(x,y,z)=x^2+2axy+2xz+z^2$ with $a \in \mathbb{R}$.

Classify the form depending on $a$.

The first thing I did was to find the matrix of the quadratic form:

Let $A$ be that matrix:

$A=\begin{pmatrix} 1 & a & 1 \\ a & 0 & 0 \\ 1 & 0 & 1 \\ \end{pmatrix}$

And then I tried to find the characteristic polynomial, and that's where I got stuck:

$|A-\lambda I|=\begin{vmatrix} 1-\lambda & a & 1 \\ a & \lambda & 0 \\ 1 & 0 & 1-\lambda \\ \end{vmatrix}$

Using Sarrus' rule I got that $\chi _A=-\lambda^3+2\lambda^2+a^2\lambda-a^2$.

I only know that when $a=0$ then $\chi _A=-\lambda^3+2\lambda^2=\lambda^2(-\lambda+2)$, so the quadratic form would be positive semidefinite, but I don't know how to manipulate the polynomial enough to classify it when $a>0$ or $a<0$.

Is there an easier way to do this?



Answers 2


Maybe you can use Principal Minors diagonal form: $$q_J(x,y,z)=D_1x^2+(D_2/D_1)y^2+(D_3/D_2)z^2.$$

In your case, $D_1=1$, $D_2=\begin{vmatrix}1&a\\a&0\end{vmatrix}=-a^2$ and $D_3={\rm det}(A)=-a^2$.

If $a\ne0$, then $q_J(x,y,z)=x^2-a^2y^2+z^2$ and, thus, the quadratic form is indefinite.

If $a=0$, $q(x,y,z)=(x+z)^2$ which is clearly positive semidefinite.

Tito Eliatron
Tito Eliatron
November 08, 2018 23:03 PM

Unnecessary.

Just observe:

$Q(x,y,z)=(x+z)^2+\dfrac{a}{2}\big( (x+y)^2-(x-y)^2 \big)$

and conclude that the for is positive semidefinite when $a = 0$ and non-definite otherwise.

Will M.
Will M.
November 08, 2018 23:04 PM

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